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    "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu",
    "title": "Nudge content generation",
    "agent": "nudge-generation",
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  "messages": [
    {
      "info": {
        "role": "user",
        "time": {
          "created": 1789696555790
        },
        "agent": "nudge-generation",
        "model": {
          "providerID": "openai",
          "modelID": "gpt-6-astra",
          "variant": "low"
        },
        "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",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 2000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'unsolvable', provide a detailed failure message to show users. Name the concrete statement/resource blocker and any relevant source lookup/access status, such as missing official tutorial, 403/Cloudflare challenge, 404, or unavailable accepted submissions. Otherwise null."
              },
              "hints": {
                "anyOf": [
                  {
                    "type": "array",
                    "minItems": 5,
                    "maxItems": 5,
                    "description": "Exactly 5 progressive hints, each building on the last.",
                    "items": {
                      "type": "object",
                      "properties": {
                        "order": {
                          "type": "integer",
                          "minimum": 1,
                          "maximum": 5,
                          "description": "Hint number, 1 through 5."
                        },
                        "content": {
                          "type": "string",
                          "pattern": "^[^\\u0000]*$",
                          "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",
                    "pattern": "^[^\\u0000]*$",
                    "minLength": 1,
                    "maxLength": 100000
                  },
                  {
                    "type": "null"
                  }
                ],
                "description": "If status is 'success', a prose editorial explaining the solution in Nudge's own voice, with no links, citations, or source/research notes. Otherwise null."
              },
              "solution": {
                "anyOf": [
                  {
                    "type": "string",
                    "pattern": "^[^\\u0000]*$",
                    "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_0b23a3b0e001TAw7JLf0euK3KS",
        "sessionID": "ses_f4dc5c501ffeNKHwWmbJhQ9yBA"
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        {
          "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 2253F: \"4-beauty\" (rated 2600).\n\nKnown problem metadata:\n- Contest ID: 2253\n- Index: F\n- Name: 4-beauty\n- Rating: 2600 (Codeforces difficulty rating)\n- Tags: bitmasks, dp, graphs, math, number theory\n- Problem URL: https://codeforces.com/contest/2253/problem/F\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\">F. 4-beauty</div><div class=\"time-limit\"><div class=\"property-title\">time limit per test</div>5 seconds</div><div class=\"memory-limit\"><div class=\"property-title\">memory limit per test</div>1024 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>For a set of integers $$$S$$$, define its <span class=\"tex-font-style-it\">divisibility characteristic</span> as the number of ordered pairs $$$(x, y)$$$ such that $$$x \\ne y$$$, $$$x$$$ and $$$y$$$ belong to $$$S$$$, and $$$x$$$ is divisible by $$$y$$$.</p><p>For a set of integers $$$A$$$, define its <span class=\"tex-font-style-it\">$$$4$$$-beauty</span> as follows:</p><ul> <li> consider all sets of four distinct numbers that belong to $$$A$$$ and form an arithmetic progression; </li><li> find the maximum divisibility characteristic among all such sets. </li></ul><p>If there are no suitable sets of four numbers, then the $$$4$$$-beauty equals $$$0$$$.</p><p>Initially, the set $$$\\{1, 2, \\ldots, n\\}$$$ is given. You may remove numbers from it. Removing the number $$$i$$$ costs $$$m_i$$$ coins.</p><p>Calculate the minimum number of coins you have to spend in order to decrease the $$$4$$$-beauty of the set.</p></div><div class=\"input-specification\"><div class=\"section-title\">Input</div><p>The first line contains one integer $$$n$$$ ($$$4 \\le n \\le 5 \\cdot 10^5$$$)&nbsp;— the number of elements in the initial set.</p><p>The second line contains $$$n$$$ integers $$$m_1, m_2, \\ldots, m_n$$$ ($$$1 \\le m_i \\le 10^9$$$), where $$$m_i$$$ is the cost of removing the number $$$i$$$.</p></div><div class=\"output-specification\"><div class=\"section-title\">Output</div><p>Print one integer&nbsp;— the minimum number of coins you have to spend in order to decrease the $$$4$$$-beauty of the set. It can be shown that it is always possible.</p></div><div class=\"sample-tests\"><div class=\"section-title\">Examples</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\">4</div><div class=\"test-example-line test-example-line-even test-example-line-0\">5 3 7 2</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>2\n</pre></div><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">5</div><div class=\"test-example-line test-example-line-even test-example-line-0\">1 100 100 100 1</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>1\n</pre></div><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">8</div><div class=\"test-example-line test-example-line-even test-example-line-0\">2 10 100 10 100 3 100 100</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>5\n</pre></div><div class=\"input\"><div class=\"title\">Input</div><pre><div class=\"test-example-line test-example-line-even test-example-line-0\">10</div><div class=\"test-example-line test-example-line-even test-example-line-0\">9 9 7 4 8 2 6 5 1 10</div></pre></div><div class=\"output\"><div class=\"title\">Output</div><pre>4\n</pre></div></div></div><div class=\"note\"><div class=\"section-title\">Note</div><p>In the first example, the only arithmetic progression of four numbers is $$$1, 2, 3, 4$$$. Its divisibility characteristic equals $$$4$$$. It is sufficient to remove the number $$$4$$$, paying $$$2$$$ coins.</p><p>In the second example, it is sufficient to remove the number $$$1$$$.</p><p>In the third example, it is optimal to remove the numbers $$$1$$$ and $$$6$$$, paying $$$2+3=5$$$ coins.</p></div>\n</problem-statement>\n\n\nSource lookup status:\n<source-lookup-status>\n- Tutorial: https://codeforces.com/blog/entry/155813 loaded (200 OK; title: \"Educational Codeforces Round 193 — Editorial - Codeforces\"; mentions 2253F)\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- Never include NUL (U+0000) characters in any field. Use literal $ or $$ for math delimiters, never control characters.\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 an object with `order` (1 through 5) and `content` (the Markdown hint text), not a plain string. The `content` must contain just the hint text, with 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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        "finish": "tool-calls",
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              "url": "https://codeforces.com/blog/entry/155813",
