{
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    "slug": "jolly-pixel",
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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
    "model": {
      "id": "gpt-5.6-luna",
      "providerID": "openai",
      "variant": "max"
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789604541334
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-5.6-luna",
          "variant": "max"
        },
        "system": "You create Codeforces learning content for Nudge. Your job is to understand the best approach for the problem and teach it clearly in your own words, using existing solutions to inform your reasoning.\n\nStart with a brief, targeted web search for existing solutions so relevant editorials are in context before substantial independent problem-solving. Use any supplied tutorial links first; otherwise search by contest ID and problem index. Prefer the official tutorial/editorial, and read relevant explanations rather than relying on search snippets. Look at accessible alternatives when useful, especially if the official explanation is missing, unclear, or unnecessarily complicated. Accepted submissions can also help clarify an approach. Respect supplied source lookup statuses and avoid repeatedly retrying blocked sources. If no usable source is accessible, proceed from the supplied statement.\n\nTreat existing solutions as a starting point, not a constraint. If an approach is clear and suitable, build on it without unnecessary re-derivation. If an explanation has gaps, seems incorrect, or uses a harder method than necessary, reason through it and derive a better or simpler approach when worthwhile. Choose a correct approach that is easy to teach and fits the constraints. Verify it against the supplied statement and edge cases before writing the hints, editorial, and C++ solution.\n\nThe editorial must be self-contained: explain the key insight, algorithm, why it works, complexity, and implementation details a learner needs, filling in steps the source skips. Keep research and verification concise; spend the output on teaching the solution, not narrating the research process. Follow the user prompt and output schema exactly.",
        "format": {
          "type": "json_schema",
          "schema": {
            "type": "object",
            "properties": {
              "status": {
                "type": "string",
                "enum": [
                  "success",
                  "unsolvable"
                ],
                "description": "Set to 'success' when the supplied statement defines a solvable programming problem. Set to 'unsolvable' only if the statement is fundamentally incomplete, contradictory, or depends on an inaccessible required resource."
              },
              "reason": {
                "anyOf": [
                  {
                    "type": "string",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "minLength": 1,
                          "maxLength": 10000,
                          "description": "Markdown hint text."
                        }
                      },
                      "required": [
                        "order",
                        "content"
                      ],
                      "additionalProperties": false
                    }
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', provide exactly 5 progressive hints. Otherwise null."
              },
              "editorial": {
                "anyOf": [
                  {
                    "type": "string",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "minLength": 1,
                    "maxLength": 200000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a complete C++ solution. Otherwise null."
              }
            },
            "required": [
              "status",
              "reason",
              "hints",
              "editorial",
              "solution"
            ],
            "additionalProperties": false,
            "title": "problem_response",
            "description": "Submit the generated content for the problem, or report if it is unsolvable."
          },
          "retryCount": 2
        },
        "id": "msg_0acbe339600157tYVAEsQoU77E",
        "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t"
      },
      "parts": [
        {
          "type": "text",
          "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2262C: \"Traveling the World\".\n\nKnown problem metadata:\n- Contest ID: 2262\n- Index: C\n- Name: Traveling the World\n- Rating: unknown / unrated\n- Tags: combinatorics, graphs, math, schedules\n- Problem URL: https://codeforces.com/contest/2262/problem/C\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n<problem-statement>\n<div class=\"header\"><div class=\"title\">C. Traveling the World</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>2 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>256 megabytes</div><div class=\"input-file input-standard\"><div class=\"property-title\">input</div>standard input</div><div class=\"output-file output-standard\"><div class=\"property-title\">output</div>standard output</div></div><div><p>There are $$$n$$$ islands, numbered from $$$1$$$ to $$$n$$$. The $$$i$$$-th island has value $$$a_i$$$. The values $$$a_1,a_2,\\ldots,a_n$$$ are strictly increasing, and $$$a_1=0$$$.</p><p>Farmer John wants to rearrange the islands. After rearranging them, their values form an array $$$b_1,b_2,\\ldots,b_n$$$, which is a rearrangement of $$$a$$$.</p><p>Bessie then travels between islands according to the following rule. Suppose she is currently on the island in position $$$i$$$ after the rearrangement. Then Bessie may travel from the island in position $$$i$$$ to the island in position $$$j$$$ if $$$$$$ b_i + b_j = \\max(b_i,b_{i+1},\\ldots,b_n). $$$$$$</p><p>A rearrangement $$$b$$$ is called good if there exists a sequence of distinct positions $$$p_1,p_2,\\ldots,p_n$$$ such that Bessie can travel from $$$p_i$$$ to $$$p_{i+1}$$$ for every $$$1\\le i \\lt n$$$. In other words, Bessie can visit all positions (and therefore all islands) exactly once by following valid travels.</p><p>Count the number of good rearrangements of the islands. Since the number can be large, output it modulo $$$10^9+7$$$.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 10^4$$$). The description of the test cases follows.</p><p>The first line of each test case contains $$$n$$$ ($$$4 \\le n \\le 2 \\cdot 10^5$$$).</p><p>The second line of each test case contains $$$n$$$ distinct integers $$$a_1,a_2,\\ldots,a_n$$$ ($$$0\\le a_i\\le 10^9$$$).</p><p>It is guaranteed that $$$a_1=0$$$ and $$$a_1 \\lt a_2 \\lt \\cdots \\lt a_n$$$.</p><p>It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2\\cdot 10^5$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>For each test case, output a single integer — the number of good rearrangements $$$b$$$ of $$$a$$$, modulo $$$10^9+7$$$.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Example</div><div class=\"sample-test\"><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">3</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">6</div><div class=\"test-example-line test-example-line-odd test-example-line-1\">0 1 2 3 4 5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">5</div><div class=\"test-example-line test-example-line-even test-example-line-2\">0 3 4 5 6</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">4</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">0 1 3 4</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre><div class=\"test-example-line test-example-line-odd test-example-line-1\">12</div><div class=\"test-example-line test-example-line-even test-example-line-2\">0</div><div class=\"test-example-line test-example-line-odd test-example-line-3\">4</div></pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>For the first testcase, one $$$b$$$ that works is $$$[2,1,0,5,3,4]$$$.</p><p>Note that Bessie can travel from $$$1 \\to 5$$$ since $$$2 + 3 = b_1 + b_5 = \\max(b_1,b_2,\\ldots,b_6) = 5$$$.</p><p>Similarly, Bessie can travel from $$$5 \\to 2$$$, $$$2\\to 6$$$, $$$6\\to 3$$$, $$$3\\to 4$$$.</p><p>Thus, starting from island $$$1$$$, Bessie can follow the path $$$1\\to 5\\to 2\\to 6\\to 3\\to 4$$$ which visits every island. Therefore, this rearrangement is good.</p><p>For the second testcase, it can be shown that there are no good rearrangements. Therefore, the answer is $$$0$$$.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial (en): https://codeforces.com/blog/entry/156688 loaded (200 OK; title: \"Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\"; mentions 2262C)\n</source-lookup-status>\n\nGenerate:\n1. Five progressive hints, from a gentle nudge to the key insight.\n2. A self-contained editorial that teaches the solution clearly, including the key insight, algorithm, correctness argument, complexity, and necessary implementation details. Explain non-obvious steps without padding or a discovery-process transcript.\n3. A complete C++26 solution that gets AC on Codeforces.\n\nFormatting & Style Rules:\n- Hints and editorial: valid Markdown with LaTeX as needed (e.g., $dp[i]$, $$\\sum_{i=1}^{n} a_i$$).\n- Hints and editorial must read like Nudge's own explanation. No research notes, source notes, citations, Markdown links, URLs, or references to editorials/submissions/posts.\n- Solution: raw C++ only, no Markdown fences.\n- Keep C++ short, clean, standard, and single-file. Comments are fine when they clarify the idea.\n\nOutput strictness:\n- Return JSON matching the provided schema exactly.\n- If the problem is solvable, return `status: \"success\"`, `reason: null`, and fill in `hints`, `editorial`, and `solution`.\n- Return `status: \"unsolvable\"`, a short `reason`, and set `hints`, `editorial`, and `solution` to null only when the supplied statement is fundamentally incomplete, contradictory, or dependent on an inaccessible resource needed to define the task. Missing research, high rating, uncertainty, or \"can't guarantee AC\" is not a reason to return unsolvable.\n- If you return `status: \"unsolvable\"`, the `reason` is shown directly to users. Be concrete: say what exact statement/resource blocker stopped you, and include relevant source lookup status if it mattered (for example: official tutorial missing, Codeforces returned 403/Cloudflare challenge, tutorial page 404, accepted submissions unavailable).\n- Each hint must be JUST the hint text. No \"Hint 1:\" or subtitles. The UI adds those automatically.\n- Do not start the editorial with an \"# Editorial\" heading. The UI already adds that section. Feel free to add other headers as needed though.\n- Use styling in the editorial if needed! Just that first header is no good but any subheaders or bolding or whatever is good!\n\nFor the C++ solution, you MUST use this template and work around it:\n\n```cpp\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n}\n```",
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        "time": {
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        "finish": "tool-calls",
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          "type": "reasoning",
          "text": "**Fetching official editorial**",
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        {
          "type": "tool",
          "tool": "webfetch",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/30864/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1120 (Div 1, Div 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n<!DOCTYPE HTML PUBLIC \"-//W3C//DTD HTML 4.01//EN\"> \n\n  \n[![Codeforces](//codeforces.com/codeforces.org/s/30864/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/156688?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/30864/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/156688?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F156688)| [Register](https://codeforces.com/register) \n\n  \n* [Home](https://codeforces.com/)\n* [Top](https://codeforces.com/top)\n* [Catalog](https://codeforces.com/catalog)\n* [Contests](https://codeforces.com/contests)\n* [Gym](https://codeforces.com/gyms)\n* [Problemset](https://codeforces.com/problemset)\n* [Groups](https://codeforces.com/groups)\n* 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Errichto\") | 139      |\n| 6  | [adamant](https://codeforces.com/profile/adamant \"Master adamant\")                       | 136      |\n| 7  | [AmShZ](https://codeforces.com/profile/AmShZ \"Grandmaster AmShZ\")                        | 134      |\n| 8  | [maroonrk](https://codeforces.com/profile/maroonrk \"Legendary Grandmaster maroonrk\")     | 133      |\n| 8  | [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 133      |\n| 10 | [DNR](https://codeforces.com/profile/DNR \"Master DNR\")                                   | 128      |\n\n|  | [View all →](https://codeforces.com/top-contributed) |\n|  | ---------------------------------------------------- |\n\n→ Find user \n\nHandle: \n\n→ Recent actions \n\n* [Down\\_bad\\_4\\_haiender288](https://codeforces.com/profile/Down%5Fbad%5F4%5Fhaiender288 \"Pupil Down_bad_4_haiender288\") → [new manmade horrors (barely within comprehension) just dropped](https://codeforces.com/blog/entry/156826) 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comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/30864/images/icons/hourglass.png \"Necropost\")\n* [plagues](https://codeforces.com/profile/plagues \"International Grandmaster plagues\") → [The most outrageous cheating in the history of Huawei Challenge](https://codeforces.com/blog/entry/156293) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\")\n* [maruf004](https://codeforces.com/profile/maruf004 \"Newbie maruf004\") → [Will Meta Hacker Cup 2026 Happen?](https://codeforces.com/blog/entry/156789) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/30864/images/icons/comment-12x12.png \"New comment(s)\")\n\n|  | [Detailed →](https://codeforces.com/recent-actions) |\n|  | --------------------------------------------------- |\n\n* [kondasujay2](https://codeforces.com/profile/kondasujay2)\n* [Blog](https://codeforces.com/blog/kondasujay2)\n* [Teams](https://codeforces.com/teams/with/kondasujay2)\n* [Submissions](https://codeforces.com/submissions/kondasujay2)\n* [Groups](https://codeforces.com/groups/with/kondasujay2)\n* [Contests](https://codeforces.com/contests/with/kondasujay2)\n* [Problemsetting](https://codeforces.com/contests/writer/kondasujay2)\n\n### [kondasujay2's blog](https://codeforces.com/blog/kondasujay2)\n\n[ Codeforces Round 1120 (Div 1, Div 2) Editorial ](https://codeforces.com/blog/entry/156688) \n\nBy [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\"), [history](https://codeforces.com/topic/157332/en14), 4 days ago, ![In English](//codeforces.com/codeforces.org/s/30864/images/flags/24/gb.png \"In English\") \n\nThank you to everyone who participated! \n\nWe are sorry for the poor testcases on C1, along with C1 and C2's solutions being so similar to the recent Div3 E. \n\n[2263A - Min Max Game](https://codeforces.com/contest/2263/problem/A \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    int cnt = 0;\n    vector<int> a(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n        cnt += a[i];\n    }\n    if(cnt >= n - cnt) {\n        cout << \"Bessie\\n\";\n    } else {\n        cout << \"Elsie\\n\";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263B - Min Matrices](https://codeforces.com/contest/2263/problem/B \"Codeforces Round 1120 (Div. 2)\")\n\nAuthor: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include<bits/stdc++.h>\n \nusing namespace std;\n \nvoid solve() {\n    int n, k; cin >> n >> k;\n    int x = k-n+1;\n    if(!(1 <= x && x <= n)) { cout << -1; return; }\n    vector<vector<int>> ans(n, vector<int>(n));\n    ans[0][0] = 1;\n    int on = 2;\n    for(int i = 1; i < x; i++) {\n        ans[0][i] = on; on++;\n        ans[i][0] = on; on++;\n    }\n    for(int i = x; i < n; i++) {\n        ans[i][i] = on; on++;\n    }\n\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(!ans[i][j]) {\n                ans[i][j] = on; on++;\n            }\n        }\n    }\n\n    for(int j = 0; j < n; j++) {\n        if(j) cout << \"\\n\";\n        for(auto &i : ans[j]) cout << i << \" \";\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) {\n        solve();\n        if(t)   cout << \"\\n\";\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2263/problem/C1 \"Codeforces Round 1120 (Div. 2)\") / [2262A1 - Floor of MEX (Easy Version)](https://codeforces.com/contest/2262/problem/A1 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [CutSandstone](https://codeforces.com/profile/CutSandstone \"Candidate Master CutSandstone\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    cout << gnums.size() << endl;\n    for(int i : gnums) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263C2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2263/problem/C2 \"Codeforces Round 1120 (Div. 2)\") / [2262A2 - Floor of MEX (Hard Version)](https://codeforces.com/contest/2262/problem/A2 \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> b(n + 1);\n    for(int i = 1; i <= n; i++) cin >> b[i];\n    vector<pair<int, int>> goodivs;\n    vector<pair<int, int>> badivs;\n    for(int i = 1; i <= n; i++) {\n        if(b[i] > (n - 1) / i + 1) {\n            cout << 0 << endl;\n            return;\n        }\n        badivs.push_back({b[i] * i, (b[i] + 1) * i - 1});\n        for(int j = 0; j < b[i]; j++) {\n            goodivs.push_back({j * i, (j + 1) * i - 1});\n        }\n    }\n    vector<int> d(n + 1);\n    for(auto [l, r] : badivs) {\n        d[min(n, l)]++;\n        d[min(n, r + 1)]--;\n    }\n    int csm = 0;\n    vector<int> gnums;\n    for(int i = 0; i < n; i++) {\n        csm += d[i];\n        if(csm == 0) {\n            gnums.push_back(i);\n        }\n    }\n    set<pair<int, int>> st;\n    sort(goodivs.begin(), goodivs.end(), [&] (pair<int, int> iv, pair<int, int> iv2) {\n        return iv.second - iv.first < iv2.second - iv2.first;\n    });\n    for(auto [l, r] : goodivs) {\n        auto it = st.lower_bound({l, 0});\n        if(it != st.end() && it->second <= r) {\n            continue;\n        }\n        st.insert({l, r});\n    }\n\n    vector<pair<int, int>> impivs(st.begin(), st.end());\n    int l = 0, r = 0;\n    vector<int> dp(impivs.size() + 1);\n    dp[0] = 1;\n    int poss = 1;\n    for(int i : gnums) {\n        while(l < impivs.size() && impivs[l].second < i) {\n            poss += MOD - dp[l];\n            poss %= MOD;\n            l++;\n        }\n        while(r < impivs.size() && impivs[r].first <= i) r++;\n        dp[r] += poss;\n        dp[r] %= MOD;\n        poss += poss;\n        poss %= MOD;\n    }\n    cout << dp[impivs.size()] << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263D - Culling Game](https://codeforces.com/contest/2263/problem/D \"Codeforces Round 1120 (Div. 2)\") / [2262B - Culling Game](https://codeforces.com/contest/2262/problem/B \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nstruct BIT {\n    vector<ll> tr;\n    BIT(int n) : tr(n + 1) {}\n    void add(int k, ll x) {\n        while(k < tr.size()) {\n            tr[k] += x;\n            k += k&-k;\n        }\n    }\n    ll sum(int k) {\n        ll res = 0;\n        while(k > 0) {\n            res += tr[k];\n            k -= k&-k;\n        }\n        return res;\n    }\n};\n\nvoid tc() {\n    int n; cin >> n;\n    vector<int> a(n), p(n);\n    for(int i = 0; i < n; i++) {\n        cin >> a[i];\n    }\n    for(int i = 0; i < n; i++) {\n        cin >> p[i]; p[i]--;\n    }\n    reverse(p.begin(), p.end());\n    BIT bit(n);\n    set<int> imp; \n    vector<int> ans(n);\n    for(int i = 0; i < n; i++) {\n        int k = p[i];\n        auto it = imp.upper_bound(k);\n        int pimp;\n        if(it == imp.begin() || bit.sum(k) - bit.sum(*prev(it)) < a[k]) {\n            imp.insert(k);\n            pimp = k;\n        } else {\n            pimp = *prev(it);\n        }\n        bit.add(k + 1, a[k]);\n        while(imp.upper_bound(k) != imp.end()) {\n            int j = *imp.upper_bound(k);\n            if(bit.sum(j) - bit.sum(pimp) >= a[j]) {\n                imp.erase(j);\n            } else {\n                break;\n            }\n        }\n        ans[i] = imp.size() - 1;\n    }\n    reverse(ans.begin(), ans.end());\n    for(int i : ans) cout << i << \" \";\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263E - Traveling the World](https://codeforces.com/contest/2263/problem/E \"Codeforces Round 1120 (Div. 2)\") / [2262C - Traveling the World](https://codeforces.com/contest/2262/problem/C \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst int MOD = 1e9 + 7;\n\nvoid tc() {\n    int n; cin >> n;\n    vector<ll> a(n);\n    for(int i = 0; i < n; i++) cin >> a[i];\n    int k = n / 2;\n\n    int val = 1;\n    for(int i = 1; i <= k; i++) {\n        val = (ll)i * val % MOD;\n    }\n    for(int i = 1; i <= k - (n % 2 == 0 ? 