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            "output": "---\nimage: https://codeforces.com/codeforces.org/s/70137/images/codeforces-telegram-square-1024x1024.png\ntitle: Educational Codeforces Round 193 — 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/70137/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/70137/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155813?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/70137/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155813?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155813)| [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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2)](https://codeforces.com/blog/entry/156653) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\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) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") → [Educational Codeforces Round 194 - Editorial](https://codeforces.com/blog/entry/156529) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [AlirezaBest](https://codeforces.com/profile/AlirezaBest \"Pupil AlirezaBest\") → [Is this thing regular?](https://codeforces.com/blog/entry/156867) ![New 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updated\")\n* [UNDERTAKER](https://codeforces.com/profile/UNDERTAKER \"Grandmaster UNDERTAKER\") → [Looking for contests to host — we'll adapt our judge to your problems](https://codeforces.com/blog/entry/156823) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Ovi\\_404](https://codeforces.com/profile/Ovi%5F404 \"Pupil Ovi_404\") → [Finally hit 1000 problems solved on Codeforces](https://codeforces.com/blog/entry/156830) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [yse](https://codeforces.com/profile/yse \"Candidate Master yse\") → [Codeforces Round 1114 (Div. 3) — Editorial](https://codeforces.com/blog/entry/155666) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [M.Bemanali](https://codeforces.com/profile/M.Bemanali \"Newbie M.Bemanali\") → [QueueForces AGAIN???](https://codeforces.com/blog/entry/156822) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\")\n* [BitByBit123](https://codeforces.com/profile/BitByBit123 \"Master BitByBit123\") → [Suspicious Master yql0991666](https://codeforces.com/blog/entry/155862) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\")\n* [lumibons](https://codeforces.com/profile/lumibons \"International Grandmaster lumibons\") → [IOI 2027 Call for Tasks](https://codeforces.com/blog/entry/156793) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\")\n* [Grizoo](https://codeforces.com/profile/Grizoo \"Candidate Master Grizoo\") → [\\[GYM\\] Aleppo Collegiate Programming Contest 2026](https://codeforces.com/blog/entry/156707) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [MikeMirzayanov](https://codeforces.com/profile/MikeMirzayanov \"Headquarters, MikeMirzayanov\") → [Rule Restricting the Use of AI \\[revision 2024-09-14\\]](https://codeforces.com/blog/entry/133941) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/70137/images/icons/hourglass.png \"Necropost\")\n* [kondasujay2](https://codeforces.com/profile/kondasujay2 \"International Master kondasujay2\") → [Codeforces Round 1120 (Div 1, Div 2) Editorial](https://codeforces.com/blog/entry/156688) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\")\n* [borgar02](https://codeforces.com/profile/borgar02 \"Master borgar02\") → [IATI 2022 editorial](https://codeforces.com/blog/entry/109545) ![New comment\\(s\\)](//codeforces.com/codeforces.org/s/70137/images/icons/comment-12x12.png \"New comment(s)\") ![Necropost](//codeforces.com/codeforces.org/s/70137/images/icons/hourglass.png \"Necropost\")\n* [Kerolloz](https://codeforces.com/profile/Kerolloz \"Newbie Kerolloz\") → [Codeforces JSON-Problem-Scraper API](https://codeforces.com/blog/entry/75074) ![Text created or updated](//codeforces.com/codeforces.org/s/70137/images/icons/x-update-12x12.png \"Text created or updated\") ![Necropost](//codeforces.com/codeforces.org/s/70137/images/icons/hourglass.png \"Necropost\")\n\n|  | [Detailed →](https://codeforces.com/recent-actions) |\n|  | --------------------------------------------------- |\n\n* [BledDest](https://codeforces.com/profile/BledDest)\n* [Blog](https://codeforces.com/blog/BledDest)\n* [Teams](https://codeforces.com/teams/with/BledDest)\n* [Submissions](https://codeforces.com/submissions/BledDest)\n* [Groups](https://codeforces.com/groups/with/BledDest)\n* [Contests](https://codeforces.com/contests/with/BledDest)\n* [Problemsetting](https://codeforces.com/contests/writer/BledDest)\n\n### [BledDest's blog](https://codeforces.com/blog/BledDest)\n\n[ Educational Codeforces Round 193 — Editorial ](https://codeforces.com/blog/entry/155813) \n\nBy [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\"), [history](https://codeforces.com/topic/156455/en1), 6 weeks ago, translation, ![In English](//codeforces.com/codeforces.org/s/70137/images/flags/24/gb.png \"In English\") \n\n[2253A - The Best Card](https://codeforces.com/contest/2253/problem/A \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (BledDest)**\n\n```\nt = int(input())\nfor i in range(t):\n    n = int(input())\n    good = True\n    for j in range(2, n + 1):\n        if (n + 1) % j == 0:\n            good = False\n    if good:\n        print('YES')\n    else:\n        print('NO')  \n\n\n```\n\n[2253B - Hypercarp and the Control Panel](https://codeforces.com/contest/2253/problem/B \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [FelixArg](https://codeforces.com/profile/FelixArg \"Master FelixArg\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (FelixArg)**\n\n```\n#include <bits/stdc++.h>\n \nusing namespace std;\n#define int long long\n\nvoid solve(){\n\tint n;\n\tcin >> n;\n\tvector<int> a(n);\n\tfor (int i = 0; i < n; i++){\n\t\tcin >> a[i];\n\t}\n\tvector<pair<int, int>> b;\n\tfor (int i = 0; i < n; i++){\n\t\tif (b.empty() || b.back().first != a[i]){\n\t\t\tb.emplace_back(a[i], 1);\n\t\t}\n\t\telse{\n\t\t\tb.back().second++;\n\t\t}\n\t}\n\tint m = b.size();\n\tfor (int i = 0; i < m - 1; i++){\n\t\tif (b[i].second > 1 && b[i + 1].second > 1){\n\t\t\tcout << m + 2 << '\\n';\n\t\t\treturn;\n\t\t}\n\t}\n\tfor (int i = 0; i < m; i++){\n\t\tif (i < m - 1 && b[i].second > 1 && (i + 2 >= m || b[i + 2].first != b[i].first)){\n\t\t\tcout << m + 1 << '\\n';\n\t\t\treturn;\n\t\t}\n\t\tif (i > 0 && b[i].second > 1 && (i - 2 < 0 || b[i - 2].first != b[i].first)){\n\t\t\tcout << m + 1 << '\\n';\n\t\t\treturn;\n\t\t}\n\t}\n\tcout << m << '\\n';\n}\n\nsigned main()\n{\n#ifdef FELIX\n\tauto _clock_start = chrono::high_resolution_clock::now();\n#endif\n\tios_base::sync_with_stdio(false);\n\tcin.tie(0);\n\tcout.tie(0);\n \n\tint tests = 1;\n\tcin >> tests;\n\twhile(tests--){\n\t\tsolve();\n\t}\n \n#ifdef FELIX\n\tcerr << \"Executed in \" << chrono::duration_cast<chrono::milliseconds>(\n\t\tchrono::high_resolution_clock::now()\n\t\t\t- _clock_start).count() << \"ms.\" << endl;\n#endif\n\treturn 0;\n}\n \n\n```\n\n[2253C - Sum of Distinct Values in a Matrix](https://codeforces.com/contest/2253/problem/C \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (BledDest)**\n\n```\ndef get_k_last(l, k):\n    if k < len(l):\n        return l[len(l)-k:]\n    else:\n        return l\n\nt = int(input())\nfor _ in range(t):\n    n, m, x, y = map(int, input().split())\n    a = list(map(int, input().split()))\n    b = list(map(int, input().split()))\n    \n    c = []\n    d = []\n    e = []\n    i = 0\n    j = 0\n    while i < x and j < y:\n        if a[i] == b[j]:\n            c.append(a[i])\n            i += 1\n            j += 1\n        elif a[i] < b[j]:\n            d.append(a[i])\n            i += 1\n        else:\n            e.append(b[j])\n            j += 1\n    d.extend(a[i:])\n    e.extend(b[j:])\n    \n    res = sorted(get_k_last(d, n) + get_k_last(e, m) + c)\n    res = get_k_last(res, n + m - 1)\n    print(sum(res))\n\n\n```\n\n[2253D - Hypercarp and Interdimensional Jumps](https://codeforces.com/contest/2253/problem/D \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [FelixArg](https://codeforces.com/profile/FelixArg \"Master FelixArg\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (FelixArg)**\n\n```\n#include <bits/stdc++.h>\n \nusing namespace std;\n#define int long long\n\nvoid solve(){\n\tint x, y;\n\tcin >> x >> y;\n\n\tint su = x + y;\n\tint k = 0;\n\twhile((k + 1) * (k + 2) / 2 <= su){\n\t\tk++;\n\t}\n\n\tint p1x = k * (k + 1) / 2 - y;\n\tint p1y = y;\n\n\tint p2x = x;\n\tint p2y = k * (k + 1) / 2 - x;\n\n\tvector<pair<int, int>> cand;\n\tif ((p1x + p2x) < 0){\n\t\tcand.emplace_back((p1x + p2x) / 2, k * (k + 1) / 2 - (p1x + p2x) / 2);\n\t\tcand.emplace_back((p1x + p2x - 1) / 2, k * (k + 1) / 2 - (p1x + p2x - 1) / 2);\n\t}\n\telse{\n\t\tcand.emplace_back((p1x + p2x) / 2, k * (k + 1) / 2 - (p1x + p2x) / 2);\n\t\tcand.emplace_back((p1x + p2x + 1) / 2, k * (k + 1) / 2 - (p1x + p2x + 1) / 2);\n\t}\n\n\tauto best = cand[0];\n\tfor (auto [p, q] : cand){\n\t\tif (p < 0){\n\t\t\tp = 0;\n\t\t}\n\t\tif (q < 0){\n\t\t\tq = 0;\n\t\t}\n\t\tif ((p - x) * (p - x) + (q - y) * (q - y) < \n\t\t\t(best.first - x) * (best.first - x) + (best.second - y) * (best.second - y)){\n\t\t\t\tbest = {p, q};\n\t\t}\n\t}\n\n\tstring ans(k, 'Y');\n\tfor (int i = 0; i < k; i++){\n\t\tif (best.first >= k - i){\n\t\t\tans[i] = 'X';\n\t\t\tbest.first -= (k - i);\n\t\t}\n\t}\n\n\tcout << ans << '\\n';\n}\n\nsigned main()\n{\n#ifdef FELIX\n\tauto _clock_start = chrono::high_resolution_clock::now();\n#endif\n\tios_base::sync_with_stdio(false);\n\tcin.tie(0);\n\tcout.tie(0);\n \n\tint tests = 1;\n\tcin >> tests;\n\twhile(tests--){\n\t\tsolve();\n\t}\n \n#ifdef FELIX\n\tcerr << \"Executed in \" << chrono::duration_cast<chrono::milliseconds>(\n\t\tchrono::high_resolution_clock::now()\n\t\t\t- _clock_start).count() << \"ms.\" << endl;\n#endif\n\treturn 0;\n}\n\n\n```\n\n[2253E - Diameter Intersections](https://codeforces.com/contest/2253/problem/E \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (BledDest)**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nconst int N = 1000043;\n\nvector<int> g[N];\nint n;\n\nvector<int> get_dist(int x)\n{\n    vector<int> d(n, -1);\n    d[x] = 0;\n    queue<int> q;\n    q.push(x);\n    while(!q.empty())\n    {\n        int k = q.front();\n        q.pop();\n        for(auto y : g[k])\n            if(d[y] == -1)\n            {\n                d[y] = d[k] + 1;\n                q.push(y);\n            }   \n    }\n    return d;\n}\n\nvoid remove_edge(int x, int y)\n{\n    int idx = -1;\n    for(int i = 0; i < g[x].size(); i++)\n        if(g[x][i] == y)\n            idx = i;\n    g[x].erase(g[x].begin() + idx, g[x].begin() + idx + 1);    \n} \n\nvector<int> process(int v)\n{\n    vector<int> d(n, -1), p(n, -1);\n    queue<int> q;\n    q.push(v);\n    d[v] = 0;\n    vector<int> visited;\n    while(!q.empty())\n    {\n        int k = q.front();\n        q.pop();\n        visited.push_back(k);\n        for(auto y : g[k])\n            if(d[y] == -1)\n            {\n                d[y] = d[k] + 1;\n                q.push(y);\n                p[y] = k;\n            }\n    }\n    int max_dist = *max_element(d.begin(), d.end());\n    vector<bool> has_end(n, false);\n    has_end[v] = true;\n    for(auto x : visited)\n        if(d[x] == max_dist)\n        {\n            int cur = x;\n            while(!has_end[cur])\n            {\n                has_end[cur] = true;\n                cur = p[cur];\n            }\n        }\n\n    vector<int> res;\n    for(auto x : visited)\n    {\n        if(!has_end[x]) continue;\n        int good_children = 0;\n        for(auto y : g[x])\n            if(p[x] != y && has_end[y]) good_children++;\n        if(good_children != 1) res.push_back(d[x]);\n    }   \n    sort(res.begin(), res.end());\n    res.erase(unique(res.begin(), res.end()), res.end());\n    return res;\n}\n\nvoid solve()\n{\n    cin >> n;\n    for(int i = 0; i < n; i++)\n        g[i].clear();\n    for(int i = 0; i < n - 1; i++)\n    {\n        int x, y;\n        cin >> x >> y;\n        --x;\n        --y;\n        g[x].push_back(y);\n        g[y].push_back(x);\n    }\n\n    auto dist0 = get_dist(0);\n    int e1 = max_element(dist0.begin(), dist0.end()) - dist0.begin();\n    auto dist1 = get_dist(e1);\n    int e2 = max_element(dist1.begin(), dist1.end()) - dist1.begin();\n    auto dist2 = get_dist(e2);\n    int d = dist1[e2];\n\n    int x = -1, y = -1;\n    for(int i = 0; i < n; i++)\n        if(dist1[i] + dist2[i] == d)\n        {\n            if(dist1[i] == dist2[i] - 1)\n                x = i;\n            else if(dist1[i] == dist2[i] + 1)\n                y = i;\n        }\n\n    remove_edge(x, y);\n    remove_edge(y, x);\n    auto ans1 = process(x);\n    auto ans2 = process(y);\n\n    vector<bool> res(n + 1);\n    for(auto x : ans1)\n        for(auto y : ans2)\n            res[x + y + 1] = true;\n    int cnt = 0;\n    for(auto x : res)\n        if(x) cnt++;\n    cout << cnt;\n    for(int i = 0; i <= n; i++)\n        if(res[i])\n            cout << \" \" << i;\n    cout << endl;\n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    cin.tie(0);\n    int t;\n    cin >> t;\n    for(int i = 0; i < t; i++)\n        solve();\n}\n\n\n```\n\n[2253F - 4-beauty](https://codeforces.com/contest/2253/problem/F \"Educational Codeforces Round 193 (Rated for Div. 2)\")\n\nIdea: [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\")\n\n**Tutorial**\n\nTutorial is loading...\n\n**Solution (BledDest)**\n\n```\n#include<bits/stdc++.h>\n\nusing namespace std;\n\nconst int N = int(5e5) + 43;\nconst long long INF64 = (long long)(1e18);\n\nint n;\nint cost[N];\n\nvoid upd(long long& x, long long y)\n{\n    if(x > y) x = y;\n}\n\nbool check_bit(int x, int y)\n{\n    return bool((x >> y) & 1);\n}\n\nlong long calc_dp(const vector<vector<int>>& c)\n{\n    int n = c.size();\n    int m = c[0].size();\n    int full = (1 << m) - 1;\n    vector<vector<vector<long long>>> dp(2, vector<vector<long long>>(m, vector<long long>(1 << m, INF64)));\n    dp[0][0][0] = 0;\n    for(int i = 0; i < n; i++)\n    {\n        int i1 = i & 1;\n        int i2 = i1 ^ 1;\n        for(int j = 0; j < m; j++)\n            for(int f = 0; f < (1 << m); f++)\n                dp[i2][j][f] = INF64;\n        for(int j = 0; j < m; j++)\n            for(int f = 0; f < (1 << m); f++)\n            {\n                if(dp[i1][j][f] == INF64) continue;\n                int ni = i1;\n                int nj = j + 1;\n                if(nj == m)\n                {\n                    ni = i2;\n                    nj = 0;\n                }\n                bool can = true;\n                if(i > 0 && j + 2 < m && check_bit(f, m - 1) && check_bit(f, m - 2) && check_bit(f, m - 3))\n                    can = false;\n                if(can)\n                {\n                    int nf = ((f << 1) & full) | 1;\n                    upd(dp[ni][nj][nf], dp[i1][j][f]); \n                }\n                int nf = (f << 1) & full;\n                upd(dp[ni][nj][nf], dp[i1][j][f] + c[i][j]);\n            }\n    }   \n    return *min_element(dp[n & 1][0].begin(), dp[n & 1][0].end());\n}\n\nlong long calc_comp(int x)\n{\n    int d2 = 0, d3 = 0;\n    int p2 = 1, p3 = 1;\n    while(x * (p2 * 2) <= n)\n    {\n        p2 *= 2;\n        d2++;\n    }\n    while(x * (p3 * 3) <= n)\n    {\n        p3 *= 3;\n        d3++;\n    }\n    d2++;\n    d3++;\n    vector<int> pow2(d2, 1), pow3(d3, 1);\n    for(int i = 1; i < d2; i++)\n        pow2[i] = pow2[i - 1] * 2;\n    for(int i = 1; i < d3; i++)\n        pow3[i] = pow3[i - 1] * 3;\n    vector<vector<int>> cur(d3, vector<int>(d2, 0));\n    for(int i = 0; i < d2; i++)\n        for(int j = 0; j < d3; j++)\n            if(x * 1ll * pow2[i] * 1ll * pow3[j] <= n)\n                cur[j][i] = cost[x * pow2[i] * pow3[j]];\n    return calc_dp(cur);\n}   \n\nvoid solve()\n{\n    cin >> n;\n    for(int i = 1; i <= n; i++)\n        cin >> cost[i];\n    long long ans = 0;\n    for(int i = 1; i <= n; i++)\n        if(i % 2 != 0 && i % 3 != 0)\n            ans += calc_comp(i);\n    cout << ans << endl; \n}\n\nint main()\n{\n    ios_base::sync_with_stdio(0);\n    int t = 1;\n    for(int i = 0; i < t; i++)\n        solve();    \n}  \n\n\n```\n\n![](//codeforces.com/codeforces.org/s/70137/images/icons/paperclip-16x16.png) Tutorial