2 : 1); i++) {\n        val = (ll)i * val % MOD;\n    }\n\n    set<ll> xposs;\n    xposs.insert(a[n - 1] - a[n - 2]);\n    xposs.insert(a[n - 1] - a[n - 3]);\n\n    int ans = 0;\n    for(ll x : xposs) {\n        vector<ll> b;\n        for(int i = 0; i < k; i++) {\n            b.push_back(i * x);\n        }\n        for(int i = 0; i < n - k; i++) {\n            b.push_back(a[n - 1] - i * x);\n        }\n        sort(b.begin(), b.end());\n        if(a == b) {\n            ans += val;\n            if(ans >= MOD) ans -= MOD;\n        }\n    }\n    cout << ans << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2263F - PLUSworld](https://codeforces.com/contest/2263/problem/F \"Codeforces Round 1120 (Div. 2)\") / [2262D - PLUSworld](https://codeforces.com/contest/2262/problem/D \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nvoid tc() {\n    int n; cin >> n;\n\n    vector<int> a(n), b(n);\n    for(int i = 0; i < n; i++) cin >> a[i], a[i]--;\n    for(int i = 0; i < n; i++) cin >> b[i], b[i]--;\n\n    int badcnt = 0;\n    for(int i = 0; i < n; i++) {\n        badcnt += a[i] != b[i];\n    }\n    if(badcnt == 0) {\n        cout << \"0 1\\n\";\n        return;\n    }\n\n    // mark everything which will ever be reached\n    vector<bool> useful(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] == b[i]) continue;\n        int x = i;\n        while(!useful[x]) {\n            useful[x] = true;\n            x = b[x];\n        }\n    }\n\n    // anything with b[i] == i once we reach we can't reach anything else, \n    // at most one of them can exit\n    vector<int> ends;\n    for(int i = 0; i < n; i++) {\n        if(useful[i] && b[i] == i) ends.push_back(i);\n    }\n\n    if(ends.size() > 1) {\n        cout << \"-1\\n\";\n        return;\n    }\n\n    vector<vector<int>> adj(n), radj(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] > b[i]) {\n            cout << \"-1\\n\";\n            return;\n        }\n        for(int j = a[i]; j <= b[i]; j++) {\n            adj[i].push_back(j);\n            radj[j].push_back(i);\n        }\n    }\n\n    // tarjan algorithm, from cp-algorithms\n    vector<int> st;    // - stack holding the unclaimed vertices\n    vector<int> roots(n, -1); // - keeps track of the SCC roots of the vertices\n    int timer = 0;         // - dfs timestamp counter\n    vector<int> t_in(n, -1);  // - keeps track of the dfs timestamp of the vertices\n    vector<int> t_low(n, -1); // - keeps track of the lowest t_in of unclaimed vertices\n    // reachable in the subtree\n\n    // implements the tarjan algorithm for strongly connected components\n    auto dfs = [&](auto&& self, int s) -> void {\n        t_low[s] = t_in[s] = timer++;\n        st.push_back(s);\n\n        for (auto u : adj[s]) {\n            if (t_in[u] == -1) { // tree-edge\n                self(self, u);\n                t_low[s] = min(t_low[s], t_low[u]);\n            } else if (roots[u] == -1) { // back-edge, cross-edge or forward-edge to an unclaimed vertex\n                t_low[s] = min(t_low[s], t_in[u]);\n            }\n        }\n\n        if (t_low[s] == t_in[s]) { // vertex is a root\n            while (true) {\n                int u = st.back();\n                st.pop_back();\n                roots[u] = s; // claims the vertex\n                if (u == s)\n                    break;\n            }\n        }\n    };\n\n    // applies the tarjan algorithm to all the vertices\n    for (int v = 0; v < n; v++) {\n        if (t_in[v] == -1) {\n            dfs(dfs, v);\n        }\n    }\n    vector<vector<int>> comps(n);\n    for(int v = 0; v < n; v++) {\n        comps[roots[v]].push_back(v);\n    }\n\n    vector<int> p(n, -1);\n    vector<int> sz(n);\n    for(int i = 0; i < n; i++) {\n        if(a[i] != b[i]) {\n            sz[roots[i]]++;\n        }\n        if(useful[i] && roots[i] != roots[b[i]]) {\n            if(p[roots[b[i]]] != -1) {\n                cout << \"-1\\n\";\n                return;\n            }\n            p[roots[b[i]]] = roots[i];\n        }\n    }\n    vector<int> chain;\n    int x = roots[ends[0]];\n    int csz = 0;\n    while(x != -1) {\n        chain.push_back(x);\n        csz += sz[x];\n        x = p[x];\n    }\n    if(csz != badcnt) {\n        cout << \"-1\\n\";\n        return;\n    }\n    reverse(chain.begin(), chain.end());\n\n    {\n        x = comps[chain[0]][0];\n        vector<int> ops;\n        auto op1 = [&] () {\n            ops.push_back(1);\n            a[x]++;\n        };\n        auto op2 = [&] () {\n            ops.push_back(2);\n            x = a[x];\n        };\n        auto solve_scc = [&] () {\n            int rt = roots[x];\n            vector<int> comp = comps[rt];\n\n            vector<int> p2(n, -1);\n            // only one thing in comp case\n            auto rdfs = [&] (auto&& self, int s) -> void {\n                for(int u : radj[s]) {\n                    if(roots[u] != rt) continue;\n                    if(p2[u] == -1) {\n                        p2[u] = s;\n                        self(self, u);\n                    }\n                }\n            };\n\n            // root a tree\n            rdfs(rdfs, comp[0]);\n\n            for(int i = 0; i < comp.size(); i++) {\n                while(x != comp[i]) {\n                    while(a[x] < max(p2[x], comp[i])) op1();\n                    op2();\n                }\n                while(a[x] < b[x]) op1();\n                op2();\n            }\n        };\n        for(int i = 0; i < chain.size(); i++) solve_scc();\n\n        cout << ops.size() << \" \" << comps[chain[0]][0] + 1 << endl;\n        for(int i : ops) cout << i << \" \";\n        cout << endl;\n\n        for(int i = 0; i < n; i++) assert(a[i] == b[i]);\n    }\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Magic**\n\nWith this, you can prove a significantly simpler solution. We will add this proof later.\n\n**Magic Solution**\n\n```\n// Note: AI generated solution\n#include <bits/stdc++.h>\nusing namespace std;\n\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n\n    int t;\n    cin >> t;\n\n    while (t--) {\n        int n;\n        cin >> n;\n\n        vector<int> a(n + 1), b(n + 1);\n        for (int i = 1; i <= n; i++) {\n            cin >> a[i];\n        }\n        for (int i = 1; i <= n; i++) {\n            cin >> b[i];\n        }\n\n        int start = 1, remaining = 0;\n        bool possible = true;\n\n        for (int i = 1; i <= n; i++) {\n            if (a[i] > b[i]) {\n                possible = false;\n            }\n            if (a[i] < b[i]) {\n                if (remaining == 0) {\n                    start = i;\n                }\n                remaining++;\n            }\n        }\n\n        if (!possible) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        vector<int> operations;\n        int position = start;\n\n        while (remaining > 0) {\n            if (a[position] < b[position]) {\n                while (a[position] < b[position] &&\n                       (a[position] >= position ||\n                        a[a[position]] == b[a[position]])) {\n                    a[position]++;\n                    operations.push_back(1);\n                }\n\n                if (a[position] == b[position]) {\n                    remaining--;\n                }\n            }\n\n            if (remaining == 0 || a[position] == position) {\n                break;\n            }\n\n            operations.push_back(2);\n            position = a[position];\n        }\n\n        if (remaining > 0) {\n            cout << -1 << '\\n';\n            continue;\n        }\n\n        cout << operations.size() << ' ' << start << '\\n';\n        for (int operation : operations) {\n            cout << operation << ' ';\n        }\n        cout << '\\n';\n    }\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262E - Paired Bracket Sequences](https://codeforces.com/contest/2262/problem/E \"Codeforces Round 1120 (Div. 1)\")\n\nAuthor: [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint MOD;\n\nvoid madd(int& a, int b) {\n    a += b;\n    if(a >= MOD) a -= MOD;\n}\n\nvoid msub(int& a, int b) {\n    a -= b;\n    if(a < 0) a += MOD;\n}\n\nint mmul(int a, int b) {\n    return 1LL * a * b % MOD;\n}\n\nint bpow(int x, int y) {\n    return y == 0 ? 1 : mmul((y % 2) ? x : 1, bpow(mmul(x, x), y / 2));\n}\n\nvoid tc() {\n    int n; cin >> n >> MOD;\n    vector<int> f(2 * n + 1), invf(2 * n + 1);\n    f[0] = 1;\n    for(int i = 1; i <= 2 * n; i++) {\n        f[i] = mmul(f[i - 1], i);\n    }\n    invf[2 * n] = bpow(f[2 * n], MOD - 2);\n    for(int i = 2 * n; i >= 1; i--) {\n        invf[i - 1] = mmul(invf[i], i);\n    }\n    auto F = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k + 1], invf[n - k]));\n    };\n    auto choose = [&] (int n, int k) -> int {\n        return mmul(f[n], mmul(invf[k], invf[n - k]));\n    };\n    vector<vector<int>> p(n + 1, vector<int>(n + 1));\n    for(int i = 0; i <= n; i++) {\n        p[0][i] = mmul(F(2 * i, i), F(2 * i, i));\n    }\n    for(int i = 1; i <= n; i++) {\n        for(int j = 0; j <= n; j++) {\n            for(int k = 0; j + k <= n; k++) {\n                madd(p[i][j + k], mmul(p[i - 1][j], p[0][k]));\n            }\n        }\n    }\n    vector<int> ans(n + 1);\n    for(int i = 0; i <= n; i++) {\n        ans[i] = mmul(F(2 * n, i), p[i][n - i]);\n    }\n    for(int i = n - 1; i >= 0; i--) {\n        for(int j = i + 1; j <= n; j++) {\n            msub(ans[i], mmul(ans[j], choose(j, i)));\n        }\n    }\n    for(int i = 0; i <= n; i++) {\n        cout << ans[i] << \" \";\n    }\n    cout << endl;\n}\n\nint main() {\n    ios::sync_with_stdio(false), cin.tie(nullptr);\n    int t; cin >> t; \n    while(t--) tc();\n}\n\n```\n\n**Rate the problem**\n\n* Amazing: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Good: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Ok: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Bad: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n* Terrible: [ ](https://codeforces.com/data/like?action=like \"I like this\")\n\n[2262F - Rank Removal](https://codeforces.com/contest/2262/problem/F \"Codeforces Round 1120 (Div. 1)\")\n\nAuthors: [sukon](https://codeforces.com/profile/sukon \"Candidate Master sukon\") & [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nconst int MXN = 400;\n\ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n\n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n\n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; j < i; j++) {\n            if(basis[j][i]) {\n                basis[j] ^= basis[i];\n                inv[j] ^= inv[i];\n            }\n        }\n    }\n    return 1;\n}\n\nbool try_removal(vector<pair<int, int>> cells, vector<bitset<MXN>>& mat, vector<bitset<MXN>>& inv, vector<int>& row_cnt) {\n    int n = mat.size(), k = cells.size();\n    for(auto [x, y] : cells) {\n        if(mat[x][y] == 0) {\n            return false;\n        }\n    }\n    assert(k < 4);\n    vector<bitset<3>> K(k); \n    for(int i = 0; i < k; i++) {\n        for(int j = 0; j < k; j++) {\n            K[i][j] = (i == j) ^ inv[cells[i].second][cells[j].first];\n        }\n    }\n    vector<bitset<3>> invK(cells.size());\n    if(!compute_inv(K, invK)) {\n        return false;\n    }\n    for(auto [x, y] : cells) {\n        mat[x][y] = 0;\n        row_cnt[x]--;\n    }\n\n    auto old = inv;\n    for(int i = 0; i < n; i++) {\n        bitset<3> coef;\n        for(int alpha = 0; alpha < k; alpha++) {\n            int r = cells[alpha].first;\n\n            if(old[i][r]) {\n                coef ^= invK[alpha];\n            }\n        }\n\n        // delta = coef * (V^T * old)\n        for(int beta = 0; beta < k; beta++) {\n            if(coef[beta]) {\n                int c = cells[beta].second;\n                inv[i] ^= old[c];\n            }\n        }\n    }\n    return true;\n}\n\nvector<int> compute_matching(const vector<bitset<MXN>>& mat) {\n    int n = mat.size();\n    vector<vector<int>> adj(n);\n    for(int i = 0; i < n; i++) {\n        for(int j = 0; j < n; j++) {\n            if(mat[i][j]) adj[i].push_back(j);\n        }\n    }\n\n    vector<int> match(n, -1);\n    vector<bool> done(n);\n\n    function<bool(int)> dfs = [&] (int s) {\n        for(int u : adj[s]) {\n            if(!done[u]) {\n                done[u] = true;\n\n                if(match[u] == -1 || dfs(match[u])) {\n                    match[u] = s;\n                    return true;\n                }\n            }\n        }\n        return false;\n    };\n\n    for(int i = 0; i < n; i++) {\n        fill(done.begin(), done.end(), false);\n        dfs(i);\n    }\n\n    for(int i = 0; i < n; i++) {\n        assert(match[i] != -1);\n    }\n    return match;\n}\n\nvoid tc() {\n    int n, m; cin >> n >> m;\n\n    vector<bitset<MXN>> mat(n);\n    vector<int> row_cnt(n);\n    for(int i = 0; i < m; i++) {\n        int x, y; cin >> x >> y;\n        x--, y--;\n        mat[x][y] = 1;\n        row_cnt[x]++;\n    }\n\n    vector<bitset<MXN>> inv(n);\n    compute_inv(mat, inv);\n\n    int sm = m;\n    int left = (sm + n - 1) / n;\n    cout << left << '\\n';\n\n    while(sm > 0) {\n        vector<bitset<MXN>> omat = mat;\n        if(sm <= n) {\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n        } else if(sm <= 2 * n) {\n            vector<int> matching = compute_matching(mat);\n            for(int i = 0; i < n; i++) {\n                for(int j = 0; j < n; j++) {\n                    mat[i][j] = 0;\n                }\n            }\n            row_cnt.assign(n, 0);\n            for(int i = 0; i < sm - n; i++) {\n                mat[matching[i]][i] = 1;\n                row_cnt[matching[i]]++;\n            }\n        } else if(sm <= 3 * n) {\n            vector<int> matching = compute_matching(mat);\n            int above = 0;\n            for(int i = 0; i < n; i++) {\n                for(int j = matching[i] + 1; j < n; j++) {\n                    above += mat[j][i];\n                }\n            }\n\n            vector<bitset<MXN>> nmat(n);\n            vector<int> nrow_cnt(n);\n            int ones_left = sm - n;\n            for(int i = 0; i < n; i++) {\n                nmat[matching[i]][i] = 1;\n                nrow_cnt[matching[i]]++;\n                ones_left--;\n            }\n            // make matrix either upper or lower triangular, whichever one\n            // removes a sufficient number of ones\n            if(above >= ones_left) {\n                for(int i = 0; i < n; i++) {\n                    for(int j = matching[i] + 1; j < n; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            } else {\n                for(int i = 0; i < n; i++) {\n                    for(int j = 0; j < matching[i]; j++) {\n                        if(mat[j][i]) {\n                            ones_left--;\n                            nmat[j][i] = 1;\n                            nrow_cnt[j]++;\n                            if(ones_left == 0) break;\n                        }\n                    }\n                    if(ones_left == 0) break;\n                }\n            }\n            mat = nmat;\n            row_cnt = nrow_cnt;\n        } else {\n            int csm = sm;\n            if(n % 2) {\n                // odd case (remove either 1 or 3)\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 2) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n                assert(dup_row != -1);\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 2);\n\n                for(int i = 0; i < n; i++) {\n                    if(try_removal({{i, cols[0]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                    if(try_removal({{i, cols[1]}}, mat, inv, row_cnt)) {\n                        csm--;\n                        break;\n                    }\n                }\n\n                if(csm == sm) {\n                    for(int i = 0; i < n; i++) {\n                        if(i != dup_row) {\n                            // remove L shape\n                            if(try_removal(\n                                    {{i, cols[0]}, \n                                     {dup_row, cols[0]}, \n                                     {dup_row, cols[1]}},\n                                mat, inv, row_cnt)) {\n                                csm -= 3;\n                                break;\n                            }\n                        }\n                    }\n                    assert(sm != csm);\n                }\n            }\n\n            while(csm > sm - n) {\n                // remove 2 at a time\n                int dup_row = -1;\n                for(int i = 0; i < n; i++) {\n                    if(row_cnt[i] >= 3) {\n                        dup_row = i;\n                        break;\n                    }\n                }\n\n                vector<int> cols;\n                for(int i = 0; i < n; i++) {\n                    if(mat[dup_row][i]) cols.push_back(i);\n                }\n\n                assert(cols.size() >= 3);\n\n                if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[1]}}, mat, inv, row_cnt)) {\n                    if(!try_removal({{dup_row, cols[1]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                        if(!try_removal({{dup_row, cols[0]}, {dup_row, cols[2]}}, mat, inv, row_cnt)) {\n                            assert(false);\n                        }\n                    }\n                }\n\n                csm -= 2;\n            }\n            assert(csm == sm - n);\n        }\n\n        vector<pair<int, int>> move;\n        for(int i = 0; i < n; i++) {\n            for(int j = 0; j < n; j++) {\n                if(omat[i][j] == 1 && mat[i][j] == 0) {\n                    move.push_back({i, j});\n                }\n            }\n        }\n        cout << move.size() << \"\\n\";\n        for(auto [x, y] : move) cout << x + 1 << \" \" << y + 1 << \"\\n\";\n        sm -= n;\n\n        int csm = 0;\n        for(int i = 0; i < n; i++) {\n            csm += row_cnt[i];\n        }\n        assert(csm == max(0, sm));\n    }\n}\n\nint main() {\n    int t; cin >> t;\n    while(t--) tc();\n}\n\n```\n\n**Tutorial 2**\n\nCredit to [BurnedChicken](https://codeforces.com/profile/BurnedChicken \"International Grandmaster BurnedChicken\") for finding this in testing!