of [Educational Codeforces Round 193 (Rated for Div. 2)](https://codeforces.com/contest/2253) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +55\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/70137/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/BledDest) [BledDest ](https://codeforces.com/profile/BledDest)\n* ![Publication date](//codeforces.com/codeforces.org/s/70137/images/blog/date_16x16.png \"Publication date\") 6 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/70137/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155813#comments) [22 ](https://codeforces.com/blog/entry/155813#comments)\n\n  \n![Comments](//codeforces.com/codeforces.org/s/70137/images/icons/comments-48x48.png \"Comments\") Comments (22) \n\n[Write comment?](#) \n\n| » [salemi1](https://codeforces.com/profile/salemi1 \"Specialist salemi1\") | 6 weeks ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------ | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4817146/avatar/2167beab4eeac4e9.jpg) ](https://codeforces.com/profile/salemi1) [salemi1](https://codeforces.com/profile/salemi1 \"Specialist salemi1\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384199 \"Link to comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) But D just needed a simple greedy [386119523](https://codeforces.com/contest/2253/submission/386119523 \"Submission 386119523 by salemi1\") **The code**void solve() {     int n,m; cin>>n>>m;       int L=0, R=20010;     while(L+1!=R) {         int cen=(L+R)/2;         int a=n, b=m;                  for(int i=cen; i>0; i--) {             if(a>b) a-= i;             else b-= i;         }           if(a<0 or b<0) {             R=cen;         } else {             L=cen;         }     }       int a=n, b=m;     for(int i=L; i>0; i--) {         if(a>b) { a-=i; cout<<'X'; }         else { b-=i; cout<<'Y'; }     } cout<<endl;   } [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n* | » » [Noobish\\_Monk](https://codeforces.com/profile/Noobish%5FMonk \"Master Noobish_Monk\") | 6 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/2080623/avatar/1db04b38e51c59b.jpg) ](https://codeforces.com/profile/Noobish%5FMonk) [Noobish\\_Monk](https://codeforces.com/profile/Noobish%5FMonk \"Master Noobish_Monk\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384215 \"Link to comment\") [^](#comment-1384199 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +3 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Mind explaining why greedy works? [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n  * | » » » [salemi1](https://codeforces.com/profile/salemi1 \"Specialist salemi1\") | 6 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------------------------- | ----------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4817146/avatar/2167beab4eeac4e9.jpg) ](https://codeforces.com/profile/salemi1) [salemi1](https://codeforces.com/profile/salemi1 \"Specialist salemi1\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384299 \"Link to comment\") [^](#comment-1384215 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +11 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) ( we move from (x,y) to (0,0) )First of all, we know the maximum operations we can perform(k). the 'i'-th change in 'a' or 'b' will finally reduce 'x' or 'y' by k+1-i if one of the numbers is much bigger and can't reach the other number by using all the operation, so it's OK!Otherwise we consider that x>y then, we have a moment that 'x' reachs 'y' and probably become smaller than 'y'. in this case, the maximum difference between 'x' and 'y' is 'k' so it compensates with k-1 (and k-1 with k-2...) and finally the maximum difference becomes 1And the minimum difference between x and y is what we need! [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n    * | » » » » [KnownAsJason](https://codeforces.com/profile/KnownAsJason \"Expert KnownAsJason\") | 6 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------------------------- | ------------------------ |\n\n| » » » » [ ![](https://codeforces.com/userpic.codeforces.org/4710940/avatar/a36a72563a88c277.jpg) ](https://codeforces.com/profile/KnownAsJason) [KnownAsJason](https://codeforces.com/profile/KnownAsJason \"Expert KnownAsJason\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384327 \"Link to comment\") [^](#comment-1384299 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) holy shoots this is brilliant [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Anush2010](https://codeforces.com/profile/Anush2010 \"Expert Anush2010\") | 6 weeks ago, [show](#) # |\n| -------------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4268424/avatar/4ff1c1e778200912.jpg) ](https://codeforces.com/profile/Anush2010) [Anush2010](https://codeforces.com/profile/Anush2010 \"Expert Anush2010\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384207 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Yeaah, finally ))) [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [Misuki](https://codeforces.com/profile/Misuki \"Master Misuki\") | 6 weeks ago, [show](#) # |\n| ----------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/1197572/avatar/e4ea7bc4debf5d23.jpg) ](https://codeforces.com/profile/Misuki) [Misuki](https://codeforces.com/profile/Misuki \"Master Misuki\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384211 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Alternative solution of D:Let $$$k$$$ be the largest integer with $$$\\\\frac{k \\\\cdot (k + 1)}{2} \\\\le x + y$$$. Consider two bag with capacity $$$x, y$$$ respectively. For $$$w = k, k - 1, \\\\ldots, 1$$$ in this order, put the item with weight $$$w$$$ into the bag with larger capacity remain. Let $$$x', y'$$$ denote the final capacity remain for two bags. We claim this process minimize $$$\\\\max(x', y')$$$.Proof.W.L.O.G. assume $$$x \\\\ge y$$$, call the bag with $$$x$$$ initial capacity the first bag, and the other be the second bag.case 1 ($$$x \\\\ge y + \\\\frac{k\\\\cdot (k + 1)}{2}$$$): trivialcase 2: items would be add into the first bag until its capacity is no more than the second bag, let $$$m$$$ be the item added to make this happen. then at this point, the difference of capacity between two bag would be no more than $$$m$$$, and in the subsequent item addition, the difference would be no more than the item added last, thus $$$|x' - y'| \\\\le 1$$$ holds at the end, which achieve the lower bound of $$$\\\\max(x', y')$$$. [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [isagiyoichi11](https://codeforces.com/profile/isagiyoichi11 \"Specialist isagiyoichi11\") | 6 weeks ago, [show (+1)](#) # |\n| ------------------------------------------------------------------------------------------ | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/5725147/avatar/c3a4b891570882b2.jpg) ](https://codeforces.com/profile/isagiyoichi11) [isagiyoichi11](https://codeforces.com/profile/isagiyoichi11 \"Specialist isagiyoichi11\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384212 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) In Problem E (2253E), how would i know number of distinct LCAs would be O($$$\\\\sqrt{n}$$$)? I was stuck with thinking it was O($$$n^2$$$) solution and won't pass. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [oelyas](https://codeforces.com/profile/oelyas \"Specialist oelyas\") | 6 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/4291105/avatar/f500eefdcb6fcd52.jpg) ](https://codeforces.com/profile/oelyas) [oelyas](https://codeforces.com/profile/oelyas \"Specialist oelyas\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384233 \"Link to comment\") [^](#comment-1384212 \"Parent comment\") \\| [←](#)Rev. 2 [→](#) [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) We have $$$n-1$$$ edges, and each edge either part of diameter or not, and each node of diameter can form just a single distinct number, and sum of nodes limited on $$$n$$$, so you wont have more than $$$sqrt(n)$$$ distinct numbers. We can prove that by summing the first k small values such that sum won't exceed $$$n$$$. [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n* | » » [mgranger22](https://codeforces.com/profile/mgranger22 \"Candidate Master mgranger22\") | 6 weeks ago, [show](#) # |\n| ----------------------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/3637499/avatar/c2e543d66773ecca.jpg) ](https://codeforces.com/profile/mgranger22) [mgranger22](https://codeforces.com/profile/mgranger22 \"Candidate Master mgranger22\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384245 \"Link to comment\") [^](#comment-1384212 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Firstly discard all nodes who's subtree does NOT have a descendant of maximal depth, as these can never be valid LCA's. For the remaining nodes, define $$$cnt\\[d\\]$$$ as the number of nodes at depth $$$d$$$. Since each path is guaranteed to continue down until the maximal depth, we know that $$$cnt\\[d + 1\\] \\\\geq cnt\\[d\\]$$$. Furthermore, when depth $$$d$$$ contains a valid LCA (i.e. some node has two or more children with max depth descendants), then $$$cnt\\[d + 1\\] \\\\geq cnt\\[d\\] + 1$$$. If there are $$$r$$$ distinct achievable depths, then the width increases at least $$$r$$$ times. After the i-th increase, the width is at least $$$i+1$$$. Therefore, among the levels containing these increases and the level immediately after each one, there are at least $$$1 + 2 + \\\\ldots + (r + 1)$$$ nodes. Since the number of nodes per valid LCA depth grows quadratically, the number of valid LCA depths per node grows at a rate of $$$O(\\\\sqrt{n})$$$. [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [mgranger22](https://codeforces.com/profile/mgranger22 \"Candidate Master mgranger22\") | 6 weeks ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/3637499/avatar/c2e543d66773ecca.jpg) ](https://codeforces.com/profile/mgranger22) [mgranger22](https://codeforces.com/profile/mgranger22 \"Candidate Master mgranger22\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384246 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I believe there's a typo in F. I think it should be x mod 2 != 0 and x mod 3 != 0 whereas it says the opposite now. [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 6 weeks ago, [show](#) # |\n| -------------------------------------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/331814/avatar/36406af67e5558a6.jpg) ](https://codeforces.com/profile/BledDest) [BledDest](https://codeforces.com/profile/BledDest \"International Grandmaster BledDest\") | 6 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155813?#comment-1384328 \"Link to comment\") [^](#comment-1384246 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/70137/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Thank you, fixed that. The editorial should update in a couple of minutes [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n| » [LNight121](https://codeforces.com/profile/LNight121 \"Master LNight121\") | 6 weeks ago, [show](#) # |\n| -------------------------------------------------------------------------- | ------------------------ |\n\n\n...23195 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0b23a4d4300174k4ndAHpxCklh\nUse the Task tool to have explore agent process this file with Grep and Read (with offset/limit). Do NOT read the full file yourself - delegate to save context.",
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            "output": "Title: N/A\nURL: https://codeforces.com/contest/2253/problem/F\nPublished: N/A\nAuthor: N/A\nHighlights:\nF. 4-beauty\n...\n- consider all sets of four distinct numbers that belong to $$ and form an arithmetic progression;\n- find the maximum divisibility characteristic among all such sets.\n...\n0\n$$\n...\nm\\_i\n$$\n\n$$\n4\n$$\n...\n$$\n4 \\\\le n \\\\le 5 \\\\cdot 10^5\n$$\n\n$$\ni\n$$\n...\n$$\n4\n$$\n...\n$$\n2\n$$\n\n$$\n...\n$$\n\n$$\n2+3=5\n$$\n\n---\n\nTitle: CF2253F 4-beauty - 洛谷\nURL: https://www.luogu.com.cn/problem/CF2253F\nPublished: N/A\nAuthor: N/A\nHighlights:\n## 题目描述\n...\n对于一个整数集合 S，定义其可整除性特征为满足 x= y， x 和 y 都属于 S，且 x 能被 y 整除的有序对 (x, y) 的数量。\n...\n给定一个整数集合 A，定义其 4-美丽度如下：\n...\n- 考虑所有属于 A 且能组成等差数列的四个不同数的集合；\n- 在所有这样的集合中，找到可整除性特征的最大值。\n...\n如果不存在满足条件的四元组，则 4-美丽度为 0。\n...\n最初给定集合 { 1, 2,…, n}。你可以从中移除若干数。移除数字 i 需要花费 m i​ 个硬币。\n...\n请计算，为了减少该集合的 4-美丽度，最少需要花费多少个硬币。\n...\n第一行包含一个整数 n（ 4≤ n≤ 5⋅ 1 0 5），表示初始集合的元素\n...\n第二行包含 n 个整数 m 1​, m 2​,…, m n​（ 1≤ m i​≤ 1 0 9），其中 m i​ 表示移除数字 i 的代价。\n...\n输出一个整数——为使集合的 4-美丽度下降，所需花费的最小硬币数。可以保证总是有办法实现。\n...\n## 说明/提示\n...\n在第一个样例中，唯一的四元等差数列是 1, 2, 3, 4，其可整除性特征等于 4。只需移除 4，花费 2 个硬币即可。\n\n在第二个样例中，只需移除 1 即可。\n\n在第三个样例中，最优操作是移除 1 和 6，共花费 2+ 3= 5 个硬币。\n\n---\n\nTitle: 【CodeForces - 255C】Almost Arithmetical Progression （dp，离散化）_牛客博客\nURL: https://blog.nowcoder.net/n/7d1c335549f44aa99c35d33516e0ee14\nPublished: 2018-11-09T00:00:00.000Z\nAuthor: N/A\nHighlights:\n【CodeForces - 255C】Almost Arithmetical Progression （dp，离散化）_牛客博客\n...\n# 【CodeForces - 255C】Almost Arithmetical Progression （dp，离散化） \n\n题干：\n...\nGena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:\n...\n- a 1 = p, where p is some integer;\n- a i = a i - 1 + ( - 1) i + 1· q (i > 1), where q is some integer.\n...\nRight now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.\n...\nSequence s 1, s 2, ..., s k is a subsequence of sequence b 1, b 2, ..., b n, if there is such increasing sequence of indexes i 1, i 2, ..., i k (1 ≤ i 1 < i 2 < ... < i k ≤ n), that b i j = s j. In other words, sequence s can be obtained from b by crossing out some elements.\n...\nPrint a single integer — the length of the required longest subsequence.\n...\n题目大意：\n\n 给出一个序列，求最长的子序列，满足隔位的两个数相等，问这个最长的子序列的长度是多少。\n...\n解题报告：\n\n 想了好久想到一个dp方程，抱着冒险的心态o(n^2)交一发，TLE41了，，，把离散化的数组提前预处理了一下，，，又莽了一发，，AC了。。小惊喜。\n...\nAC代码：\n...\nconst int MAX = 1e6+5;\n...\nint dp[4004][4004];\n...\nint a[4004],b[4004];\nint cnt[MAX];\nint pos[MAX];\n...\nint len,n;\n...\nint get(int x) {\n\treturn lower_bound(b+1,b+len+1,x) - b;\n}\n...\nint main()\n{\n\tcin>>n;\n\tfor(int i = 1; i<=n; i++) scanf(\"%d\",a+i),b[i] = a[i],cnt[a[i]]++;\n\tsort(b+1,b+n+1);\n\tlen = unique(b+1,b+n+1) - b - 1;\n\tfor(int i = 1; i<=n; i++) pos[a[i]] = get(a[i]);\n\tfor(int i = 1; i<=n; i++) \n\t\tfor(int j = 1; j<=n; j++)\n\t\t\tdp[pos[a[i]]][pos[a[j]]] = 1;\n\tfor(int i = 1; i<=n; i++) {\n\t\tfor(int j = 1; j<i; j++) {\n\t\t\tif(a[i]!=a[j])\n\t\t\tdp[pos[a[i]]][pos[a[j]]] = dp[pos[a[j]]][pos[a[i]]]+1;\n\t\t}\n\t}\n\tint maxx = -1;\n\tfor(int i = 1; i<=n; i++) {\n\t\tfor(int j = 1; j<=n; j++) {\n\t\t\tmaxx = max(maxx,dp[pos[a[i]]][pos[a[j]]]);\n\t\t}\n\t}\n\tfor(int i = 1; i<=n; i++) {\n\t\tmaxx = max(maxx,cnt[a[i]]);\n\t}\n\tprintf(\"%d\\n\",maxx);\n\treturn 0 ;\n }\n...\n总结：\n\n 思维过程是这样的，，刚开始想暴力（用vector存权值啊之类的乱搞），，一想肯定不行，，有太多重复操作了。。然后想各种优化后来发现就是要求个序列，，所以跟最长上升子序列联系起来了，，后来发现还真可以写出来个递推式，，但是是o(n^2)的不知道会不会T，，而且1600W的数组大小不知道会不会MLE，，交一发，WA，，造样例，，发现全是同一个数的时候就会爆炸，，我就想，，再多也不可能超过数组长度啊，，所以就像把元素相等的时候特判一下（并没想过正确性），改代码，，测样例，，没啥问题，，再交一发，TLE41，，，失望（但是还是有点小惊喜的因为跑起来了最起码说明算法的问题不大）。再想优化，开个数组预处理一波，去掉了时间复杂度的log，再交，AC。",
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          "text": "The key reduction is that only progressions $\\{x,2x,3x,4x\\}$ achieve the original maximum beauty. Splitting numbers by their factors other than $2$ and $3$ then turns the removal problem into small grid DPs.",