\n\nWe will first prove that for any row/column of a rank $$$n$$$ matrix, we can zero out everything in that row/column except one, and maintain rank $$$n$$$. This is because rank $$$n$$$ means the determinant is $$$1$$$, so there are exactly an odd number of perfect matchings in total (read Solution 1 for more understanding). Therefore at least $$$1$$$ of the ones in the row/column must be in an odd number of perfect matchings (since if they were all even then the total number would be even). This means if we delete all other ones but that one, then we would keep rank $$$n$$$.\n\nThis actually all you need to prove to solve the problem. Note that after we clear out one row except one number, we can clear out that column except that one number (even partially) and the determinant will not change (and the same works when clearing out a column with the leftover row). \n\nFor every turn, we want to try to find $$$n$$$ things to remove without changing the rank. Therefore, we begin by finding the one in the biggest row/column we can keep while removing everything else. Then we remove everything in the column/row of that one, and the everything in the row/column into the queue. If the queue + the current removals is big enough to reach $$$n$$$ removals, then we can just remove the ones in the queue. If the queue is too small, then we repeat one the next biggest row/column. Because the queue contains the row/column with the most ones, when the current removals will never exceed $$$n$$$ if the the queue + current removals is less than $$$n$$$. Once we run out of row/columns to clear, we can start removing the ones we excluded removing, since we have to reduce the rank at this point.\n\nTo find the one we must keep, you have to utilize the inverse, since we can clear out the row/column on $$$M\\_{i, j}$$$ if $$$M^{-1}\\_{j, i} = 1$$$ (proof is by Cramer's Rule). We can keep track of the inverse after this clear out operation in $$$O(n^2 / w)$$$ if done properly (using something similar to Woodbury Matrix Identity, look at Solution 1 for details). If implemented properly, this results in a $$$O(n^3 / w)$$$ solution.\n\n**Solution 2**\n\n```\n#include <bits/stdc++.h>\nusing namespace std;\n \nconst int SZ = 400;\n \n \ntemplate<size_t SZ> int compute_inv(const vector<bitset<SZ>>& mat, vector<bitset<SZ>>& inv) {\n    int n = mat.size();\n    vector<bool> empty(n, true);\n    vector<bitset<SZ>> basis(n);\n \n    for(int i = 0; i < n; i++) {\n        bitset<SZ> cur = mat[i];\n        bitset<SZ> curinv;\n        curinv[i] = 1;\n        for(int j = 0; j < n; j++) {\n            if(cur[j]) {\n                if(empty[j]) {\n                    basis[j] = cur;\n                    inv[j] = curinv;\n                    empty[j] = false;\n                    break;\n                } else {\n                    cur ^= basis[j];\n                    curinv ^= inv[j];\n                }\n            }\n        }\n    }\n    for(int i = 0; i < n; i++) {\n        if(empty[i]) return 0;\n    }\n \n    for(int i = n - 1; i >= 1; i--) {\n        for(int j = 0; 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              "prompt": "The official Codeforces editorial fetch was saved at /Users/zayd/.local/share/opencode/tool-output/tool_0acbe4fee001SDIwptXW0OVT9t. Use Grep and Read on that file (do not read the whole thing) to locate the section for 2263E / 2262C, especially the Tutorial and any solution code/explanation. Return a concise but technically complete summary of the exact algorithm, formula, necessary conditions, candidate construction, combinatorial count, and complexity. Also flag any ambiguity or likely transcription issues. Do not edit files. This is research only.",
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            "output": "<task id=\"ses_f534173b3ffeUD7eeu1KGP5SxD\" state=\"completed\">\n<task_result>\nInspected `/Users/zayd/.local/share/opencode/tool-output/tool_0acbe4fee001SDIwptXW0OVT9t` around lines 478–538.\n\n### Editorial contents\n\nThe Tutorial is unavailable in the fetch: lines 482–484 contain only `Tutorial is loading...`. The only material is the solution code.\n\n### Algorithm\n\nLet:\n\n- `k = floor(n/2)`\n- `M = a[n−1]` (largest value)\n- `x` be the common difference.\n\nA rearrangement can be good only if its values are exactly\n\n\\[\nL_x=\\{0,x,2x,\\ldots,(k-1)x\\}\n\\]\n\nand\n\n\\[\nH_x=\\{M,M-x,M-2x,\\ldots,M-(n-k-1)x\\}.\n\\]\n\nThus, necessary and sufficient multiset validation is:\n\n\\[\n\\operatorname{sort}(L_x\\cup H_x)=a.\n\\]\n\nThe only possible `x` values are:\n\n```cpp\nM - a[n-2]\nM - a[n-3]\n```\n\nThat is, `M−x` must be either the second- or third-largest input value. At most one low value, `(k−1)x`, can lie above `M−x`; all smaller low values are below it. The `set` removes duplicate candidates.\n\nFor every candidate, the code constructs:\n\n```text\n0, x, ..., (k−1)x,\nM, M−x, ..., M−(n−k−1)x\n```\n\nsorts it, and compares it with the already-sorted input `a`. This checks nonnegativity, absence of overlaps/duplicates, and exact equality of the multisets.\n\n### Count for each valid `x`\n\nEvery good arrangement has:\n\n- positions `1..k`: an arbitrary permutation of `L_x` — `k!` choices;\n- position `k+1`: fixed to `M`;\n- position `n`: fixed to `M−x`;\n- positions `k+2..n−1`: arbitrary permutation of the remaining high values — `(n−k−2)!` choices.\n\nTherefore:\n\n\\[\n\\text{ways}(x)=k!(n-k-2)!.\n\\]\n\nEquivalently, the code computes:\n\n- even `n`: `k!(k−2)!`;\n- odd `n`: `k!(k−1)!`.\n\nAll arithmetic is modulo \\(10^9+7\\). Accepted distinct `x` values are added separately; this is intentional because they produce different positional families.\n\nA witnessing Hamiltonian path is, using \\(L_r=rx\\) and \\(H_r=M-rx\\),\n\n\\[\nL_{k-1}\\to H_{k-1}\\to L_{k-2}\\to H_{k-2}\\to\\cdots\\to L_1\\to H_1\\to L_0\\to H_0.\n\\]\n\n### Complexity\n\nAt most two candidates are tested. Each requires sorting `n` values:\n\n- Time: \\(O(n\\log n)\\) per test case\n- Space: \\(O(n)\\)\n\n### Ambiguities / transcription notes\n\n- The saved editorial has no Tutorial explanation, only the lazy-loading placeholder; the structural proof is therefore reconstructed from the code and problem definition.\n- `a[n-3]` is intentional, not an obvious off-by-one: it is the third-largest value and covers the case where `(k−1)x` lies between `M` and `M−x`.\n- The constructed vector named `b` is only a multiset-checking witness; the actual positional arrangements are counted implicitly by the factorial formula.\n</task_result>\n</task>",
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            "output": "Title: [CF666B]World Tour 题解 | KSkun's Blog\nURL: https://ksmeow.moe/wt_cf666b_sol/\nPublished: 2018-03-10T14:43:14.000Z\nAuthor: KSkun\nHighlights:\nA famous sculptor Cicasso goes to a world tour! Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn’t he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland. Cicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has “favourites”. Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn’t like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help. There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest. Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: 1., 2, 3, 4, 5., 4 ,3, 2., 3, 4.. Four cities in the order of visiting marked as overlines: [1, 5, 2, 4]. Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do. 一张n个点m条边的有向图,每条边的权值相同.你要找4个点a,b,c,d,使得a->b->c->d的最短路最长(a,b,c,d之间要有路),输出一组解.\n...\n输入格式： In the first line there is a pair of integers n and m (4 ≤ n ≤\n...\n000\n...\n3 ≤ m\n...\n000) —\n...\ncities and one-way roads in\n...\nEach of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi\n...\nNote that ui and vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.\n...\nasso will visit according to optimal choice of the route.\n...\nthe order that Cicasso will visit them\n...\nmultiple solutions, print any\n...\n# 题解\n...\n既然是4个点，我们就考虑枚举的方法来做。枚举4个点复杂度太大，那么我们只枚举中间两个点，然后再处理出从其中一个点出发最远能到达的点以及从一个点出发最远能到这个点的点。但是万一前面的两个最远会与之前已经枚举了的点重复了怎么办？那我们考虑记下最远、次远、第三远三个点，这样总不会重了吧。因为A→B→C→D如果ABC的枚举已经完成，D只可能跟AB重，也就是说，最多可以和2个点重复，那么需要记下的点最少就是三个。 说白了就是个大力BFS大力枚举。\n...\nque.\n...\nif(!\n...\nvis[v] = true;\n...\n, v);\n...\nint main() {\n    n = readint();\n    m = readint();\n    for(int i = 1; i <= m; i++) {\n        ut = readint();\n        vt = readint();\n        gra[ut].push_back(vt);\n    }\n    for(int i = 1; i <= n; i++) bfs(i);\n    int ans = 0, ansc[4];\n    for(int i = 1; i <= n; i++) {\n        for(int j = 1; j <= n; j++) {\n            if(i == j || dis[i][j] == 0) continue;\n            for(int k1 = 0; k1 < 3; k1++) {\n                for(int k2 = 0; k2 < 3; k2++) {\n                    if(!check(lt[i][k1], i, j, lf[j][k2])) continue;\n                    if(dis[lt[i][k1]][i] + dis[i][j] + dis[j][lf[j][k2]] > ans) {\n                        ans = dis[lt[i][k1]][i] + dis[i][j] + dis[j][lf[j][k2]];\n                        ansc[0] = lt[i][k1];\n                        ansc[1] = i;\n                        ansc[2] = j;\n                        ansc[3] = lf[j][k2];\n                    }\n                }\n            }\n        }\n    }\n    for(int i = 0; i < 4; i++) {\n        printf(\"%d \", ansc[i]);\n    }\n    return 0;\n}\n\n---\n\nTitle: [Thought process] Solving 1503C — Travelling Salesman Problem - Codeforces\nURL: https://codeforces.com/topic/92222/en4\nPublished: N/A\nAuthor: N/A\nHighlights:\nprocess] Solving 1503C — Travelling Salesman Problem - Codeforces\n...\n[Thought process] Solving 1503C — Travelling Salesman Problem\n...\nthe problem came out, someone\n...\nfor an explanation of how people come up with\n...\neven before that,\n...\nI've\n...\nAnyway now, two months later, I finally had the time to put this\n...\nBut the reason I really wanted to write this blog is that the solution to 1503C - Travelling Salesman Problem has a very much non-obvious greedy solution. From reading beginners' comments, it seems that a lot of people have this implicit belief that greedy algorithms are invented by guessing: for example, guessing an order in which to process things, guessing how to process things and then maybe proving correctness if you have time (but really skipping it, because who cares, right?). I think some contest sites even perpetuate this idea by presenting a greedy algorithm in the editorial without any kind of proof or explanation where this comes from. Even here, you may have tutorials that say \"here's the algorithm\" and after that \"here's the proof\". While it is convenient (and completely fine!) to present the solution in this form (algorithm and then proof), it's not necessarily how the solution was invented.\n...\nWhile you can solve problems this way (guessing and then proving (or not proving)) and I have solved problems this way, it is quite unusual for me. If someone were to think (perhaps implicitly) that greedy algorithms are invented like that, they'd be shocked when they read the solution to 1503C - Travelling Salesman Problem: how could anyone guess that, how could anyone come up with ordering the cities like this and then doing things like that?\n...\nAnd the answer is that you really can't. It would be very hard to guess this solution unless you already have some insights to the problem. That is roughly how I feel I solve problems: try to really understand the problem, gather insights about it. And at some point, you have found enough observations that you can piece together a solution. The solution isn't proved afterwards — the proof develops alongside the solution. It isn't always like this, but very often it is. (Of course, there is some guesswork involved, but it is more \"I guess I should think about this thing\", not \"I guess the entire algorithm\").\n...\n#### The problem\n...\nGiven $$$n$$$ cities and two arrays $$$a$$$ and $$$c$$$ of length $$$n$$$, you want to visit every city once, starting from city $$$1$$$ and returning to city $$$1$$$. The cost of going from $$$i$$$ to $$$j$$$ is\n...\n$$$\\max(c_i, a_j - a_i),$$$\n...\nand you must minimize the total cost. We have $$$2 \\le n \\le 10^5$$$ and $$$0 \\le a_i, c_i \\le 10^9$$$. $$$~$$$\n...\n#### The path to the solution\n...\nI tried to keep the following as true to my thought process as possible, so this is my path to my solution and not necessarily the shortest path to the cleanest solution. And maybe even not the clearest explanation because I really did try to mirror what was going on in my head.\n...\nWe have to pay $$$\\max(c_i, a_j - a_i)$$$ every time. If you ignore the $$$c_i$$$ terms, then the $$$a_j - a_i$$$ terms cancel out. So, let's think about the extra $$$c_i$$$ that we may have to pay sometimes.\n...\nActually, let's do it the other way around. We always pay at least $$$c_i$$$ for each town, right? So it would make sense to think of $$$\\sum_i c_i$$$ as the baseline. And to think only about the extra $$$\\max(0, a_j - a_i - c_i)$$$ that we have to pay sometimes.\n...\nActually, $$$\\max(0, a_j - a_i - c_i) = \\max(0, a_j - (a_i + c_i))$$$. So from city $$$i$$$, flying in the direction of decreasing beauty is always free. Flying in the direction of increasing beauty is free to a point, and after that we are hit with a tariff of 1 dollar per beauty. Visually, I imagine it something like this:\n...\nThe idea of cities as points on a line seemed useful, let's do more of that. Draw from $$$a_i$$$ an arc to $$$a_i + c_i$$$ to denote the area of free travel from city $$$i$$$. If I do this for many steps, \"connected components\" (sorta) arise:\n...\nExperience tells me that within a component, I can move from anywhere to anywhere, free of charge. (See for example DFS-tree blog, observations 5 and 6 for a proof that can be adapted and the source of this intuititon).\n...\nOf course, that's just normal shortest paths on graphs, without all this stuff about visiting every vertex once and only once. But I'm hopeful. You see these red bits in this image:\n...\nThose are the bits that I absolutely have to pay for. Do I have to pay for anything else? If I enter each component through the leftmost vertex, find some Hamiltonian path within the component from the leftmost vertex to the \"most powerful one\" (i.e. the one with highest $$$a_i + c_i$$$) and exit through the most powerful vertex, then I will only need to pay for the red bits. Can I always find such a path in a component?\n...\nOh, I think I can't. For example, here\n...\nI have to enter the component through the vertex $$$u$$$, and exit through $$$u$$$ too. There's no opporturnity for me to visit vertices $$$v$$$, $$$w$$$ and $$$x$$$.\n...\nBut wait, I don't actually need to visit all the vertices of the component. It is enough for me to find some free path from the leftmost vertex of the component to the most powerful one, not visiting any vertex twice. Then, I'll just visit all the missed vertices on my way back from right to left. Can I always do that? Yeah, it looks like I can: if I am at the most powerful vertex of the component, there will always be some vertex to the left that can reach this vertex. And there will always be some vertex to the left of that vertex that can reach that vertex, and so on, until we reach the beginning of the component.\n...\nOk! Now I have constructed a Hamiltonian circuit that will take me through each city and will always pay for the \"red bits\" and nothing else. It's also clear that this is the best solution because it's easy to see we absolutely have to pay for the red bits. So all that's left to do is implement code that will calculate the total length of the red segments. It looks like this:\n...\n```\nsort(arr, arr + n); // sorts by a\n\nll ans = 0;\nll mxac = 0;\nfor (int i = 0; i < n; i++) {\n  if (i != 0 && mxac < arr[i].a) {\n    ans += arr[i].a - mxac;\n  }\n  mxac = max(mxac, arr[i].a + arr[i].c);\n}\n\n```\n...\nAdd to`ans` the sum of $$$c_i$$$ and output that. That's it!\n\n---\n\nTitle: CodeForces 666B World Tour(spfa+枚举)\nURL: https://www.365seal.com/y/wWperYXPnG.html\nPublished: N/A\nAuthor: N/A\nHighlights:\nA famous sculptor Cicasso goes to a world tour!\n...\nWell, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.\n...\nCicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has \"favourites\". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.\n...\nThere are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.\n...\nNote that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that:. Four cities in the order of visiting marked as overlines:[1, 5, 2, 4].\n...\nNote that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.\n...\nIn the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.\n...\nEach of the next m lines contains a pair of integers ui, vi (1 ≤ ui, vi ≤ n) — a one-way road from the city ui to the city vi. Note that uiand vi are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.\n...\nPrint four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.\n...\nFloyed求最短路回超时，用spfa然后再枚举4个点中的中间两个点，头尾两个点\n...\n只需要取最大的就好了，当然不能有重复\n...\n```\n#include <iostream>\n#include <string.h>\n#include <stdlib.h>\n#include <string>\n#include <math.h>\n#include <stdio.h>\n#include <vector>\n#include <queue>\n#include <algorithm>\n...\nusing namespace std;\n...\nconst int INF=1000000;\n...\n#define MAX 3000\n...\nvector < pair<int,int> > a[MAX+5];\nvector < pair<int,int> > b[MAX+5];\nvector <int> c[MAX+5];\nint d[MAX+5][MAX+5];\nbool vis[MAX+5];\nint res[5];\n...\nint n,m;\n...\nvoid spfa(int i)\n{\n   memset(vis,false,sizeof(vis));\n   queue<int> q;\n   q.push(i);vis[i]=1;\n   while(!q.empty())\n   {\n       int x=q.front();q.pop();\n       vis[x]=0;\n       for(int j=0;j<c[x].size();j++)\n       {\n           int next=c[x][j];\n           if(d[i][next]>d[i][x]+1)\n           {\n               d[i][next]=d[i][x]+1;\n               if(!vis[next])\n               {\n                   vis[next]=1;\n                   q.push(next);\n               }\n           }\n       }\n   }\n}\n...\nint main()\n{\n    scanf(\"%d%d\",&n,&m);\n    int x,y;\n    for(int i=1;i<=n;i++)\n        for(int j=1;j<=n;j++)\n        {\n              if(i==j) d[i][j]=0;\n              else d[i][j]=INF;\n        }\n    for(int i=1;i<=m;i++)\n    {\n        scanf(\"%d%d\",&x,&y);\n        c[x].push_back(y);\n    }\n    for(int i=1;i<=n;i++)\n        spfa(i);\n    for(int i=1;i<=n;i++)\n    {\n        for(int j=1;j<=n;j++)\n        {\n            if(i==j) continue;\n            if(d[i][j]!=INF) a[i].push_back(make_pair(d[i][j],j));\n            if(d[j][i]!=INF) b[i].push_back(make_pair(d[j][i],j));\n        }\n        sort(a[i].begin(),a[i].end());\n        sort(b[i].begin(),b[i].end());\n    }\n    int ans=0;\n    for(int i=1;i<=n;i++)\n    {\n        int nn=b[i].size();\n        for(int j=1;j<=n;j++)\n        {\n            if(i==j) continue;\n            if(d[i][j]==INF) continue;\n            int mm=a[j].size();\n            for(int k=nn-1;k>=0&&k>=nn-3;k--)\n            {\n                int kk=b[i][k].second;\n                if(kk==j||kk==i) continue;\n                for(int p=mm-1;p>=0&&p>=mm-3;p--)\n                {\n                    int pp=a[j][p].second;\n                    if(pp==i||pp==j||pp==kk) continue;\n                    if(ans<d[kk][i]+d[i][j]+d[j][pp])\n                    {\n                        ans=d[kk][i]+d[i][j]+d[j][pp];\n                        res[1]=kk;res[2]=i;res[3]=j;res[4]=pp;\n                    }\n                }\n            }\n        }\n    }\n    //printf(\"%d\\n\",ans);\n    for(int i=1;i<=4;i++)\n    {\n        if(i==4)\n            printf(\"%d\\n\",res[i]);\n        else\n            printf(\"%d \",res[i]);\n    }\n    return 0;\n}\n```\n\n---\n\nTitle: Codeforces Round #349 (Div. 1) B. World Tour 暴力最短路 - qscqesze - 博客园\nURL: https://www.cnblogs.com/qscqesze/p/5449684.html\nPublished: 2016-04-30T00:00:00.000Z\nAuthor: N/A\nHighlights:\nA famous sculptor Cicasso goes to a world tour!\n...\nWell, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.\n...\nCicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has \"favourites\". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.\n...\nThere are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.\n...\nNote that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: . Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].\n...\nNote that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.\n...\n— a one\n...\nway road from\n...\ncity ui to\n...\nNote that ui and vi are not required\n...\n, it can be several one-way roads between\n...\nsame pair of cities\n...\n## 题意\n...\n给你一个有向图，你需要选出四个点出来，使得dis[i][j]+dis[j][t]+dis[t][k]最大\n...\n当然这四个点一定得是可达的\n...\n## 题解：\n...\n暴力枚举j,t这两个点，然后我们去选择i和k这两个点就好了\n...\n这个我们可以一开始n^2logn去预处理两两之间的最短路\n...\n然后去暴力枚举就好了，i和k这俩显然就只可能是最大的那3个中的一个。\n...\n[maxn]; vector<pair<int,int> >fi[\n...\n]; vector<pair<int,int> >se[maxn]; void spfa(int x) { memset(inq,0,sizeof(inq)); for(int i=1;i<=n;i++) d[x][i]=1e9; d[x][x]=0; queue\n\n Q; Q.push(x); inq[x]=1; while(!Q.empty()) { int now = Q.front(); Q.pop(); inq[now]=0; for(int i=0;i<E[now].size();i++) { int next = E[now][i]; if(d[x][next]>d[x][now]+1) { d[x][next]=d[x][now]+1; if(!inq[next]) { inq[next]=1; Q.push(next); } } } } } int main() { scanf(\"%d%d\",&n,&m); for(int i=1;i<=m;i++) { int x,y; scanf(\"%d%d\",&x,&y); E[x].push_back(y); } for(int i=1;i<=n;i++)spfa(i); for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i j)continue; if(d[i][j]!=1e9)fi[i].push_back(make_pair(d[i][j],j)); if(d[j][i]!=1e9)se[i].push_back(make_pair(d[j][i],j)); } sort(fi[i].begin(),fi[i].end()); reverse(fi[i].begin(),fi[i].end()); sort(se[i].begin(),se[i].end()); reverse(se[i].begin(),se[i].end()); } long long ans=0,ans1=0,ans2=0,ans3=0,ans4=0; for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(i j)continue; if(d[j][i] 1e9)continue; for(int k=0;k<3&&k<fi[i].size();k++) { if(i fi[i][k].second)continue; if(j fi[i][k].second)continue; for(int t=0;t<3&&t<se[j].size();t++) { if(fi[i][k].second se[j][t].second)continue; if(i se[j][t].second)continue; if(j se[j][t].second)continue; long long tmp = d[j][i]+fi[i][k].first+se[j][t].first; if(tmp>ans) { ans = tmp; ans1 = se[j][t].second; ans2 = j; ans3 = i; ans4 = fi[i][k].second; } } } } } cout<<ans1<<\" \"<<ans2<<\" \"<<ans3<<\" \"<<ans4<<endl; }\n\n---\n\nTitle: World Tour CodeForces - 667D [暴力+bfs求解最短路]_牛客博客\nURL: https://blog.nowcoder.net/n/9ed477414248475d949b73e58ce62ca8\nPublished: 2018-03-08T00:00:00.000Z\nAuthor: N/A\nHighlights:\n# World Tour CodeForces - 667D [暴力+bfs求解最短路] \n\n time limit per test \n\n output \n\n standard output \n\n A famous sculptor Cicasso goes to a world tour!\n\n Well, it is not actually a world-wide. But not everyone should have the opportunity to see works of sculptor, shouldn't he? Otherwise there will be no any exclusivity. So Cicasso will entirely hold the world tour in his native country — Berland.\n...\nCicasso is very devoted to his work and he wants to be distracted as little as possible. Therefore he will visit only four cities. These cities will be different, so no one could think that he has \"favourites\". Of course, to save money, he will chose the shortest paths between these cities. But as you have probably guessed, Cicasso is a weird person. Although he doesn't like to organize exhibitions, he likes to travel around the country and enjoy its scenery. So he wants the total distance which he will travel to be as large as possible. However, the sculptor is bad in planning, so he asks you for help.\n\n There are n cities and m one-way roads in Berland. You have to choose four different cities, which Cicasso will visit and also determine the order in which he will visit them. So that the total distance he will travel, if he visits cities in your order, starting from the first city in your list, and ending in the last, choosing each time the shortest route between a pair of cities — will be the largest.\n\n Note that intermediate routes may pass through the cities, which are assigned to the tour, as well as pass twice through the same city. For example, the tour can look like that: . Four cities in the order of visiting marked as overlines: [1, 5, 2, 4].\n\n Note that Berland is a high-tech country. So using nanotechnologies all roads were altered so that they have the same length. For the same reason moving using regular cars is not very popular in the country, and it can happen that there are such pairs of cities, one of which generally can not be reached by car from the other one. However, Cicasso is very conservative and cannot travel without the car. Choose cities so that the sculptor can make the tour using only the automobile. It is guaranteed that it is always possible to do.\n\n Input \n\n In the first line there is a pair of integers n and m (4 ≤ n ≤ 3000, 3 ≤ m ≤ 5000) — a number of cities and one-way roads in Berland.\n\n Each of the next m lines contains a pair of integers u i, v i (1 ≤ u i, v i ≤ n) — a one-way road from the city u i to the city v i. Note that u i and v i are not required to be distinct. Moreover, it can be several one-way roads between the same pair of cities.\n\n Output \n\n Print four integers — numbers of cities which Cicasso will visit according to optimal choice of the route. Numbers of cities should be printed in the order that Cicasso will visit them. If there are multiple solutions, print any of them.\n...\n题意:相邻点的边权都是1,选择4个点,使得A->B->C->D的路径和最大并且保证A->B,B->C,C->D都是最短路\n...\n思路:n<=3000 . O(n^2)枚举第二个点i,第三个点j. 再用O(n^2)的bfs预处理出所有到其他点的最长长度. 至于为什么用BFS,因为权值都相等且为1. 变相的dijkstra,但是dijkstra的复杂度要到O(n^3). 因为四个点一定要不同,所以对于每个点要处理出前三远的点,包括反向和顺向\n...\n复杂度O(9*n^2)\n...\nvoid updates(int u,int v){\n    shun[u][3]=shun[u][2];\n    shun[u][2]=shun[u][1];\n    shun[u][1]=make_pair(v,dis[u][v]);\n    return ;\n}\n...\nvoid updaten(int u,int v){\n    int t=dis[u][v];\n    if(t>ni[v][1].second){\n        ni[v][3]=ni[v][2];\n        ni[v][2]=ni[v][1];\n        ni[v][1]=make_pair(u,dis[u][v]);\n    }\n    else if(t>ni[v][2].second){\n        ni[v][3]=ni[v][2];\n        ni[v][2]=make_pair(u,dis[u][v]);\n    }\n    else if(t>ni[v][3].second){\n        ni[v][3]=make_pair(u,dis[u][v]);\n    }\n    return ;\n}\n...\nvoid bfs(int st){\n    int vis[MAX_N];\n    memset(vis,0,sizeof(vis));\n    dis[st][st]=0;\n    queue <int>q;\n    q.push(st);\n    while(!q.empty()){\n        int v=q.front();q.pop();\n        vis[v]=1;\n        for(int i=0;i<edge[v].size();++i){\n            int u=edge[v][i];\n            if(vis[u])  continue;\n            dis[st][u]=dis[st][v]+1;\n            updates(st,u);\n            updaten(st,u);\n            vis[u]=1;\n            q.push(u);\n        }\n    }\n    return ;\n}\n...\nbool check(int i,int j,int k,int l){\n    int node[5];\n    node[1]=i,node[2]=j,node[3]=ni[i][k].first,node[4]=shun[j][l].first;\n    if(node[3]==0 || node[4]==0)    return false;\n    for(int i=1;i<=4;i++){\n        for(int j=i+1;j<=4;j++){\n            if(node[i]==node[j])    return false;\n        }\n    }\n    return true;\n}\n...\nint main(void){\n    cin >> n >>m;\n    for(int i=1;i<=m;i++){\n        int u,v;\n        scanf(\"%d%d\",&u,&v);\n        edge[u].push_back(v);\n    }\n    memset(dis,INF,sizeof(dis));\n    for(int i=1;i<=n;i++)   bfs(i);\n\n//    for(int i=1;i<=n;i++){\n//        printf(\"shun[%d][%d].second=%d\\n\",i,1,shun[i][1].second);\n//    printf(\"~%d\\n\",dis[1][i]);\n//    }\n\n    int node[50];\n    int mx=-INF;\n    for(int i=1;i<=n;i++){\n        for(int j=1;j<=n;j++){\n            if(i==j)    continue;\n            for(int k=1;k<=3;k++){\n                for(int l=1;l<=3;l++){\n                    if(!check(i,j,k,l)) continue;\n                    int ans=dis[i][j]+ni[i][k].second+shun[j][l].second;\n                    if(ans>INF) continue;\n                    if(ans>mx){\n                        mx=ans;\n                        node[1]=ni[i][k].first;\n                        node[2]=i;\n                        node[3]=j;\n                        node[4]=shun[j][l].first;\n//                         printf(\"%d %d %d %d ~%d\\n\",node[1],node[2],node[3],node[4],ans);\n                    }\n                }\n            }\n        }\n    }\n    printf(\"%d %d %d %d\\n\",node[1],node[2],node[3],node[4]);\n    return 0;\n}\n\n---\n\nTitle: Problem - 1662C - Codeforces\nURL: https://mirror.codeforces.com/problemset/problem/1662/C\nPublished: N/A\nAuthor: N/A\nHighlights:\nThe map of Europe can be represented by a set of $$$n$$$ cities, numbered from $$$1$$$ through $$$n$$$, which are connected by $$$m$$$ bidirectional roads, each of which connects two distinct cities. A trip of length $$$k$$$ is a sequence of $$$k+1$$$ cities $$$v_1, v_2, \\ldots, v_{k+1}$$$ such that there is a road connecting each consecutive pair $$$v_i$$$, $$$v_{i+1}$$$ of cities, for all $$$1 \\le i \\le k$$$. A special trip is a trip that does not use the same road twice in a row, i.e., a sequence of $$$k+1$$$ cities $$$v_1, v_2, \\ldots, v_{k+1}$$$ such that it forms a trip and $$$v_i \\neq v_{i + 2}$$$, for all $$$1 \\le i \\le k - 1$$$.\n...\nGiven an integer $$$k$$$, compute the number of distinct special trips of length $$$k$$$ which begin and end in the same city. Since the answer might be large, give the answer modulo $$$998\\,244\\,353$$$.\n...\nThe first line contains three integers $$$n$$$, $$$m$$$ and $$$k$$$ ($$$3 \\le n \\le 100$$$, $$$1 \\le m \\le n(n - 1) / 2$$$, $$$1 \\le k \\le 10^4$$$) — the number of cities, the number of roads and the length of trips to consider.\n...\nEach of the following $$$m$$$ lines contains a pair of distinct integers $$$a$$$ and $$$b$$$ ($$$1 \\le a, b \\le n$$$) — each pair represents a road connecting cities $$$a$$$ and $$$b$$$. It is guaranteed that the roads are distinct (i.e., each pair of cities is connected by at most one road).\n...\nPrint the number of special trips of length $$$k$$$ which begin and end in the same city, modulo $$$998\\,244\\,353$$$.\n...\nIn the first sample, we are looking for special trips of length $$$2$$$, but since we cannot use the same road twice once we step away from a city we cannot go back, so the answer is $$$0$$$.\n...\nIn the second sample, we have the following $$$12$$$ special trips of length $$$3$$$ which begin and end in the same city: $$$(1, 2, 4, 1)$$$, $$$(1, 3, 4, 1)$$$, $$$(1, 4, 2, 1)$$$, $$$(1, 4, 3, 1)$$$, $$$(2, 1, 4, 2)$$$, $$$(2, 4, 1, 2)$$$, $$$(3, 1, 4, 3)$$$, $$$(3, 4, 1, 3)$$$, $$$(4, 1, 3, 4)$$$, $$$(4, 3, 1, 4)$$$, $$$(4, 1, 2, 4)$$$, and $$$(4, 2, 1, 4)$$$.\n\n---\n\nTitle: Codeforces Round #896 Travel Plan Solution (Java/C++)\nURL: https://www.xloypaypa.pub/codeforces-round-896-travel-plan-solution-java-c/\nPublished: 2023-10-05T00:00:00.000Z\nAuthor: N/A\nHighlights:\nCodeforces Round #896 Travel Plan Solution (Java/C++)\n...\nRound #89\n...\nTravel Plan Solution (Java/C++)\n...\ni,j\n...\nnumber of nodes visited from node $i\n...\nj$ along the entire path\n...\nFor example, in Sample 1, $len(2,3\n...\n3$, $len(1,2\n...\n$max\\_deep$,\n...\nthe height of the entire tree. For example, when $n=2$, $max\\_deep=2$. When $n=9$, $max\\_deep=4$.\n...\ndeep(node)$ calculates the depth of a specific node.\n...\nlayer(layer\n...\n### Overall Approach\n...\nThe problem can be broken down into three main steps:\n...\nStep 1: Maintain a $dp\\_s[i][j]$, representing all possibilities of a path of length $j$ where the maximum value $s$ is $i$. Here, $1\\leq i\\leq m$，$1\\leq j \\leq 2 \\cdot max\\_deep - 1$. (In other words, the longest possible path in this tree contains at most $2 \\cdot max\\_deep - 1$ nodes)\n...\nStep 2: Maintain a $dp\\_len[i]$, representing the total number of paths with length $i$ in this tree. For instance, when $n=4$, $dp\\_len[2]=3$, $dp\\_len[1]=4$.\n...\nStep 3: Calculate the final result. The final result equals $\\sum_{len=1}^{2 \\cdot max\\_deep - 1}dp\\_len[len]\\cdot m^{n-len} \\cdot (\\sum_{s=1}^{m} s\\cdot dp\\_s[s][len])$. This means, for each specified length of the path $len$: There are $dp\\_len[i]$ different paths. Nodes outside of this path can be chosen arbitrarily, leading to $m^{n-len}$ possibilities. Nodes within this path, for each possible $s$, offer $dp\\_s[s][len]$ choices. Finally, the result needs to be multiplied by $s$ in the calculation.\n...\n### Why this idea?\n...\n1. The expression $\\sum_{i=1}^n\\sum_{j=i}^n s_{i,j}$ involves summation, and the final result also requires a double summation over this. It is natural to consider rearranging $s_{i,j}$. Unify the summation.\n2. Since $dp\\_s[i][j]$ is relatively easy to obtain, the only thing missed from the final result is $dp\\_len[i]$. This problem seemed to resemble a classic one, and it was assumed there should be an existing solution.\n...\n### Calculation of $dp\\_s[i][j]$\n...\nFirstly, a straightforward combinatorial method came to mind. It's clear that $dp\\_s[i][j]=\\sum_{k=1}^{j} C_j^k\\cdot (i-1)^{j-k}$. In other words, select $k$ out of $j$ nodes and set their value to $i$, while the values of other nodes can be any value between $1$ and $i-1$.\n...\nHowever, this method has an issue with its time complexity: $O(m\\cdot max\\_deep^2)\\approx 10^5 \\cdot {(10^2)}^2 \\approx 10^9$. This is because for each $i,j$, we need to loop through $k$.\n...\nSo, I tried to observe the relationship between $dp\\_s[i][j]$ and other nodes. I noticed that $dp\\_s[i][j]=i\\cdot dp\\_s[i][j-1] + \\sum_{k=1}^{i-1} dp\\_s[k][j-1]$. This means either the maximum value of the previous $j-1$ nodes was already $i$, allowing the new node to be any value between $1$ and $i$. Or the previous maximum value was less than $i$, and the new node must be $i$.\n...\nThe term $\\sum_{k=1}^i dp\\_s[k][j-1]$ can be maintained as a prefix sum, so the complexity is $O(m\\cdot max\\_deep)$.\n...\n### Calculation of $dp\\_len[i]$\n...\nThere are various methods to solve this, and I chose a relatively straightforward approach.\n...\nFor a given path, if we know its starting and ending points, we can determine the length of this path and all the nodes it passes through. Furthermore, if we know the starting point, the topmost point, and the depth of the ending point of this path, we can calculate: 1. The length of this path; 2. The total number of such paths. Let's denote the starting point as $from$, the depth of the topmost point as $father$, and the depth of the ending point as $to$. It's not hard to find that for each pair $[from, father, to]$, the length of this path is $[father-deep(from)] + (father - to) + 1$, and there are $2^{to - father -1}$ possible paths. So, $dp\\_len[father-deep(from) + (father - to) + 1]+=number\\_of\\_nodes\\_at\\_layer(deep(from))\\cdot 2^{to - father -1}$. In other words, we iterate through $deep(from)$, for each starting point with depth $deep(from)$, we iterate through the depth of the topmost point and the depth of the ending point, and add the result to $dp\\_len[i]$ one by one. (Note that this idea chooses a fixed starting point, not the depth of the starting point.)