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              "patchText": "*** Begin Patch\n*** Add File: main.cpp\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int n;\n+    cin >> n;\n+    vector<ll> m(n + 1);\n+    for (int i = 1; i <= n; ++i) cin >> m[i];\n+    const ll INF = 1LL << 60;\n+    ll answer = 0;\n+    for (int b = 1; b <= n / 4; ++b) {\n+        if (b % 2 == 0 || b % 3 == 0) continue;\n+        int w = 0;\n+        for (ll x = b; x <= n; x *= 2) ++w;\n+        int states = 1 << w, half = states / 2;\n+        int triple = 7 << (w - 3);\n+        vector<ll> dp(states, INF), next(states);\n+        dp[0] = 0;\n+        int row = 0;\n+        for (ll first = b; first <= n; first *= 3, ++row) {\n+            ll value = first;\n+            for (int col = 0; col < w; ++col, value *= 2) {\n+                bool exists = value <= n;\n+                ll cost = exists ? m[value] : 0;\n+                bool check = row > 0 && col + 2 < w;\n+                for (int low = 0; low < half; ++low) {\n+                    ll a = dp[low], c = dp[low + half];\n+                    next[2 * low] = min(a, c) + cost;\n+                    if (check && ((low + half) & triple) == triple)\n+                        c = INF;\n+                    next[2 * low + 1] = exists ? min(a, c) : INF;\n+                }\n+                dp.swap(next);\n+            }\n+        }\n+        answer += *min_element(dp.begin(), dp.end());\n+    }\n+    cout << answer << '\\n';\n+}\n*** End Patch"
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            "output": "Success. Updated the following files:\nA private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp",
            "metadata": {
              "diff": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n@@ -0,0 +1,47 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int n;\n+    cin >> n;\n+    vector<ll> m(n + 1);\n+    for (int i = 1; i <= n; ++i) cin >> m[i];\n+    const ll INF = 1LL << 60;\n+    ll answer = 0;\n+    for (int b = 1; b <= n / 4; ++b) {\n+        if (b % 2 == 0 || b % 3 == 0) continue;\n+        int w = 0;\n+        for (ll x = b; x <= n; x *= 2) ++w;\n+        int states = 1 << w, half = states / 2;\n+        int triple = 7 << (w - 3);\n+        vector<ll> dp(states, INF), next(states);\n+        dp[0] = 0;\n+        int row = 0;\n+        for (ll first = b; first <= n; first *= 3, ++row) {\n+            ll value = first;\n+            for (int col = 0; col < w; ++col, value *= 2) {\n+                bool exists = value <= n;\n+                ll cost = exists ? m[value] : 0;\n+                bool check = row > 0 && col + 2 < w;\n+                for (int low = 0; low < half; ++low) {\n+                    ll a = dp[low], c = dp[low + half];\n+                    next[2 * low] = min(a, c) + cost;\n+                    if (check && ((low + half) & triple) == triple)\n+                        c = INF;\n+                    next[2 * low + 1] = exists ? min(a, c) : INF;\n+                }\n+                dp.swap(next);\n+            }\n+        }\n+        answer += *min_element(dp.begin(), dp.end());\n+    }\n+    cout << answer << '\\n';\n+}\n\n",
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                {
                  "filePath": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp",
                  "relativePath": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp",
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                  "patch": "Index: /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n===================================================================\n--- /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n+++ /private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-G3bwUu/main.cpp\n@@ -0,0 +1,47 @@\n+#include <bits/stdc++.h>\n+using namespace std;\n+\n+using ll = long long;\n+\n+void setIO() {\n+    ios::sync_with_stdio(false);\n+    cin.tie(nullptr);\n+}\n+\n+int main() {\n+    setIO();\n+    int n;\n+    cin >> n;\n+    vector<ll> m(n + 1);\n+    for (int i = 1; i <= n; ++i) cin >> m[i];\n+    const ll INF = 1LL << 60;\n+    ll answer = 0;\n+    for (int b = 1; b <= n / 4; ++b) {\n+        if (b % 2 == 0 || b % 3 == 0) continue;\n+        int w = 0;\n+        for (ll x = b; x <= n; x *= 2) ++w;\n+        int states = 1 << w, half = states / 2;\n+        int triple = 7 << (w - 3);\n+        vector<ll> dp(states, INF), next(states);\n+        dp[0] = 0;\n+        int row = 0;\n+        for (ll first = b; first <= n; first *= 3, ++row) {\n+            ll value = first;\n+            for (int col = 0; col < w; ++col, value *= 2) {\n+                bool exists = value <= n;\n+                ll cost = exists ? m[value] : 0;\n+                bool check = row > 0 && col + 2 < w;\n+                for (int low = 0; low < half; ++low) {\n+                    ll a = dp[low], c = dp[low + half];\n+                    next[2 * low] = min(a, c) + cost;\n+                    if (check && ((low + half) & triple) == triple)\n+                        c = INF;\n+                    next[2 * low + 1] = exists ? min(a, c) : INF;\n+                }\n+                dp.swap(next);\n+            }\n+        }\n+        answer += *min_element(dp.begin(), dp.end());\n+    }\n+    cout << answer << '\\n';\n+}\n",
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                  "deletions": 0
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              "command": "python3 -c 'import random,subprocess,time\nrandom.seed(42)\nfor n in range(4,17):\n aps=[]\n for a in range(1,n+1):\n  for d in range(1,(n-a)//3+1):\n   v=[a+i*d for i in range(4)]; score=sum(v[j]%v[i]==0 for i in range(4) for j in range(i+1,4))\n   if score>=4:\n    assert score==4 and a==d\n    aps.append(sum(1<<(x-1) for x in v))\n for _ in range(8):\n  costs=[random.randint(1,100) for i in range(n)]\n  best=sum(costs)\n  for mask in range(1<<n):\n   if all(mask&p for p in aps):\n    best=min(best,sum(costs[i] for i in range(n) if mask>>i&1))\n  got=int(subprocess.check_output([\"./main\"],input=(str(n)+\"\\n\"+\" \".join(map(str,costs))+\"\\n\").encode()))\n  assert got==best,(n,costs,best,got)\nprint(\"104 random exhaustive comparisons passed\")\nn=500000; start=time.monotonic(); result=subprocess.check_output([\"./main\"],input=(str(n)+\"\\n\"+\"1000000000 \"*n+\"\\n\").encode()); print(\"Maximum-size run:\",result.decode().strip(),\"time:\",round(time.monotonic()-start,3),\"seconds\")'",
              "timeout": 120000
            },
            "output": "104 random exhaustive comparisons passed\nMaximum-size run: 67398000000000 time: 0.681 seconds\n",
            "metadata": {
              "output": "104 random exhaustive comparisons passed\nMaximum-size run: 67398000000000 time: 0.681 seconds\n",
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            "title": "python3 -c 'import random,subprocess,time\nrandom.seed(42)\nfor n in range(4,17):\n aps=[]\n for a in range(1,n+1):\n  for d in range(1,(n-a)//3+1):\n   v=[a+i*d for i in range(4)]; score=sum(v[j]%v[i]==0 for i in range(4) for j in range(i+1,4))\n   if score>=4:\n    assert score==4 and a==d\n    aps.append(sum(1<<(x-1) for x in v))\n for _ in range(8):\n  costs=[random.randint(1,100) for i in range(n)]\n  best=sum(costs)\n  for mask in range(1<<n):\n   if all(mask&p for p in aps):\n    best=min(best,sum(costs[i] for i in range(n) if mask>>i&1))\n  got=int(subprocess.check_output([\"./main\"],input=(str(n)+\"\\n\"+\" \".join(map(str,costs))+\"\\n\").encode()))\n  assert got==best,(n,costs,best,got)\nprint(\"104 random exhaustive comparisons passed\")\nn=500000; start=time.monotonic(); result=subprocess.check_output([\"./main\"],input=(str(n)+\"\\n\"+\"1000000000 \"*n+\"\\n\").encode()); print(\"Maximum-size run:\",result.decode().strip(),\"time:\",round(time.monotonic()-start,3),\"seconds\")'",
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          "hints": [
            {
              "order": 1,
              "content": "First figure out what beauty you are trying to destroy. For a progression $a,a+d,a+2d,a+3d$, inspect divisibility among the last three numbers. How many such pairs can exist?"