\n...\nThere are two issues here: 1. When $father=to$, there is only 1 possible path. We handle this as a special case. 2. When $deep(from)=to$, many paths are counted twice. This is more complicated to handle.\n...\nFor the second issue, I chose to handle it separately. So, first, in the calculation process mentioned above, when enumerating $to$, limit its range to $deep(from)-1$. This ensures that at least the starting and ending points have different depths, making our results correct. It also handles non-complete binary trees.\n...\nFor the case where the depths of the starting and ending points are the same, we consider the situation of the topmost point of the path. Because two leaf nodes must be in the left and right subtrees of the topmost node. So, For non-complete binary trees, there are three possible scenarios for these left and right subtrees: 1. Both subtrees are complete; 2. At least one of subtree is not complete; 3. At least one of subtrees don't reach the required depth.\n...\nLet's denote the depth of the starting and ending points as $target$ and the depth of the topmost point as $father$. For the first scenario, for each $father$, there are $(2^{target-father -1})^2$ possible selections. There are $\\lfloor\\frac {number\\_of\\_nodes\\_at\\_layer(target)} {2^{target-father}}\\rfloor$ fathers satisfying this condition. For the third scenario, obviously, there are 0 possibilities. For the second scenario, there is only one father that satisfies this condition. To calculate its selection count, we need to find out how many leaf nodes are in its right subtree. It can be observed that the right subtree has $\\max([number\\_of\\_nodes\\_at\\_layer(target) \\% (2^{target-father})] - (2^{target-father -1}),0)$ leaf nodes, denoted as $num\\_right$ (subtracting left subtree and first scenario). So, its selection count is $num\\_right \\cdot (2^{target-father -1})$. Finally, add up all the above results to $dp\\_len[i]$. Which is $dp\\_len[(target-father) * 2 + 1]+=(2^{target-father -1})^2\\cdot \\lfloor\\frac {number\\_of\\_nodes\\_at\\_layer(target)} {2^{target-father}}\\rfloor +$$(2^{target-father -1}) \\cdot \\max([number\\_of\\_nodes\\_at\\_layer(target) \\% (2^{target-father})] - (2^{target-father -1}),0) $\n\n---\n\nTitle: 【题解】Codeforces 2062F Traveling Salescat_doscommand的技术博客_51CTO博客\nURL: https://blog.51cto.com/u_12935/14299847\nPublished: 2025-11-02T22:58:53.000Z\nAuthor: N/A\nHighlights:\nTraveling Salescat\n...\n## 题目大意\n...\n给定一张图含 \\(N\\) 个点，第 \\(i\\) 点有属性 \\(a_i\\) 和 \\(b_i\\)，任意两点间都有无向边，边权为 \\(max(a_i + b_j, a_j + b_i)\\) ，对于特定整数 \\(k\\) ，求一条恰经过 \\(k\\)\n...\n## 思路\n...\n看到 \\(max\\) 内部和多个点有关的任务，如果暴力处理求 \\(max\\) 肯定会多一个 \\(N\\) 的复杂度，第一想法就是想办法在 \\(max\\) 内部把 \\(i\\) 和 \\(j\\)\n...\n基于这个思路，很容易想到如下公式：\n\n\\[w_{ij} = max(a_i + b_j, a_j + b_i) = max(a_i - b_i, a_j - b_j) + b_i + b_j \\]\n...\n其中 \\(w_{ij}\\) 为 \\(i\\) 和 \\(j\\)\n...\n此时很显然的，如果我先把 \\(N\\) 个点按照 \\(a_i - b_i\\)\n...\n\\[w_{ij} = \\begin{cases} a_j + b_i, i<j\\\\ a_i + b_j, j ```plain\n...\nclass City{\npublic:\n    long long a, b;\n\n    bool operator<(const City &o) const {\n        return a-b<o.a-o.b;\n    }\n};\n...\nCity city[N];\n\nlong long ans[N];\n\nlong long dp[N][3];\nlong long tdp[N][3];\n...\nint main() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n    cout.tie(nullptr);\n\n    int T;\n    cin>>T;\n    while(T--) {\n        int n;\n        cin>>n;\n        for(int i=1;i<=n;i++) {\n            cin>>city[i].a>>city[i].b;\n            ans[i] = INF;\n            for(int j=0;j<3;j++) {\n                tdp[i][j] = INF;\n            }\n        }\n\n        sort(city+1, city+1+n);\n\n        for(int i=1;i<=n;i++) {\n            for(int j=0;j<=n;j++) {\n                for(int k=0;k<3;k++) {\n                    dp[j][k] = tdp[j][k];\n                }\n            }\n\n            tdp[1][0] = min(tdp[1][0], city[i].b*2); // vs > v1 时，v1的贡献\n            tdp[1][1] = min(tdp[1][1], city[i].b); // vs < v1 时，vs的贡献\n\n            for(int j=1;j<=n;j++) {\n                for(int k=0;k<3;k++) {\n                    tdp[j+1][k] = min(tdp[j+1][k], dp[j][k] + city[i].a + city[i].b); // 一般点的贡献\n                }\n\n                tdp[j+1][1] = min(tdp[j+1][1], dp[j][0] + city[i].a); // vs > v1 时，vs的贡献\n                tdp[j+1][2] = min(tdp[j+1][2], dp[j][1] + city[i].b); // vt < v_k-2 时，v_t的贡献\n\n                ans[j+1] = min(ans[j+1], dp[j][1] + city[i].a); // vt > v_k-2 时，vt的贡献\n                ans[j+1] = min(ans[j+1], dp[j][2] + city[i].a*2); // vt < v_k-2 时，v_k-2的贡献\n            }\n        }\n\n        for(int i=2;i<=n;i++) {\n            cout<<ans[i]<<\" \";\n        }\n        cout<<endl;\n    }\n}",
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            }
          },
          "id": "prt_0acc5aeb4001IKTALOUWyL4dym",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc49c0e001XqRD4zUPJgs6F5"
        },
        {
          "type": "tool",
          "tool": "websearch",
          "callID": "call_BB8cU9PVrgi468xVMm5K1tCA",
          "state": {
            "status": "completed",
            "input": {
              "query": "\"2262C\" \"Traveling the World\" x factorial"
            },
            "output": "Title: Top\nURL: https://codeforces.com/\nPublished: N/A\nAuthor: N/A\nHighlights:\ntasks.\n\nThe problems were authored and prepared by me, CutSandstone, and sukon.\n\nWe would like to thank:\n\n- abc864197532 for being the coordinator.\n- Um_nik for preliminary reviewing the round.\n- Alexdat2000 for translating the statements to Russian.\n- baluteshih, dlu, kevinyu, MattTheNub, n685, nadya271, yunz_qiao, _istil, alvingogo, avnithv, awoo, BurnedChicken, Fysty, Misuki, passwordisa, pbj2006, snowythecat, WiwiHo, and YuukiS for testing the round and providing valuable feedback.\n- KAN and MikeMirzayanov for the great Codeforces and Polygon platforms.\n- And finally, You, for participating!\n\nThe scoring distribution is:\n\n1. Div. 1: $$ — $$ — $$ — $$ — $$ — $$\n2. Div. 2: $$ — $$ — $$ — $$ — $$ — $$\n\nWe hope you enjoy the problems.\n\nGood luck and have fun!\n\nUPD:\n\nCongratulations to the winners (subject to change):\n\nDiv. 1:\n\n1. Kevin114514\n2. XorGhost\n3. strapple\n4. PiGoat\n5. zjy2008\n\nDiv. 2 (trusted participants):\n\n1. gausset\n2. Dream_Sky\n3. Guddu_06\n4. Saransh_Gupta_\n5. zhudaxia\n\nUPD 2:\n\nSorry for the late tutorial, its posted now!\n\nFull text and comments »\n\n Announcement of Codeforces Round 1120 (Div. 1)\n\n Announcement of Codeforces Round 1120 (Div. 2)\n\n- +278\n- kondasujay2 \n- 5 days ago\n- 123 \n\n Educational Codeforces Round 194 [rated for Div. 2] \n\nBy BledDest, 8 days ago, translation, \n\nHello Codeforces!\n\nThe series of Educational Rounds continues thanks to the support of the Neapolis University Pafos.\n\nEducational Codeforces Round 194 (Rated for Div. 2) will start on Sep/08/2026 17:35 (Moscow time).\n\nThis round will be rated for the participants with rating lower than 2100. It will be held on extended ICPC rules. The penalty for each incorrect submission until the submission with a full solution is 10 minutes. After the end of the contest, you will have 12 hours to hack any solution you want. You will have access to copy any solution and test it locally.\n\nYou will be given 7 problems and 2 hours to solve them. I hope everyone participating finds interesting problems for them to solve.\n\nI would like to express my gratitude to Mike MikeMirzayanov Mirzayanov for creating Codeforces and Polygon, two platforms which made conducting programming contests and preparing problems for them much more convenient. Also big thanks to testers: awoo, FelixArg, Alenochka.\n\nOur friends at Neapolis University Pafos have a message for you:\n\nPlanning to apply for the Computer Science and Artificial Intelligence (CSAI) program at Neapolis University Pafos?\n\nJetBrains Youth Clubs can help you prepare.\n\n🤖 AI Club is a chance to explore one of the core areas you’ll study in the CSAI program. You’ll work with AI topics, strengthen your understanding, and get more experience before university.\n\n➗ Math Club helps you sharpen the mathematical problem-solving skills you’ll need for the program. There’s also a direct application benefit: participation in the Math Club can give you 30% of your entrance test mark for the Computer Science and Artificial Intelligence BSc supported by JetBrains Foundation.\n\nBoth clubs are designed for high school students with a strong interest in STEM who want to challenge themselves and prepare for the next step.\n\n👉 Join JetBrains Youth Clubs: AI and Math\n\nGood luck in solving the problems! See you at the contest.\n\nUPD: The editorial has been published.\n\nFull text and comments »\n\n Announcement of Educational Codeforces Round 194 (Rated for Div. 2)\n\n- +214\n- BledDest \n- 8 days ago\n- 86 \n\n Codeforces Round 1119 (Div. 3) \n\nBy nik_exists, 2 weeks ago, \n\n`#include \"codeforces.h\"\n#include\n#include\n\nusing namespace std;\nusing namespace std::chrono_literals;\nusing namespace cf;\n\nint main() {\nBlog announcement(\"Codeforces Round 1119 (Div. 3)\");\n\n```\nannouncement << \"Hello Codeforces!\\n\";\nannouncement << \"I am delighted to invite you to participate in \" << `[Codeforces Round 1119 (Div. 3)](https://codeforces.com/contest/2259 \"Codeforces Round 1119 (Div. 3)\")` << \"!\\n\";\nannouncement << \"The problems were written and prepared by \" << `[nik_exists](https://codeforces.com/profile/nik%5Fexists \"Expert nik_exists\")` << \".\\n\\n\";\n\nannouncement << ContestInfo {\n    .start_time = `[Sep/05/2026 17:45 (Moscow time)](https://www.timeanddate.com/worldclock/fixedtime.html?day=5&month=9&year=2026&hour=17&min=45&sec=0&p1=166)`,\n    .duration = 2h + 15min,\n    .number_of_problems = 7,\n    .subtasks = true // possibly not adjacent\n};\n\nannouncement << RatingRules {\n    .rated_for_rating_below = 1600,\n    .unrated_registration_allowed = true,\n    .trusted_max_rating_below = 1900,\n    .trusted_min_rated_rounds = 5\n};\n\nannouncement << JudgingRules {\n    .scoring = codeforces::ICPC,\n    .wrong_answer_penalty = 10min,\n    .open_hacking_phase = 12h,\n    .final_tests = after_open_hacks\n};\n\nauto is_cheater = [](User user) {\n    // https://codeforces.com/blog/entry/133941\n    if (user.used_ai_in_contest()) return true;\n    if (user.discussed_problems_online_before_contest_end()) return true;\n    if (user.violated_other_rules_im_too_lazy_to_list()) return true;\n    return false;\n};\n\nfor (auto user : get_users(2259)) {\n    if (is_cheater(user)) user.send_to_cry_basement();\n}\n\nannouncement << Acknowledgments {\n    .coordinator = `[cry](https://codeforces.com/profile/cry \"International Master cry\")`,\n    .russian_translation = `[Vladosiya](https://codeforces.com/profile/Vladosiya \"Expert Vladosiya\")`,\n    .red_testers = {`[__baozii__](https://codeforces.com/profile/%5F%5Fbaozii%5F%5F \"International Grandmaster __baozii__\")`, `[AksLolCoding](https://codeforces.com/profile/AksLolCoding \"Grandmaster AksLolCoding\")`, `[awesomeguy856](https://codeforces.com/profile/awesomeguy856 \"Grandmaster awesomeguy856\")`, `[nifeshe](https://codeforces.com/profile/nifeshe \"Grandmaster nifeshe\")`},\n    .golden_testers = {`[Arpa](https://codeforces.com/profile/Arpa \"Master Arpa\")`, `[Intellegent](https://codeforces.com/profile/Intellegent \"International Master Intellegent\")`, `[Proof_by_QED](https://codeforces.com/profile/Proof%5Fby%5FQED \"Master Proof_by_QED\")`, `[temporary1](https://codeforces.com/profile/temporary1 \"Master temporary1\")`},\n    .purple_testers = {`[Argentum47](https://codeforces.com/profile/Argentum47 \"Candidate Master Argentum47\")`, `[Euclid73](https://codeforces.com/profile/Euclid73 \"Candidate Master Euclid73\")`, `[Jteh](https://codeforces.com/profile/Jteh \"Candidate Master Jteh\")`, `[Lilypad](https://codeforces.com/profile/Lilypad \"Candidate Master Lilypad\")`, `[SpyrosAliv](https://codeforces.com/profile/SpyrosAliv \"Candidate Master SpyrosAliv\")`, `[nimoxide](https://codeforces.com/profile/nimoxide \"Candidate Master nimoxide\")`, `[omsincoconut](https://codeforces.com/profile/omsincoconut \"Candidate Master omsincoconut\")`, `[wakanda-forever](https://codeforces.com/profile/wakanda-forever \"Candidate Master wakanda-forever\")`, `[yse](https://codeforces.com/profile/yse \"Candidate Master yse\")`},\n    .blue_testers = {`[CatsAreCool](https://codeforces.com/profile/CatsAreCool \"Expert CatsAreCool\")`, `[Vladosiya](https://codeforces.com/profile/Vladosiya \"Expert Vladosiya\")`, `[chromate00](https://codeforces.com/profile/chromate00 \"Expert chromate00\")`, `[fatespeaker](https://codeforces.com/profile/fatespeaker \"Expert fatespeaker\")`, `[linearspace](https://codeforces.com/profile/linearspace \"Expert linearspace\")`, `[reirugan](https://codeforces.com/profile/reirugan \"Expert reirugan\")`, `[simplelife](https://codeforces.com/profile/simplelife \"Expert simplelife\")`},\n    .green_testers = {`[hannah12345](https://codeforces.com/profile/hannah12345 \"Pupil hannah12345\")`},\n    .codeforces_and_polygon = {`[KAN](https://codeforces.com/profile/KAN \"Legendary Grandmaster KAN\")`, `[MikeMirzayanov](https://codeforces.com/profile/MikeMirzayanov \"Headquarters, MikeMirzayanov\")`}\n};\n\nannouncement << \"\\nBest of luck, and thank y'all so much for competing!\\n\";\n\n```\n\n}\n`\n\nEDITORIAL\n\nFull text and comments »\n\n Announcement of Codeforces Round 1119 (Div. 3)\n\n- +2123\n-\n\n---\n\nTitle: Travel around the world | HackerRank\nURL: https://www.hackerrank.com/challenges/travel-around-the-world/problem\nPublished: N/A\nAuthor: N/A\nHighlights:\nThere are N cities and N directed roads in Steven's world. The cities are numbered from 0 to N - 1. Steven can travel from city i to city (i + 1) % N, ( 0-> 1 -> 2 -> .... -> N - 1 -> 0).\n...\nSteven wants to travel around the world by car. The capacity of his car's fuel tank is C gallons. There are a[i] gallons he can use at the beginning of city i and the car takes b[i] gallons to travel from city i to (i + 1) % N.\n...\nHow many cities can Steven start his car from so that he can travel around the world and reach the same city he started?\n...\nThe first line contains two integers (separated by a space): city number N and capacity C. The second line contains N space-separated integers: a[0], a[1], … , a[N - 1]. The third line contains N space-separated integers: b[0], b[1], … , b[N - 1].\n...\n2 ≤ N ≤ 105 1 ≤ C ≤ 1018 0 ≤ a[i], b[i] ≤ 109\n...\nThe number of cities which can be chosen as the start city.\n...\nSteven starts from city 0, fills his car with 3 gallons of fuel, and use 2 gallons of fuel to travel to city 1. His fuel tank now has 1 gallon of fuel. On refueling 1 gallon of fuel at city 1, he then travels to city 2 by using 2 gallons of fuel. His fuel tank is now empty. On refueling 2 gallon of fuel at city 2, he then travels back to city 0 by using 2 gallons of fuel.\n...\nHere is the second possible solution. Steven starts from city 2, fill his car with 2 gallons, and travels to city 0. On refueling 3 gallons of fuel from city 0, he then travels to city 1, and exhausts 2 gallons of fuel. His fuel tank contains 1 gallon of fuel now. He can then refuel 1 gallon of fuel at City 1, and increase his car's fuel to 2 gallons and travel to city 2.\n...\nHowever, Steven cannot start from city 1, because he is given only 1 gallon of fuel, but travelling to city 2 requires 2 gallons.\n...\nHence the answer 2.\n\n---\n\nTitle: Problem - 1662C - Codeforces\nURL: https://codeforces.com/problemset/problem/1662/C\nPublished: N/A\nAuthor: N/A\nHighlights:\nProblem - 1662C - Codeforces\n...\nC. European Trip\n...\nThe map of Europe can be represented by a set of n cities, numbered from 1 through n, which are connected by m bidirectional roads, each of which connects two distinct cities. A trip of length k is a sequence of k+1 cities v1,v2,…,vk+1 such that there is a road connecting each consecutive pair vi, vi+1 of cities, for all 1≤i≤k. A special trip is a trip that does not use the same road twice in a row, i.e., a sequence of k+1 cities v1,v2,…,vk+1 such that it forms a trip and vi≠vi+2, for all 1≤i≤k−1.\n...\nGiven an integer k, compute the number of distinct special trips of length k which begin and end in the same city. Since the answer might be large, give the answer modulo 998244353.\n...\nThe first line contains three integers n, m and k (3≤n≤100, 1≤m≤n(n−1)/2, 1≤k≤104) — the number of cities, the number of roads and the length of trips to consider.\n...\nEach of the following m lines contains a pair of distinct integers a and b (1≤a,b≤n) — each pair represents a road connecting cities a and b. It is guaranteed that the roads are distinct (i.e., each pair of cities is connected by at most one road).\n...\nPrint the number of special trips of length k which begin and end in the same city, modulo 998244353.\n...\nIn the first sample, we are looking for special trips of length 2, but since we cannot use the same road twice once we step away from a city we cannot go back, so the answer is 0.\n...\nIn the second sample, we have the following 12 special trips of length 3 which begin and end in the same city: (1,2,4,1), (1,3,4,1), (1,4,2,1), (1,4,3,1), (2,1,4,2), (2,4,1,2), (3,1,4,3), (3,4,1,3), (4,1,3,4), (4,3,1,4), (4,1,2,4), and (4,2,1,4).\n\n---\n\nTitle: HackerRank Travel around the world problem solution\nURL: https://programmingoneonone.com/hackerrank-travel-around-the-world-problem-solution.html\nPublished: 2024-07-31T14:09:06.000Z\nAuthor: YASH PAL\nHighlights:\nIn this HackerRank Travel around the world problem solution There are N cities and N directed roads in Steven’s world. The cities are numbered from 0 to N – 1. Steven can travel from city i to city (i + 1) % N, ( 0-> 1 -> 2 -> …. -> N – 1 -> 0).\n...