            },
            {
              "order": 2,
              "content": "Neither of the last two adjacent pairs can contribute: their difference is $d$, smaller than either possible divisor. The only remaining pair among the last three requires $(a+d)\\mid 2d$, which forces $a=d$. Thus only progressions $\\{x,2x,3x,4x\\}$ have characteristic $4$; every other progression has characteristic at most $3$. You must remove at least one number from each of these special progressions."
            },
            {
              "order": 3,
              "content": "Write every number uniquely as $b\\cdot2^u3^v$, where $b$ is divisible by neither $2$ nor $3$. All four numbers $x,2x,3x,4x$ have the same $b$, so different values of $b$ can be solved independently. Put $b2^u3^v$ at grid position $(v,u)$. What shape does one forbidden progression make?"
            },
            {
              "order": 4,
              "content": "The forbidden shape consists of $(v,u),(v,u+1),(v,u+2)$ and $(v+1,u)$. Scan the grid row by row, left to right. When processing its bottom cell, all three top cells have already been processed. A bitmask recording which of the previous $W$ cells were kept is enough, where $W=1+\\lfloor\\log_2(n/b)\\rfloor\\le19$."
            },
            {
              "order": 5,
              "content": "Use minimum-cost profile DP. A mask bit is $1$ for a kept cell and $0$ for a removed cell. Removing the current number adds its cost and shifts in $0$. Keeping it shifts in $1$, but is forbidden when the three relevant cells in the preceding row were all kept; these are bits $W-1,W-2,W-3$. Check this only after the first row and when the column has two successors. Pad nonexistent grid cells with forced zero bits at zero cost. Take the minimum final state for each $b$ and sum the results."
            }
          ],
          "editorial": "## 1. Which progressions actually matter?\n\nWe only need to make the maximum characteristic smaller. We do **not** need to destroy every four-term arithmetic progression—that would be a much nastier problem.\n\nConsider\n$$\na,\\ a+d,\\ a+2d,\\ a+3d,\\qquad a,d>0.\n$$\nBecause all numbers are positive and distinct, only the larger number in a pair can be divisible by the smaller one.\n\nLook at pairs entirely within the last three terms:\n\n- $a+d$ cannot divide $a+2d$: it would have to divide their difference $d$, but $a+d>d$.\n- Likewise, $a+2d$ cannot divide $a+3d$.\n- If $a+d$ divides $a+3d$, then it divides $2d$. Since $a+d>d$, this is possible only when $a+d=2d$, or $a=d$.\n\nTherefore, if $a\\ne d$, every contributing pair must involve the first term, giving characteristic at most $3$.\n\nIf $a=d=x$, the progression is\n$$\n\\{x,2x,3x,4x\\},\n$$\nand its characteristic is exactly $4$: $x$ divides the other three numbers, and $2x$ divides $4x$.\n\nSince $n\\ge4$, the initial set contains $\\{1,2,3,4\\}$, so its beauty is exactly $4$. The task is now:\n\n> Minimize the total removal cost while removing at least one number from every set $\\{x,2x,3x,4x\\}$ with $4x\\le n$.\n\nRemoving numbers cannot create new progressions. Once all these sets are broken, the beauty is at most $3$.\n\n## 2. Split the numbers into independent groups\n\nEvery positive integer has a unique representation\n$$\nb\\cdot2^u3^v,\n$$\nwhere $u,v\\ge0$ and $b$ is divisible by neither $2$ nor $3$.\n\nMultiplying by $2$, $3$, or $4$ does not change $b$. Consequently, every forbidden set lies entirely within one group with a fixed $b$.\n\nThe groups share neither numbers nor constraints, so their optimal costs can be added.\n\nIf $b>n/4$, its group has no forbidden set: any possible starting number $x$ in that group is at least $b$, so $4x>n$. Such groups contribute zero and can be skipped.\n\n## 3. Turn one group into a grid\n\nFor a fixed $b$, put the number $b2^u3^v$ at row $v$, column $u$.\n\nUse a rectangular grid with\n$$\nW=1+\\left\\lfloor\\log_2\\frac nb\\right\\rfloor,\n\\qquad\nH=1+\\left\\lfloor\\log_3\\frac nb\\right\\rfloor.\n$$\nSome cells represent numbers greater than $n$. Treat those cells as permanently absent.\n\nFor $x=b2^u3^v$, the forbidden set occupies\n$$\n(v,u),\\quad(v,u+1),\\quad(v,u+2),\\quad(v+1,u).\n$$\nVisually:\n\n```text\nx    2x    4x\n3x\n```\n\nSo we must avoid keeping all four cells of this shape.\n\nThe grid width is at most $19$. That is small enough for a bitmask, even though the original set has half a million numbers.\n\n## 4. Scan cells with a profile DP\n\nProcess cells row by row, from left to right.\n\nLet $dp[mask]$ be the minimum cost after processing the current prefix, where the mask describes the last $W$ processed cells:\n\n- bit $0$ describes the most recently processed cell;\n- bit $W-1$ describes the oldest remembered cell;\n- a bit is $1$ if that cell was kept, and $0$ otherwise.\n\nInitially, $dp[0]=0$ and every other state is infinity. The initial zero bits are just padding before the grid starts.\n\n### Removing a cell\n\nFor a real cell containing number $z$, pay $m_z$ and shift in a zero:\n$$\nnewMask=(mask\\ll1)\\mathbin{\\&}(2^W-1).\n$$\n\nFor a nonexistent cell, this is the only permitted choice and costs zero.\n\n### Keeping a cell\n\nSuppose the current position is $(r,c)$. The only forbidden shape that can become complete now has this cell as its bottom cell. Its other three cells are\n$$\n(r-1,c),\\quad(r-1,c+1),\\quad(r-1,c+2).\n$$\nBefore processing the current cell, their statuses are stored in bits\n$$\nW-1,\\quad W-2,\\quad W-3.\n$$\nThis shape needs checking only when $r>0$ and $c+2<W$.\n\nIf all three bits are $1$, keeping the current cell is forbidden. Otherwise, keep it at no additional cost and shift in a one:\n$$\nnewMask=((mask\\ll1)\\mathbin{\\&}(2^W-1))\\mathbin{|}1.\n$$\n\nAfter the last cell, the group's answer is the minimum over all masks.\n\n### Why only one row of history is enough\n\nEvery forbidden shape spans two consecutive rows. When its bottom cell is reached, its three upper cells are still in the mask. Every shape is checked exactly at that moment, so older decisions can safely be forgotten.\n\n## 5. Implementation details\n\nUse two DP arrays and swap them after every cell. All costs and DP values need `long long`: the total removal cost can reach $5\\cdot10^{14}$.\n\nThe implementation groups the two predecessor masks that differ only in their oldest bit. Write them as\n$$\nlow\\quad\\text{and}\\quad low+2^{W-1},\n\\qquad 0\\le low<2^{W-1}.\n$$\nBoth shift to $2\\cdot low$ when removing the current cell, or to $2\\cdot low+1$ when keeping it.\n\nThe first predecessor has its oldest bit zero, so it can never complete the forbidden shape. Only the second predecessor needs the three-bit test. This fills each entry of the next array exactly once and avoids clearing it between cells.\n\nBecause we process only $b\\le n/4$, every processed grid has $W\\ge3$, so the three-bit mask is always well-defined. Generate powers using integer multiplication and 64-bit intermediates rather than floating-point logarithms.\n\n## Correctness proof\n\n**Lemma 1.** The initial beauty is $4$, and the progressions with characteristic $4$ are exactly $\\{x,2x,3x,4x\\}$.\n\nAs shown above, unless the first term equals the common difference, none of the last three terms divides another, leaving at most three contributing pairs. When the first term equals the difference, exactly four pairs contribute. The initial set contains $\\{1,2,3,4\\}$.\n\n**Lemma 2.