\nSteven wants to travel around the world by car. The capacity of his car’s fuel tank is C gallons. There are a[i] gallons he can use at the beginning of city I and the car takes b[i] gallons to travel from city I to (i + 1) % N.\n...\nHow many cities can Steven start his car from so that he can travel around the world and reach the same city he started?\n...\n#!/bin/python3 import os import sys # # Complete the travelAroundTheWorld function below. # def travelAroundTheWorld(a, b, c): total = 0 n = len(a) city_num_to_validate = n for index in range(n-1, -1, -1): tank = 0 is_valid = True for i in range(city_num_to_validate): curr_index = (index + i) % n tank += a[curr_index] if tank > c: tank = c tank -= b[curr_index] if tank < 0: is_valid = False break if is_valid: total += 1 city_num_to_validate = 1 elif city_num_to_validate < n: city_num_to_validate += 1 return total if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], '\n...\n') nc = input().split() n = int(nc[0]) c = int(nc[1]) a = list(map(int, input().rstrip().split())) b = list(map(int, input().rstrip().split())) result = travelAroundTheWorld(a, b, c) fptr.write(str(result) + 'n') fptr.close()\n...\n{\n...\nint n =\n...\nlong c = scanner.nextLong();\n...\nint a[] =\n...\nint[n\n...\nint b[] =\n...\nint[n\n...\nfor (int i = 0; i < n;\n...\na[\n...\nscanner.nextInt();\n...\nfor (int\n...\n] = scanner.nextInt();\n...\nint city = 0;\n    long fuel = 0;\n\n    for (int i = 0; i < n; i++) {\n\n        int j = 0;\n        while (j < n) {\n\n            fuel += a[i % n];\n            fuel = Math.min(fuel, c);\n\n            if (fuel >= b[i % n])\n                fuel -= b[i % n];\n            else {\n                fuel = 0;\n                break;\n            }\n            i++;\n            j++;\n        }\n\n        if (j == n)\n            city = i % n;\n        else city = -1;\n    }\n\n    int res = 0;\n    if (city >= 0) {\n        res = 1;\n        long ans[] = new long[n];\n        ans[city] = 0;\n        for (int j = 0; j < n - 1; j++) {\n            int i = (city - j - 1 + n) % n;\n            if (Math.min(a[i], c) - b[i] >= ans[(i + 1) % n]) {\n                ans[i] = 0;\n                res++;\n            } else {\n                ans[i] = ans[(i + 1) % n] - (Math.min(a[i], c) - b[i]);\n            }\n\n        }\n    }\n\n    System.out.println(res);\n}\n}\n...\nvol;\n    r =\n...\n4*n + 2;\n...\n<n; ++i\n...\n+n] = a[i\n...\nfor (\n...\n0;\n...\ntank = 0;\n    for (int i=0; i<2*n; ++i){\n        tank += a[i];\n        tank = min(tank, vol);\n        tank -= b[i];\n        if (tank < 0){\n            tank = 0;\n            s = i+1;\n        }\n    }\n    lld ans;\n    if (s >= n){\n        ans = 0;\n    }else {\n        ans = 1;\n        need[s+n] = 0;\n        for (int i=1; i<n; ++i){\n            int id = s+n-i;\n            need[id] = max((long long int)0, need[id+1] + b[id] - min((long long int)a[id], vol));\n            if (need[id] == 0) ans ++;\n        }\n    }\n...\nint main() {\n    static int aa[N * 2 + 1], qu[N * 2 + 1];\n    static long long pp[N * 2 + 1];\n    static char ok[N];\n    int n, i, j, head, cnt;\n    long long c;\n\n    scanf(\"%d%lld\", &n, &c);\n    for (i = 0; i < n; i++) {\n        scanf(\"%d\", &aa[i]);\n        aa[i + n] = aa[i];\n    }\n    for (i = 0; i < n; i++) {\n        int b;\n\n        scanf(\"%d\", &b);\n        pp[i + 1] = pp[i + n + 1] = aa[i] - b;\n    }\n    for (i = 1; i <= n * 2; i++)\n        pp[i] += pp[i - 1];\n...\nhead = cnt = 0;\n    for (i = n * 2, j = n * 2; i >= 0; i--) {\n        while (cnt && pp[qu[head + cnt - 1]] >= pp[i])\n            cnt--;\n        qu[head + cnt++] = i;\n        while (cnt && pp[qu[head]] - pp[i] < aa[i] - c)\n            if (qu[head] == j--)\n                head++, cnt--;\n        if (i < n)\n            ok[i] = j > i + n;\n    }\n    cnt = 0;\n    for (i = n * 2; i >= 0; i--) {\n        while (cnt && pp[qu[cnt - 1]] >= pp[i])\n            cnt--;\n        if (i < n && cnt && qu[cnt - 1] <= i + n)\n            ok[i] = 0;\n        qu[cnt++] = i;\n    }\n    cnt = 0;\n    for (i = 0; i < n; i++)\n        if (ok[i])\n            cnt++;\n    printf(\"%dn\", cnt);\n    return 0;\n}\n\n---\n\nTitle: C - Traveling Plan\nURL: https://atcoder.jp/contests/arc093/tasks/arc093_a?lang=en\nPublished: N/A\nAuthor: AtCoder Inc.\nHighlights:\nC - Traveling Plan\n\n Contest Duration: 2018-03-25 21:00:00+0900 - 2018-03-25 22:40:00+0900 (local time) (100 minutes) Back to Home \n\n Time Limit: 2 sec / Memory Limit: 256 MiB \n\n配点\n...\nx 軸上に N 個の観光スポットがあり、 1, 2, ..., N の番号がついています。 観光スポット i は座標 A_i の点にあります。 また、 x 軸上を座標 a の点から座標 b の点まで移動するには |a - b| 円かかります。\n...\nあなたは x 軸上を旅行する計画を立てました。 計画では、最初に座標 0 の点を出発し、 N 個の観光スポットを番号順に訪れ、最後に座標 0 の点に戻ってくることになっています。\n...\nところが、旅行の直前に急用が入り、 N 個すべての観光スポットを訪れる時間的余裕がなくなってしまいました。 そこで、ある i を選び、観光スポット i を訪れるのを取りやめることにしました。 それ以外の観光スポットは予定通り番号順に訪れます。 また、最初に座標 0 の点を出発し、最後に座標 0 の点に戻ってくることについても、予定に変更はありません。\n...\ni = 1, 2, ..., N それぞれについて、観光スポット i を訪れるのを取りやめたときの、旅行中の移動にかかる金額の総和を求めてください。\n...\n- 2 \\leq N \\leq 10^5\n- -5000 \\leq A_i \\leq 5000 (1 \\leq i \\leq N)\n- 入力値はすべて整数である。\n...\nN 行出力せよ。 このうち i 行目には、観光スポット i を訪れるのを取りやめたときの、旅行中の移動にかかる金額の総和を出力せよ。\n...\n観光スポット 1, 2, 3 はそれぞれ座標 3, 5, -1 の点にあります。 各 i について、観光スポット i を訪れるのを取りやめた場合の移動経路および移動にかかる金額は以下のようになります。\n...\n- i = 1 のとき、移動経路は 0 \\rightarrow 5 \\rightarrow -1 \\rightarrow 0 となり、移動にかかる金額は 5 + 6 + 1 = 12 円となります。\n- i = 2 のとき、移動経路は 0 \\rightarrow 3 \\rightarrow -1 \\rightarrow 0 となり、移動にかかる金額は 3 + 4 + 1 = 8 円となります。\n...\n- i = 3 のとき、移動経路は 0 \\rightarrow 3 \\rightarrow 5 \\rightarrow 0 となり、移動にかかる金額は 3 + 2 + 5 = 10 円となります。\n...\n### Problem Statement\n...\nThere are N sightseeing spots on the x-axis, numbered 1, 2, ..., N. Spot i is at the point with coordinate A_i. It costs |a - b| yen (the currency of Japan) to travel from a point with coordinate a to another point with coordinate b along the axis.\n...\nYou planned a trip along the axis. In this plan, you first depart from the point with coordinate 0, then visit the N spots in the order they are numbered, and finally return to the point with coordinate 0.\n...\nHowever, something came up just before the trip, and you no longer have enough time to visit all the N spots, so you decided to choose some i and cancel the visit to Spot i. You will visit the remaining spots as planned in the order they are numbered. You will also depart from and return to the point with coordinate 0 at the beginning and the end, as planned.\n...\nFor each i = 1, 2, ..., N, find the total cost of travel during the trip when the visit to Spot i is canceled.\n...\n- 2 \\leq N \\leq 10^5\n- -5000 \\leq A_i \\leq 5000 (1 \\leq i \\leq N)\n- All input values are integers.\n...\nPrint N lines. In the i-th line, print the total cost of travel during the trip when the visit to Spot i is canceled.\n...\nSpot 1, 2 and 3 are at the points with coordinates 3, 5 and -1, respectively. For each i, the course of the trip and the total cost of travel when the visit to Spot i is canceled, are as follows:\n...\n- For i = 1, the course of the trip is 0 \\rightarrow 5 \\rightarrow -1 \\rightarrow 0 and the total cost of travel is 5 + 6 + 1 = 12 yen.\n- For i = 2, the course of the trip is 0 \\rightarrow 3 \\rightarrow -1 \\rightarrow 0 and the total cost of travel is 3 + 4 + 1 = 8 yen.\n- For i = 3, the course of the trip is 0 \\rightarrow 3 \\rightarrow 5 \\rightarrow 0 and the total cost of travel is 3 + 2 + 5 = 10 yen.\n\n---\n\nTitle: JZOJ 5628. 【NOI2018模拟4.4】Travel - 灰信网（软件开发博客聚合）\nURL: https://www.freesion.com/article/6173330397/\nPublished: N/A\nAuthor: N/A\nHighlights:\n## JZOJ 5628. 【NOI2018模拟4.4】Travel\n...\n有N个人出去旅行，第i个人去A国有Ai种游玩方式，去B国有Bi种游玩方式，问至少有C个人去A国的情况下，所有人的游玩方式有多少种不同的可能。 两种所有人的游玩方式不同当且仅当存在一个人选择的游玩方式不同，或选择去的国家不同。 接下来有P次修改，每次修改一个人的Ai和Bi。\n...\n第一行两个正整数，表示N,C，含义如题所示。 接下来一行N个整数，第i个整数表示Ai。 接下来一行N个整数，第i个整数表示Bi。 接下来一行一个正整数表示P。 接下来P行，每行三个整数i,x,y，表示修改Ai为x，Bi为y。\n...\n对每次修改输出一行一个整数，表示总方案数 mod 10007。\n...\n对于100%的数据满足：Ai,Bi,x,y在int范围内。\n...\n直接线段树……\n...\n由于 c 特别的小，考虑正难则反。\n...\n即用总方案数 tot ，减去有 0 到 c−1 个人到A国的方案数 ans 。\n...\n易得：\n\ntot=∏i=1n(ai+bi)\n...\n开 c 棵线段树，代表一个区间代表恰好有 i 个人去A国的方案数。\n...\n那么有：\n\nans=∑i=0c−1f[root][i]\n...\n每次转移时枚举左右子树的 i ，O(c2) 转移即可。\n...\n总时间复杂度 O(Nc2 logN) 。\n...\ninline void update(int v)\n{\n    int ls=v<<1,rs=ls|1;\n    for(int i=0;i<c;i++) f[v][i]=0;\n    for(int i=0;i<c;i++)\n        for(int j=0;i+j<c;j++)\n            f[v][i+j]=(f[v][i+j]+f[ls][i]*f[rs][j])%mo;\n}\n...\nvoid make(int v,int l,int r)\n{\n    if(l==r)\n    {\n        f[v][0]=b[l];\n        f[v][1]=a[l];\n        return;\n    }\n    int mid=l+r>>1;\n    make(v<<1,l,mid);\n    make(v<<1|1,mid+1,r);\n    update(v);\n}\n...\nvoid change(int v,int l,int r)\n{\n    if(l==r)\n    {\n        f[v][0]=b[qi];\n        f[v][1]=a[qi];\n        return;\n    }\n    int mid=l+r>>1;\n    if(qi<=mid) change(v<<1,l,mid); else change(v<<1|1,mid+1,r);\n    update(v);\n}\n...\nint main()\n{\n    n=read(),c=read();\n    for(int i=1;i<=n;i++) a[i]=read()%mo;\n    for(int i=tot=1;i<=n;i++)\n    {\n        b[i]=read()%mo;\n        tot=tot*(a[i]+b[i])%mo;\n    }\n    make(1,1,n);\n    int p=read();\n    while(p--)\n    {\n        qi=read();\n        int x=read()%mo,y=read()%mo;\n        tot=tot*ksm(a[qi]+b[qi],mo-2)%mo;\n        tot=tot*(x+y)%mo;\n        a[qi]=x,b[qi]=y;\n        change(1,1,n);\n        int ans=0;\n        for(int i=0;i<c;i++) ans=(ans+f[1][i])%mo;\n        ans=(tot-ans+mo)%mo;\n        write(ans),putchar('\\n');\n    }\n    return 0;\n}\n```\n\n---\n\nTitle: 2262C Distributor, 2262C Price, 2262C Supplier - IC2world.com\nURL: https://ic2world.com/product/1138-2262C.html\nPublished: N/A\nAuthor: N/A\nHighlights:\n62C\n...\n62C\n...\nHome>> Product>> 2262C\n...\n| Part No.: | 2262C | Manufacturer: | TI |\n| --- | --- | --- | --- |\n| Package: | SOP8 | Date Code: |\n| Quantity: | 1409 | Price: |\n...\n# Get 2262C price, Buy 2262C\n...\nWe specialize in sourcing parts like 2262C for hard-to-find, long lead time, obsolete electronic components and ICs for customers everyday. We are professional 2262C supplier and 2262C distributor with good credit standing in china. Please Send RFQ to get 2262C price and other information. Buy electronic parts 2262C from Rantle East Electronic - Your Source for Quality Electronic Components in China.\n\n---\n\nTitle: Allen-Bradley 20-750-2262C-2R Industrial Control Module - newdcsplc Automation\nURL: https://www.newdcsplc.com/product/allen-bradley-20-750-2262c-2r-industrial-control-module/\nPublished: 2024-11-03T00:00:00.000Z\nAuthor: N/A\nHighlights:\nAllen-Bradley 20-750-2262C-2R Industrial Control Module - newdcsplc Automation\n...\n# Allen-Bradley 20-750-2262C-2R Industrial Control Module\n\n In stock \n\nThe Allen-Bradley 20-750-2262C-2R is a high-performance control module designed for industrial automation applications. It offers robust performance and reliability in various control system setups.\n...\nModel:20-750-2262C-2R\n...\nType:Programmable Logic Controller Module\n...\nInputs, 16\n...\nWeight:0\n...\n5 kg\n\n The Allen-Bradley 20-750-2262C-2R is designed for industrial environments where reliability and durability are paramount. Its robust construction ensures it can withstand harsh conditions, making it ideal for a wide range of applications including manufacturing, process control, and building automation.\n\n This PLC controller module integrates seamlessly with other Allen-Bradley products, providing a comprehensive solution for complex automation tasks. With its versatile communication interfaces and ample I/O points, it can handle a variety of input and output signals, ensuring efficient data processing and control.\n\n The 20-750-2262C-2R offers a broad operating temperature range, allowing it to function reliably in both cold and hot environments. Its compact size and lightweight design make installation easy and flexible, fitting into tight spaces without compromising performance.\n...\nThe module’s digital inputs and outputs provide precise control over various processes, enabling users to monitor and manage their operations efficiently. Its advanced features, combined with Allen-Bradley’s renowned reliability, make it an essential component for modern industrial automation systems.\n...\nInvesting in the Allen-Bradley 20-750-2262C-2R ensures long-term performance and minimal downtime. Its compatibility with existing systems and ease of maintenance make it a cost-effective choice for businesses looking to upgrade or expand their automation capabilities.",
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          "id": "prt_0acc5d3ab001fb1I4RckF6wMDa",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5b3ae001i7S4JiF0INXBK6"
        },
        {
          "type": "text",
          "text": "The official page exposes the accepted construction but not its prose tutorial, so I’m checking the structural claim directly on small exhaustive cases. The key risk is not the factorial count; it’s proving that no other value pattern can sneak through.",
          "time": {
            "start": 1789605041085,
            "end": 1789605042655
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          "metadata": {
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              "itemId": "msg_02fd12533bf62b7e016aab34b105c487d1afb93d1badb54748",
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        {
          "type": "tool",
          "tool": "bash",
          "callID": "call_LkjVLDrSQkRJVgGI5ccBhanD",
          "state": {
            "status": "completed",
            "input": {
              "command": "python3 -c 'import itertools\n\ndef good(b):\n n=len(b)\n f=[]\n for i,x in enumerate(b):\n  m=max(b[i:])\n  targets=[j for j,y in enumerate(b) if x+y==m]\n  f.append(targets)\n for p in itertools.permutations(range(n)):\n  if all(p[i+1] in f[p[i]] for i in range(n-1)):\n   return True,p\n return False,None\nfor n in range(4,8):\n print(\"n\",n)\n for b in itertools.permutations(range(n)):\n  ok,p=good(b)\n  if ok:\n   print(b,\"path\",p)\n print()'",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "n 4\n(0, 1, 3, 2) path (1, 3, 0, 2)\n(0, 2, 3, 1) path (1, 3, 0, 2)\n(1, 0, 3, 2) path (0, 3, 1, 2)\n(2, 0, 3, 1) path (0, 3, 1, 2)\n\nn 5\n(0, 1, 4, 2, 3) path (3, 1, 4, 0, 2)\n(1, 0, 4, 2, 3) path (3, 0, 4, 1, 2)\n\nn 6\n(0, 1, 2, 5, 3, 4) path (2, 4, 1, 5, 0, 3)\n(0, 2, 1, 5, 3, 4) path (1, 4, 2, 5, 0, 3)\n(0, 2, 4, 5, 1, 3) path (2, 4, 1, 5, 0, 3)\n(0, 4, 2, 5, 1, 3) path (1, 4, 2, 5, 0, 3)\n(1, 0, 2, 5, 3, 4) path (2, 4, 0, 5, 1, 3)\n(1, 2, 0, 5, 3, 4) path (1, 4, 0, 5, 2, 3)\n(2, 0, 1, 5, 3, 4) path (0, 4, 2, 5, 1, 3)\n(2, 0, 4, 5, 1, 3) path (2, 4, 0, 5, 1, 3)\n(2, 1, 0, 5, 3, 4) path (0, 4, 1, 5, 2, 3)\n(2, 4, 0, 5, 1, 3) path (1, 4, 0, 5, 2, 3)\n(4, 0, 2, 5, 1, 3) path (0, 4, 2, 5, 1, 3)\n(4, 2, 0, 5, 1, 3) path (0, 4, 1, 5, 2, 3)\n\nn 7\n(0, 1, 2, 6, 3, 4, 5) path (4, 2, 5, 1, 6, 0, 3)\n(0, 1, 2, 6, 4, 3, 5) path (5, 2, 4, 1, 6, 0, 3)\n(0, 2, 1, 6, 3, 4, 5) path (4, 1, 5, 2, 6, 0, 3)\n(0, 2, 1, 6, 4, 3, 5) path (5, 1, 4, 2, 6, 0, 3)\n(1, 0, 2, 6, 3, 4, 5) path (4, 2, 5, 0, 6, 1, 3)\n(1, 0, 2, 6, 4, 3, 5) path (5, 2, 4, 0, 6, 1, 3)\n(1, 2, 0, 6, 3, 4, 5) path (4, 1, 5, 0, 6, 2, 3)\n(1, 2, 0, 6, 4, 3, 5) path (5, 1, 4, 0, 6, 2, 3)\n(2, 0, 1, 6, 3, 4, 5) path (4, 0, 5, 2, 6, 1, 3)\n(2, 0, 1, 6, 4, 3, 5) path (5, 0, 4, 2, 6, 1, 3)\n(2, 1, 0, 6, 3, 4, 5) path (4, 0, 5, 1, 6, 2, 3)\n(2, 1, 0, 6, 4, 3, 5) path (5, 0, 4, 1, 6, 2, 3)\n\n",
            "metadata": {
              "output": "n 4\n(0, 1, 3, 2) path (1, 3, 0, 2)\n(0, 2, 3, 1) path (1, 3, 0, 2)\n(1, 0, 3, 2) path (0, 3, 1, 2)\n(2, 0, 3, 1) path (0, 3, 1, 2)\n\nn 5\n(0, 1, 4, 2, 3) path (3, 1, 4, 0, 2)\n(1, 0, 4, 2, 3) path (3, 0, 4, 1, 2)\n\nn 6\n(0, 1, 2, 5, 3, 4) path (2, 4, 1, 5, 0, 3)\n(0, 2, 1, 5, 3, 4) path (1, 4, 2, 5, 0, 3)\n(0, 2, 4, 5, 1, 3) path (2, 4, 1, 5, 0, 3)\n(0, 4, 2, 5, 1, 3) path (1, 4, 2, 5, 0, 3)\n(1, 0, 2, 5, 3, 4) path (2, 4, 0, 5, 1, 3)\n(1, 2, 0, 5, 3, 4) path (1, 4, 0, 5, 2, 3)\n(2, 0, 1, 5, 3, 4) path (0, 4, 2, 5, 1, 3)\n(2, 0, 4, 5, 1, 3) path (2, 4, 0, 5, 1, 3)\n(2, 1, 0, 5, 3, 4) path (0, 4, 1, 5, 2, 3)\n(2, 4, 0, 5, 1, 3) path (1, 4, 0, 5, 2, 3)\n(4, 0, 2, 5, 1, 3) path (0, 4, 2, 5, 1, 3)\n(4, 2, 0, 5, 1, 3) path (0, 4, 1, 5, 2, 3)\n\nn 7\n(0, 1, 2, 6, 3, 4, 5) path (4, 2, 5, 1, 6, 0, 3)\n(0, 1, 2, 6, 4, 3, 5) path (5, 2, 4, 1, 6, 0, 3)\n(0, 2, 1, 6, 3, 4, 5) path (4, 1, 5, 2, 6, 0, 3)\n(0, 2, 1, 6, 4, 3, 5) path (5, 1, 4, 2, 6, 0, 3)\n(1, 0, 2, 6, 3, 4, 5) path (4, 2, 5, 0, 6, 1, 3)\n(1, 0, 2, 6, 4, 3, 5) path (5, 2, 4, 0, 6, 1, 3)\n(1, 2, 0, 6, 3, 4, 5) path (4, 1, 5, 0, 6, 2, 3)\n(1, 2, 0, 6, 4, 3, 5) path (5, 1, 4, 0, 6, 2, 3)\n(2, 0, 1, 6, 3, 4, 5) path (4, 0, 5, 2, 6, 1, 3)\n(2, 0, 1, 6, 4, 3, 5) path (5, 0, 4, 2, 6, 1, 3)\n(2, 1, 0, 6, 3, 4, 5) path (4, 0, 5, 1, 6, 2, 3)\n(2, 1, 0, 6, 4, 3, 5) path (5, 0, 4, 1, 6, 2, 3)\n\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'import itertools\n\ndef good(b):\n n=len(b)\n f=[]\n for i,x in enumerate(b):\n  m=max(b[i:])\n  targets=[j for j,y in enumerate(b) if x+y==m]\n  f.append(targets)\n for p in itertools.permutations(range(n)):\n  if all(p[i+1] in f[p[i]] for i in range(n-1)):\n   return True,p\n return False,None\nfor n in range(4,8):\n print(\"n\",n)\n for b in itertools.permutations(range(n)):\n  ok,p=good(b)\n  if ok:\n   print(b,\"path\",p)\n print()'",
            "time": {
              "start": 1789605046470,
              "end": 1789605050846
            }
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          "reason": "tool-calls",
          "type": "step-finish",
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            "reasoning": 213,
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    },
    {
      "info": {
        "parentID": "msg_0acbe339600157tYVAEsQoU77E",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "max",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3",
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        },
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        "tokens": {
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        "modelID": "gpt-5.6-luna",
        "providerID": "openai",
        "time": {
          "created": 1789605050850,
          "completed": 1789605263315
        },
        "finish": "tool-calls",
        "id": "msg_0acc5f9e2001cUYiuJBQ1RlNvF",
        "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t"
      },
      "parts": [
        {
          "type": "step-start",
          "id": "prt_0acc5feb3001nD5kcXmXyGH0BO",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5f9e2001cUYiuJBQ1RlNvF"
        },
        {
          "type": "reasoning",
          "text": "**Analyzing valid set patterns for permutations**",
          "time": {
            "start": 1789605052088,
            "end": 1789605066010
          },
          "metadata": {
            "openai": {
              "itemId": "rs_0a0f24ab9a8a786f016aab34bc0a7487d180849658ea8ca406"
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          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5f9e2001cUYiuJBQ1RlNvF"
        },
        {
          "type": "reasoning",
          "text": "**Confirming valid x candidates and permutation counts**",
          "time": {
            "start": 1789605066012,
            "end": 1789605066067
          },
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          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5f9e2001cUYiuJBQ1RlNvF"
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          "type": "tool",
          "tool": "bash",
          "callID": "call_B0xHy9OmBP7qtU8Tbf3wzpVg",
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            "status": "completed",
            "input": {
              "command": "python3 -c 'import itertools\n\ndef good(b):\n n=len(b); f=[]\n for i,x in enumerate(b):\n  m=max(b[i:]); f.append(next((j for j,y in enumerate(b) if x+y==m),-1))\n for p in itertools.permutations(range(n)):\n  if all(f[p[i]]==p[i+1] for i in range(n-1)): return p\nfor n in range(4,7):\n found=[]\n for vals in itertools.combinations(range(1,2*n+2),n-1):\n  b0=(0,)+vals\n  for b in itertools.permutations(b0):\n   p=good(b)\n   if p is not None and p[-1]==n-1: found.append((b,p)); break\n  if found: break\n print(n, found[:1])'",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "4 []\n5 []\n6 []\n",
            "metadata": {
              "output": "4 []\n5 []\n6 []\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'import itertools\n\ndef good(b):\n n=len(b); f=[]\n for i,x in enumerate(b):\n  m=max(b[i:]); f.append(next((j for j,y in enumerate(b) if x+y==m),-1))\n for p in itertools.permutations(range(n)):\n  if all(f[p[i]]==p[i+1] for i in range(n-1)): return p\nfor n in range(4,7):\n found=[]\n for vals in itertools.combinations(range(1,2*n+2),n-1):\n  b0=(0,)+vals\n  for b in itertools.permutations(b0):\n   p=good(b)\n   if p is not None and p[-1]==n-1: found.append((b,p)); break\n  if found: break\n print(n, found[:1])'",
            "time": {
              "start": 1789605152901,
              "end": 1789605263311
            }
          },
          "metadata": {
            "openai": {
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          },
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          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5f9e2001cUYiuJBQ1RlNvF"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 52939,
            "input": 2660,
            "output": 265,
            "reasoning": 3550,
            "cache": {
              "write": 0,
              "read": 46464
            }
          },
          "cost": 0,
          "id": "prt_0acc937d0001rqyt44Xdtg840N",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc5f9e2001cUYiuJBQ1RlNvF"
        }
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    },
    {
      "info": {
        "parentID": "msg_0acbe339600157tYVAEsQoU77E",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "max",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3",
          "root": "/"
        },
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          "reasoning": 12637,
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          }
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        "providerID": "openai",
        "time": {
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          "completed": 1789605706362
        },
        "finish": "tool-calls",
        "id": "msg_0acc937d6001TWbu4hD9njjX4W",
        "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t"
      },
      "parts": [
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          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acc937d6001TWbu4hD9njjX4W"
        },
        {
          "type": "reasoning",
          "text": "**Planning theorem derivation and proof articulation**",
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            "end": 1789605275197
          },
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            "openai": {
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        },
        {
          "type": "reasoning",
          "text": "**Analyzing structural properties of functional graph**",
          "time": {
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        },
        {
          "type": "reasoning",
          "text": "**Deriving value sets from functional equations**",
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          "messageID": "msg_0acc937d6001TWbu4hD9njjX4W"
        },
        {
          "type": "reasoning",
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              "command": "python3 -c 'import itertools,math\n\ndef good(b):\n n=len(b); f=[]\n for i,x in enumerate(b):\n  m=max(b[i:]); js=[j for j,y in enumerate(b) if x+y==m]\n  f.append(js[0] if len(js)==1 else -1)\n for p in itertools.permutations(range(n)):\n  if all(f[p[i]]==p[i+1] for i in range(n-1)): return True\n return False\n\ndef formula(a):\n n=len(a); k=n//2; M=a[-1]; ans=0\n for x in {M-a[-2],M-a[-3]}:\n  b=[i*x for i in range(k)]+[M-i*x for i in range(n-k)]\n  if sorted(b)==list(a): ans+=math.factorial(k)*math.factorial(n-k-2)\n return ans\nfor n in range(4,8):\n bad=[]; total=0\n for a in itertools.combinations(range(0,2*n+4),n):\n  if a[0]!=0: continue\n  c=0\n  for b in itertools.permutations(a): c+=good(b)\n  f=formula(a)\n  if c!=f: bad.append((a,c,f)); break\n  total+=1\n print(n,\"checked\",total,\"bad\",bad[:1])'",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "(no output)\n\n<shell_metadata>\nshell tool terminated command after exceeding timeout 120000 ms. If this command is expected to take longer and is not waiting for interactive input, retry with a larger timeout value in milliseconds.\n</shell_metadata>",
            "metadata": {
              "output": "(no output)\n\n<shell_metadata>\nshell tool terminated command after exceeding timeout 120000 ms. If this command is expected to take longer and is not waiting for interactive input, retry with a larger timeout value in milliseconds.\n</shell_metadata>",
              "exit": null,
              "truncated": false
            },
            "title": "python3 -c 'import itertools,math\n\ndef good(b):\n n=len(b); f=[]\n for i,x in enumerate(b):\n  m=max(b[i:]); js=[j for j,y in enumerate(b) if x+y==m]\n  f.append(js[0] if len(js)==1 else -1)\n for p in itertools.permutations(range(n)):\n  if all(f[p[i]]==p[i+1] for i in range(n-1)): return True\n return False\n\ndef formula(a):\n n=len(a); k=n//2; M=a[-1]; ans=0\n for x in {M-a[-2],M-a[-3]}:\n  b=[i*x for i in range(k)]+[M-i*x for i in range(n-k)]\n  if sorted(b)==list(a): ans+=math.factorial(k)*math.factorial(n-k-2)\n return ans\nfor n in range(4,8):\n bad=[]; total=0\n for a in itertools.combinations(range(0,2*n+4),n):\n  if a[0]!=0: continue\n  c=0\n  for b in itertools.permutations(a): c+=good(b)\n  f=formula(a)\n  if c!=f: bad.append((a,c,f)); break\n  total+=1\n print(n,\"checked\",total,\"bad\",bad[:1])'",
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              "end": 1789605706357
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        "path": {
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          "type": "reasoning",
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                  "status": "in_progress",
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                {
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            "title": "3 todos",
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          "id": "prt_0acd1cbeb001WcCqdxdJZksx9t",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acd19fee001d3SYHtRvBg67L3"
        },
        {
          "type": "tool",
          "tool": "bash",
          "callID": "call_hEexetZsnYuGbwiQpX6IGfWK",
          "state": {
            "status": "completed",
            "input": {
              "command": "g++ -std=c++20 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\" - <<'CPP'\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\nCPP\n\"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\" <<'EOF'\n3\n6\n0 1 2 3 4 5\n5\n0 3 4 5 6\n4\n0 1 3 4\nEOF",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "ld: open() failed, errno=20 (Not a directory) for '/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c'\nclang++: error: linker command failed with exit code 1 (use -v to see invocation)\nzsh:58: not a directory: /var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\n",
            "metadata": {
              "output": "ld: open() failed, errno=20 (Not a directory) for '/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c'\nclang++: error: linker command failed with exit code 1 (use -v to see invocation)\nzsh:58: not a directory: /var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\n",
              "exit": 127,
              "truncated": false
            },
            "title": "g++ -std=c++20 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\" - <<'CPP'\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\nCPP\n\"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/solution_check/2262c\" <<'EOF'\n3\n6\n0 1 2 3 4 5\n5\n0 3 4 5 6\n4\n0 1 3 4\nEOF",
            "time": {
              "start": 1789605839563,
              "end": 1789605840240
            }
          },
          "metadata": {
            "openai": {
              "itemId": "fc_06f82a0020ec5065016aab37c16ce087d1870ab4f56ec3910e"
            }
          },
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          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acd19fee001d3SYHtRvBg67L3"
        },
        {
          "reason": "tool-calls",
          "type": "step-finish",
          "tokens": {
            "total": 71330,
            "input": 1393,
            "output": 637,
            "reasoning": 308,
            "cache": {
              "write": 0,
              "read": 68992
            }
          },
          "cost": 0,
          "id": "prt_0acd20571001oore1NByQ7buN9",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acd19fee001d3SYHtRvBg67L3"
        }
      ]
    },
    {
      "info": {
        "parentID": "msg_0acbe339600157tYVAEsQoU77E",
        "role": "assistant",
        "mode": "nudge-generation",
        "agent": "nudge-generation",
        "variant": "max",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3",
          "root": "/"
        },
        "cost": 0,
        "tokens": {
          "total": 72233,
          "input": 1440,
          "output": 633,
          "reasoning": 144,
          "cache": {
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            "read": 70016
          }
        },
        "modelID": "gpt-5.6-luna",
        "providerID": "openai",
        "time": {
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          "completed": 1789605856640
        },
        "finish": "tool-calls",
        "id": "msg_0acd20574001SjcsWbAiaVf6ZY",
        "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t"
      },
      "parts": [
        {
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          "messageID": "msg_0acd20574001SjcsWbAiaVf6ZY"
        },
        {
          "type": "reasoning",
          "text": "**Verifying directory and file structure**",
          "time": {
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          },
          "id": "prt_0acd20b9f001jKEQu1lCYifDV1",
          "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t",
          "messageID": "msg_0acd20574001SjcsWbAiaVf6ZY"
        },
        {
          "type": "reasoning",
          "text": "**Planning immediate tool invocation**",
          "time": {
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            "end": 1789605846312
          },
          "metadata": {
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              "command": "g++ -std=c++20 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/2262c_check\" - <<'CPP'\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\nCPP\n\"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/2262c_check\" <<'EOF'\n3\n6\n0 1 2 3 4 5\n5\n0 3 4 5 6\n4\n0 1 3 4\nEOF",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "12\n0\n4\n",
            "metadata": {
              "output": "12\n0\n4\n",
              "exit": 0,
              "truncated": false
            },
            "title": "g++ -std=c++20 -O2 -pipe -x c++ -o \"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/2262c_check\" - <<'CPP'\n#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\nCPP\n\"/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/opencode/2262c_check\" <<'EOF'\n3\n6\n0 1 2 3 4 5\n5\n0 3 4 5 6\n4\n0 1 3 4\nEOF",
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    {
      "info": {
        "parentID": "msg_0acbe339600157tYVAEsQoU77E",
        "role": "assistant",
        "mode": "nudge-generation",
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        "variant": "max",
        "path": {
          "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3",
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        "providerID": "openai",
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        "id": "msg_0acd24583001YJs6w8a1ZHLcZq",
        "sessionID": "ses_f5341cc79ffem0U3wCYsZqkI1t"
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        {
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          "id": "prt_0acd24f8d001kru1BJHlr3GSlc",
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          "messageID": "msg_0acd24583001YJs6w8a1ZHLcZq"
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        {
          "type": "reasoning",
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              "command": "python3 -c 'import itertools,math\n\ndef isgood(b):\n n=len(b); nxt=[]\n for i,v in enumerate(b):\n  s=max(b[i:]); q=[j for j,w in enumerate(b) if v+w==s and j!=i]\n  nxt.append(q[0] if len(q)==1 else -1)\n for st in range(n):\n  seen=set(); u=st\n  while u!=-1 and u not in seen:\n   seen.add(u); u=nxt[u]\n  if len(seen)==n:return True\n return False\n\ndef formula(a):\n n=len(a);k=n//2;M=a[-1];w=math.factorial(k)*math.factorial(n-k-2);ans=0\n for x in {M-a[-2],M-a[-3]}:\n  v=[i*x for i in range(k)]+[M-i*x for i in range(n-k)]\n  if sorted(v)==list(a):ans+=w\n return ans\nfor n in (4,5,6):\n checked=0\n for rest in itertools.combinations(range(1,10),n-1):\n  a=(0,)+rest\n  actual=sum(isgood(b) for b in itertools.permutations(a))\n  expected=formula(a)\n  assert actual==expected,(n,a,actual,expected)\n  checked+=1\n print(n,checked,\"sets verified\")'",
              "timeout": 120000,
              "workdir": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-IqMqa3"
            },
            "output": "4 84 sets verified\n5 126 sets verified\n6 126 sets verified\n",
            "metadata": {
              "output": "4 84 sets verified\n5 126 sets verified\n6 126 sets verified\n",
              "exit": 0,
              "truncated": false
            },
            "title": "python3 -c 'import itertools,math\n\ndef isgood(b):\n n=len(b); nxt=[]\n for i,v in enumerate(b):\n  s=max(b[i:]); q=[j for j,w in enumerate(b) if v+w==s and j!=i]\n  nxt.append(q[0] if len(q)==1 else -1)\n for st in range(n):\n  seen=set(); u=st\n  while u!=-1 and u not in seen:\n   seen.add(u); u=nxt[u]\n  if len(seen)==n:return True\n return False\n\ndef formula(a):\n n=len(a);k=n//2;M=a[-1];w=math.factorial(k)*math.factorial(n-k-2);ans=0\n for x in {M-a[-2],M-a[-3]}:\n  v=[i*x for i in range(k)]+[M-i*x for i in range(n-k)]\n  if sorted(v)==list(a):ans+=w\n return ans\nfor n in (4,5,6):\n checked=0\n for rest in itertools.combinations(range(1,10),n-1):\n  a=(0,)+rest\n  actual=sum(isgood(b) for b in itertools.permutations(a))\n  expected=formula(a)\n  assert actual==expected,(n,a,actual,expected)\n  checked+=1\n print(n,checked,\"sets verified\")'",
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              "end": 1789605869042
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            "openai": {
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        "role": "assistant",
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        "variant": "max",
        "path": {
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                {
                  "content": "Extract and verify the official 2262C solution idea",
                  "status": "completed",
                  "priority": "high"
                },
                {
                  "content": "Derive the counting formula and prove necessary/sufficient conditions",
                  "status": "completed",
                  "priority": "high"
                },
                {
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                  "status": "completed",
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                {
                  "content": "Check samples, edge cases, and implementation complexity",
                  "status": "completed",
                  "priority": "high"
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            "metadata": {
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                {
                  "content": "Check samples, edge cases, and implementation complexity",
                  "status": "completed",
                  "priority": "high"
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              "truncated": false
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            "title": "0 todos",
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          "hints": [
            {
              "order": 1,
              "content": "For a position $i$, let $s_i$ be the maximum value in the suffix starting at $i$. A destination must contain the single value $s_i-b_i$. Since all values are distinct, each position has at most one possible successor. Think of the arrangement as a functional directed graph, not a general graph monster."
            },
            {
              "order": 2,
              "content": "Call $i$ a suffix record when $b_i=s_i$. Every such position can only go to the position containing $0$, so a Hamiltonian path can have at most one record before its last vertex. Locate the global maximum and use this restriction to show it splits the array into a prefix and a suffix with only two suffix records in total."
            },
            {
              "order": 3,
              "content": "Let $P$ be the values before the global maximum, let $T$ be the values after it, and let $Y$ be the last value. If $r=|P|$ and $s=|T|$, complementing around the two suffix maxima gives $r-1\\le s\\le r$. Therefore $r=\\lfloor n/2\\rfloor$. Now set $x=M-Y$: repeatedly applying the two complement rules sends every positive value $u$ in $P$ to $u-x$, forcing $P$ to be an arithmetic progression."
            },
            {
              "order": 4,
              "content": "With $k=\\lfloor n/2\\rfloor$ and $M=a_n$, the only possible value sets are $L=0,x,2x,\\ldots,(k-1)x$ and $H=M,M-x,M-2x,\\ldots$. The value $M-x$ is either the second- or third-largest value of the whole set, so only test $x=M-a_{n-1}$ and $x=M-a_{n-2}$."
            },
            {
              "order": 5,
              "content": "For a tested $x$, build the proposed low and high multisets and compare them with $a$. If they match, the first $k$ positions can contain the low values arbitrarily, position $k+1$ must be $M$, position $n$ must be $M-x$, and the other high values are arbitrary. That contributes $k!(n-k-2)!$ arrangements; the alternating complement chain proves all of them are good."
            }
          ],
          "editorial": "Let $M=a_n$, which is also the largest value in every rearrangement. For a position $i$, define\n\n$$s_i = max(b_i,b_{i+1},...,b_n).$$\n\nA valid destination from $i$ must contain the value $s_i-b_i$. Because all values are distinct, position $i$ has at most one possible destination. Call it $next(i)$ when that value exists.\n\nSo a good arrangement is a Hamiltonian path in a graph with outdegree at most one. Every position except the last one in the path must use its unique successor, and those successors must be different.\n\n**The Structural Shape**\n\nCall $i$ a suffix record if $b_i=s_i$. Its required destination has value $s_i-b_i=0$. Thus all suffix records point to the same position, the one containing $0$. At most one suffix record can be non-last in the Hamiltonian path, so there are at most two suffix records. Position $n$ is always one of them.\n\nLet $q$ be the position of $M$. It cannot be $n$. If it were, every earlier position would have suffix maximum $M$, so the only possible edges involving the values $0$ and $M$ would be $0 -> M$ and $M -> 0$. If $M$ were the path's last vertex, $0$ would have to be first and the path would stop after two vertices. If $M$ were not last, then $0$ would have to be last to avoid a 2-cycle; consequently $M$ would have no predecessor and would have to be first. Again, only two vertices could be visited. This contradicts $n >= 4$.\n\nTherefore $q<n$. The only suffix records are now $q$ and $n$. Let $Y=b_n$; it is the maximum value in the suffix after $q$.\n\nThe position $q$ must be the last vertex of the Hamiltonian path. Indeed, suppose $n$ were last. If $0$ were before $q$, then $q$ and $0$ would point to each other. If $0$ were after $q$, no edge could enter $q$: a position before $q$ can enter $M$ only when its value is $0$, while a position after $q$ points to a value smaller than $M$. Thus $q$ would have to be first, after which the path can reach only $0$ and $n$, still leaving vertices unvisited. Hence $q$ is last. Since $n$ is then non-last, $0$ must be before $q`; otherwise $n$ and $0$ form a 2-cycle.\n\nLet $P$ be the set of values in positions before $q$, and $T$ the set of values in positions after $q$. Write $r=|P|$ and $s=|T|$.\n\nFor every positive $u$ in $P$, the suffix maximum is $M$, so its successor must have value $M-u$. That value cannot also be in $P$, because the two positions would point to each other. Therefore $M-u$ is in $T$, giving $s >= r-1$.\n\nFor every $v$ in $T$, the suffix maximum is $Y$, so its successor has value $Y-v$. This value must be in $P$: if it were in $T$, the two positions would again form a 2-cycle, and it cannot be $M$ because $Y<M$. Hence $s <= r$.\n\nSince $n=r+1+s$, these inequalities force either $n=2r$ or $n=2r+1$. In both cases\n\n$$r = k = floor(n/2),$$\n\nso the maximum is exactly at position $k+1$.\n\nNow define\n\n$$x=M-Y > 0.$$ \n\nFor a positive $u$ in $P$, the value $M-u$ lies in $T$, so $M-u <= Y$ and therefore $u >= x$. From that tail position, the next value is\n\n$$Y-(M-u)=u-x,$$\n\nwhich is again in $P$. Repeatedly subtracting $x$ eventually reaches $0. Thus every value in $P$ is a multiple of $x$, and closure under subtraction means\n\n$$P = 0,x,2x,...,(k-1)x.$$\n\nThe values $M-u$ for positive $u$ in $P$ must be in $T$, so $T$ contains\n\n$$M-x,M-2x,...,M-(k-1)x.$$\n\nConversely, every $v$ in $T$ has the form $v=Y-u=M-x-u$ for some $u$ in $P$, so it is among $M-x,M-2x,...,M-kx$. The size of $T$ decides the endpoint:\n\n- If $n=2k$, then $T$ is $M-x,...,M-(k-1)x$.\n- If $n=2k+1$, then $T$ is $M-x,...,M-kx$.\n\nIncluding $M$ itself, every good arrangement therefore uses the multiset\n\n$$0,x,...,(k-1)x$$\n\nand\n\n$$M,M-x,...,M-(n-k-1)x.$$\n\n**Finding the Difference**\n\nThe high value immediately below $M$ is $H_1=M-x$. All later high values are smaller. In the even case, nonnegativity of $M-(k-1)x$ gives $H_1 >= (k-2)x$, so at most the single low value $(k-1)x$ can be above $H_1$. In the odd case, nonnegativity of $M-kx$ gives $H_1 >= (k-1)x$, so no low value is above it.\n\nConsequently $H_1$ is either the second- or third-largest input value. The only possible differences are therefore\n\n$$x=M-a_{n-1}$$\n\nand\n\n$$x=M-a_{n-2}.$$\n\nFor each of these at most two candidates, construct the two progressions above, sort the constructed values, and compare them with the sorted input. This single comparison also rejects negative values, duplicate values, and every wrong pattern.\n\n**Counting and Sufficiency**\n\nFor a candidate that passes the multiset check, the positions are forced into this layout:\n\n- Positions $1$ through $k$ contain the low progression in any order: $k!$ choices.\n- Position $k+1$ contains $M$.\n- Position $n$ contains $M-x$.\n- The remaining $n-k-2$ positions contain the other high values in any order: $(n-k-2)!$ choices.\n\nEvery such arrangement is good. Let $L_i=ix$ and $H_i=M-ix$. In the low prefix, the suffix maximum is $M$, so $L_i -> H_i$. At position $k+1$, $H_0=M$ goes to $L_0=0$. In the high suffix, $H_1$ is placed at the very end and is larger than every other high value, so every earlier high position has suffix maximum $H_1$ and $H_i -> L_{i-1}$ for $i>=2$; also $H_1 -> L_0$.\n\nThus a spanning path is\n\n$$L_{k-1} -> H_{k-1} -> L_{k-2} -> H_{k-2} -> ... -> L_0 -> H_0$$\n\nwhen $n$ is even, and\n\n$$H_k -> L_{k-1} -> H_{k-1} -> ... -> L_0 -> H_0$$\n\nwhen $n$ is odd. It visits every value exactly once, regardless of the arbitrary orders inside the low prefix and the middle of the high suffix.\n\nTherefore each valid difference contributes\n\n$$k!(n-k-2)!$$\n\narrangements. Different differences cannot count the same arrangement because $x=M-b_n$ is determined by its last value. Add the contribution of the distinct candidates modulo $10^9+7$.\n\nThe time complexity is $O(n log n)$ per test case, since at most two candidates are checked. The extra space is $O(n)$; factorials are precomputed once.",
          "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\n"
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        {
          "type": "tool",
          "tool": "StructuredOutput",
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              "hints": [
                {
                  "order": 1,
                  "content": "For a position $i$, let $s_i$ be the maximum value in the suffix starting at $i$. A destination must contain the single value $s_i-b_i$. Since all values are distinct, each position has at most one possible successor. Think of the arrangement as a functional directed graph, not a general graph monster."
                },
                {
                  "order": 2,
                  "content": "Call $i$ a suffix record when $b_i=s_i$. Every such position can only go to the position containing $0$, so a Hamiltonian path can have at most one record before its last vertex. Locate the global maximum and use this restriction to show it splits the array into a prefix and a suffix with only two suffix records in total."
                },
                {
                  "order": 3,
                  "content": "Let $P$ be the values before the global maximum, let $T$ be the values after it, and let $Y$ be the last value. If $r=|P|$ and $s=|T|$, complementing around the two suffix maxima gives $r-1\\le s\\le r$. Therefore $r=\\lfloor n/2\\rfloor$. Now set $x=M-Y$: repeatedly applying the two complement rules sends every positive value $u$ in $P$ to $u-x$, forcing $P$ to be an arithmetic progression."
                },
                {
                  "order": 4,
                  "content": "With $k=\\lfloor n/2\\rfloor$ and $M=a_n$, the only possible value sets are $L=0,x,2x,\\ldots,(k-1)x$ and $H=M,M-x,M-2x,\\ldots$. The value $M-x$ is either the second- or third-largest value of the whole set, so only test $x=M-a_{n-1}$ and $x=M-a_{n-2}$."
                },
                {
                  "order": 5,
                  "content": "For a tested $x$, build the proposed low and high multisets and compare them with $a$. If they match, the first $k$ positions can contain the low values arbitrarily, position $k+1$ must be $M$, position $n$ must be $M-x$, and the other high values are arbitrary. That contributes $k!(n-k-2)!$ arrangements; the alternating complement chain proves all of them are good."
                }
              ],
              "editorial": "Let $M=a_n$, which is also the largest value in every rearrangement. For a position $i$, define\n\n$$s_i = max(b_i,b_{i+1},...,b_n).$$\n\nA valid destination from $i$ must contain the value $s_i-b_i$. Because all values are distinct, position $i$ has at most one possible destination. Call it $next(i)$ when that value exists.\n\nSo a good arrangement is a Hamiltonian path in a graph with outdegree at most one. Every position except the last one in the path must use its unique successor, and those successors must be different.\n\n**The Structural Shape**\n\nCall $i$ a suffix record if $b_i=s_i$. Its required destination has value $s_i-b_i=0$. Thus all suffix records point to the same position, the one containing $0$. At most one suffix record can be non-last in the Hamiltonian path, so there are at most two suffix records. Position $n$ is always one of them.\n\nLet $q$ be the position of $M$. It cannot be $n$. If it were, every earlier position would have suffix maximum $M$, so the only possible edges involving the values $0$ and $M$ would be $0 -> M$ and $M -> 0$. If $M$ were the path's last vertex, $0$ would have to be first and the path would stop after two vertices. If $M$ were not last, then $0$ would have to be last to avoid a 2-cycle; consequently $M$ would have no predecessor and would have to be first. Again, only two vertices could be visited. This contradicts $n >= 4$.\n\nTherefore $q<n$. The only suffix records are now $q$ and $n$. Let $Y=b_n$; it is the maximum value in the suffix after $q$.\n\nThe position $q$ must be the last vertex of the Hamiltonian path. Indeed, suppose $n$ were last. If $0$ were before $q$, then $q$ and $0$ would point to each other. If $0$ were after $q$, no edge could enter $q$: a position before $q$ can enter $M$ only when its value is $0$, while a position after $q$ points to a value smaller than $M$. Thus $q$ would have to be first, after which the path can reach only $0$ and $n$, still leaving vertices unvisited. Hence $q$ is last. Since $n$ is then non-last, $0$ must be before $q`; otherwise $n$ and $0$ form a 2-cycle.\n\nLet $P$ be the set of values in positions before $q$, and $T$ the set of values in positions after $q$. Write $r=|P|$ and $s=|T|$.\n\nFor every positive $u$ in $P$, the suffix maximum is $M$, so its successor must have value $M-u$. That value cannot also be in $P$, because the two positions would point to each other. Therefore $M-u$ is in $T$, giving $s >= r-1$.\n\nFor every $v$ in $T$, the suffix maximum is $Y$, so its successor has value $Y-v$. This value must be in $P$: if it were in $T$, the two positions would again form a 2-cycle, and it cannot be $M$ because $Y<M$. Hence $s <= r$.\n\nSince $n=r+1+s$, these inequalities force either $n=2r$ or $n=2r+1$. In both cases\n\n$$r = k = floor(n/2),$$\n\nso the maximum is exactly at position $k+1$.\n\nNow define\n\n$$x=M-Y > 0.$$ \n\nFor a positive $u$ in $P$, the value $M-u$ lies in $T$, so $M-u <= Y$ and therefore $u >= x$. From that tail position, the next value is\n\n$$Y-(M-u)=u-x,$$\n\nwhich is again in $P$. Repeatedly subtracting $x$ eventually reaches $0. Thus every value in $P$ is a multiple of $x$, and closure under subtraction means\n\n$$P = 0,x,2x,...,(k-1)x.$$\n\nThe values $M-u$ for positive $u$ in $P$ must be in $T$, so $T$ contains\n\n$$M-x,M-2x,...,M-(k-1)x.$$\n\nConversely, every $v$ in $T$ has the form $v=Y-u=M-x-u$ for some $u$ in $P$, so it is among $M-x,M-2x,...,M-kx$. The size of $T$ decides the endpoint:\n\n- If $n=2k$, then $T$ is $M-x,...,M-(k-1)x$.\n- If $n=2k+1$, then $T$ is $M-x,...,M-kx$.\n\nIncluding $M$ itself, every good arrangement therefore uses the multiset\n\n$$0,x,...,(k-1)x$$\n\nand\n\n$$M,M-x,...,M-(n-k-1)x.$$\n\n**Finding the Difference**\n\nThe high value immediately below $M$ is $H_1=M-x$. All later high values are smaller. In the even case, nonnegativity of $M-(k-1)x$ gives $H_1 >= (k-2)x$, so at most the single low value $(k-1)x$ can be above $H_1$. In the odd case, nonnegativity of $M-kx$ gives $H_1 >= (k-1)x$, so no low value is above it.\n\nConsequently $H_1$ is either the second- or third-largest input value. The only possible differences are therefore\n\n$$x=M-a_{n-1}$$\n\nand\n\n$$x=M-a_{n-2}.$$\n\nFor each of these at most two candidates, construct the two progressions above, sort the constructed values, and compare them with the sorted input. This single comparison also rejects negative values, duplicate values, and every wrong pattern.\n\n**Counting and Sufficiency**\n\nFor a candidate that passes the multiset check, the positions are forced into this layout:\n\n- Positions $1$ through $k$ contain the low progression in any order: $k!$ choices.\n- Position $k+1$ contains $M$.\n- Position $n$ contains $M-x$.\n- The remaining $n-k-2$ positions contain the other high values in any order: $(n-k-2)!$ choices.\n\nEvery such arrangement is good. Let $L_i=ix$ and $H_i=M-ix$. In the low prefix, the suffix maximum is $M$, so $L_i -> H_i$. At position $k+1$, $H_0=M$ goes to $L_0=0$. In the high suffix, $H_1$ is placed at the very end and is larger than every other high value, so every earlier high position has suffix maximum $H_1$ and $H_i -> L_{i-1}$ for $i>=2$; also $H_1 -> L_0$.\n\nThus a spanning path is\n\n$$L_{k-1} -> H_{k-1} -> L_{k-2} -> H_{k-2} -> ... -> L_0 -> H_0$$\n\nwhen $n$ is even, and\n\n$$H_k -> L_{k-1} -> H_{k-1} -> ... -> L_0 -> H_0$$\n\nwhen $n$ is odd. It visits every value exactly once, regardless of the arbitrary orders inside the low prefix and the middle of the high suffix.\n\nTherefore each valid difference contributes\n\n$$k!(n-k-2)!$$\n\narrangements. Different differences cannot count the same arrangement because $x=M-b_n$ is determined by its last value. Add the contribution of the distinct candidates modulo $10^9+7$.\n\nThe time complexity is $O(n log n)$ per test case, since at most two candidates are checked. The extra space is $O(n)$; factorials are precomputed once.",
              "solution": "#include <bits/stdc++.h>\nusing namespace std;\n\nusing ll = long long;\n\nconst ll MOD = 1'000'000'007;\nconst int MAXN = 200000;\n\nvoid setIO() {\n    ios::sync_with_stdio(false);\n    cin.tie(nullptr);\n}\n\nint main() {\n    setIO();\n\n    vector<ll> fact(MAXN + 1, 1);\n    for (int i = 1; i <= MAXN; ++i) {\n        fact[i] = fact[i - 1] * i % MOD;\n    }\n\n    int t;\n    cin >> t;\n    while (t--) {\n        int n;\n        cin >> n;\n        vector<ll> a(n);\n        for (ll &x : a) cin >> x;\n\n        int k = n / 2;\n        ll ways = fact[k] * fact[n - k - 2] % MOD;\n        set<ll> candidates = {\n            a[n - 1] - a[n - 2],\n            a[n - 1] - a[n - 3]\n        };\n\n        ll answer = 0;\n        for (ll x : candidates) {\n            vector<ll> values;\n            values.reserve(n);\n            for (int i = 0; i < k; ++i) {\n                values.push_back(1LL * i * x);\n            }\n            for (int i = 0; i < n - k; ++i) {\n                values.push_back(a[n - 1] - 1LL * i * x);\n            }\n            sort(values.begin(), values.end());\n            if (values == a) {\n                answer += ways;\n                if (answer >= MOD) answer -= MOD;\n            }\n        }\n        cout << answer << '\\n';\n    }\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
              "valid": true
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            "title": "Structured Output",
            "time": {
              "start": 1789606078808,
              "end": 1789606078811
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