** Different groups indexed by $b$ can be optimized independently.\n\nAll numbers in any forbidden set have the same value after removing all factors of $2$ and $3$. Thus every constraint belongs to one group, and the groups contain disjoint numbers. Their minimum costs therefore add.\n\n**Lemma 3.** The profile DP computes the minimum valid removal cost within one group.\n\nInduct on the number of processed cells. The initial state correctly describes an empty prefix. At a real cell, the transitions consider both possible decisions: remove it at its cost or keep it for free. Removing cannot complete a forbidden shape. Keeping is rejected exactly when the three already processed cells of its forbidden shape were all kept. No other shape can become complete at this step. Nonexistent cells are correctly forced absent.\n\nThe mask retains every earlier decision needed by a future check. Therefore, for each mask, retaining only the cheapest valid prefix loses no useful solution. The invariant follows, and the minimum final state is optimal for the group.\n\n**Theorem.** The algorithm returns the minimum cost required to decrease the beauty.\n\nBy Lemma 1, decreasing the beauty is equivalent to breaking every forbidden set. Lemma 3 optimally breaks those sets inside each group, and Lemma 2 allows their costs to be summed. Groups with $b>n/4$ contain no constraints and require no removals. Hence the returned total is optimal.\n\n## Complexity\n\nFor a group $b$, the DP takes\n$$\nO(H_bW_b2^{W_b})\n$$\ntime and $O(2^{W_b})$ auxiliary space.\n\nSince $2^{W_b}\\le2n/b$ and both dimensions are $O(\\log(n/b))$, summing over groups gives\n$$\nO\\!\\left(n\\sum_{b=1}^{\\lfloor n/4\\rfloor}\n\\frac{\\log^2(n/b)}b\\right)\n=O(n\\log^3 n)\n$$\ntime.\n\nThe largest DP has $2^{19}$ states, and groups are processed one at a time. Including the input costs, total space is $O(n)$.",
          "solution": "#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    int n;\n    cin >> n;\n    vector<ll> m(n + 1);\n    for (int i = 1; i <= n; ++i) cin >> m[i];\n\n    const ll INF = 1LL << 60;\n    ll answer = 0;\n\n    for (int b = 1; b <= n / 4; ++b) {\n        if (b % 2 == 0 || b % 3 == 0) continue;\n\n        int w = 0;\n        for (ll x = b; x <= n; x *= 2) ++w;\n        int states = 1 << w, half = states / 2;\n        int triple = 7 << (w - 3);\n\n        vector<ll> dp(states, INF), next(states);\n        dp[0] = 0;\n\n        int row = 0;\n        for (ll first = b; first <= n; first *= 3, ++row) {\n            ll value = first;\n            for (int col = 0; col < w; ++col, value *= 2) {\n                bool exists = value <= n;\n                ll cost = exists ? m[value] : 0;\n                bool check = row > 0 && col + 2 < w;\n\n                // Pair predecessors differing only in the oldest bit.\n                for (int low = 0; low < half; ++low) {\n                    ll a = dp[low], c = dp[low + half];\n                    next[2 * low] = min(a, c) + cost;\n\n                    if (check && ((low + half) & triple) == triple)\n                        c = INF;\n                    next[2 * low + 1] = exists ? min(a, c) : INF;\n                }\n                dp.swap(next);\n            }\n        }\n        answer += *min_element(dp.begin(), dp.end());\n    }\n\n    cout << answer << '\\n';\n}\n"
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          "text": "**Verifying the normalization proof**",
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                  "order": 1,
                  "content": "First figure out what beauty you are trying to destroy. For a progression $a,a+d,a+2d,a+3d$, inspect divisibility among the last three numbers. How many such pairs can exist?"
                },
                {
                  "order": 2,
                  "content": "Neither of the last two adjacent pairs can contribute: their difference is $d$, smaller than either possible divisor. The only remaining pair among the last three requires $(a+d)\\mid 2d$, which forces $a=d$. Thus only progressions $\\{x,2x,3x,4x\\}$ have characteristic $4$; every other progression has characteristic at most $3$. You must remove at least one number from each of these special progressions."
                },
                {
                  "order": 3,
                  "content": "Write every number uniquely as $b\\cdot2^u3^v$, where $b$ is divisible by neither $2$ nor $3$. All four numbers $x,2x,3x,4x$ have the same $b$, so different values of $b$ can be solved independently. Put $b2^u3^v$ at grid position $(v,u)$. What shape does one forbidden progression make?"
                },
                {
                  "order": 4,
                  "content": "The forbidden shape consists of $(v,u),(v,u+1),(v,u+2)$ and $(v+1,u)$. Scan the grid row by row, left to right. When processing its bottom cell, all three top cells have already been processed. A bitmask recording which of the previous $W$ cells were kept is enough, where $W=1+\\lfloor\\log_2(n/b)\\rfloor\\le19$."
                },
                {
                  "order": 5,
                  "content": "Use minimum-cost profile DP. A mask bit is $1$ for a kept cell and $0$ for a removed cell. Removing the current number adds its cost and shifts in $0$. Keeping it shifts in $1$, but is forbidden when the three relevant cells in the preceding row were all kept; these are bits $W-1,W-2,W-3$. Check this only after the first row and when the column has two successors. Pad nonexistent grid cells with forced zero bits at zero cost. Take the minimum final state for each $b$ and sum the results."
                }
              ],
              "editorial": "## 1. Which progressions actually matter?\n\nWe only need to make the maximum characteristic smaller. We do **not** need to destroy every four-term arithmetic progression—that would be a much nastier problem.\n\nConsider\n$$\na,\\ a+d,\\ a+2d,\\ a+3d,\\qquad a,d>0.\n$$\nBecause all numbers are positive and distinct, only the larger number in a pair can be divisible by the smaller one.\n\nLook at pairs entirely within the last three terms:\n\n- $a+d$ cannot divide $a+2d$: it would have to divide their difference $d$, but $a+d>d$.\n- Likewise, $a+2d$ cannot divide $a+3d$.\n- If $a+d$ divides $a+3d$, then it divides $2d$. Since $a+d>d$, this is possible only when $a+d=2d$, or $a=d$.\n\nTherefore, if $a\\ne d$, every contributing pair must involve the first term, giving characteristic at most $3$.\n\nIf $a=d=x$, the progression is\n$$\n\\{x,2x,3x,4x\\},\n$$\nand its characteristic is exactly $4$: $x$ divides the other three numbers, and $2x$ divides $4x$.\n\nSince $n\\ge4$, the initial set contains $\\{1,2,3,4\\}$, so its beauty is exactly $4$. The task is now:\n\n> Minimize the total removal cost while removing at least one number from every set $\\{x,2x,3x,4x\\}$ with $4x\\le n$.\n\nRemoving numbers cannot create new progressions. Once all these sets are broken, the beauty is at most $3$.\n\n## 2. Split the numbers into independent groups\n\nEvery positive integer has a unique representation\n$$\nb\\cdot2^u3^v,\n$$\nwhere $u,v\\ge0$ and $b$ is divisible by neither $2$ nor $3$.\n\nMultiplying by $2$, $3$, or $4$ does not change $b$. Consequently, every forbidden set lies entirely within one group with a fixed $b$.\n\nThe groups share neither numbers nor constraints, so their optimal costs can be added.\n\nIf $b>n/4$, its group has no forbidden set: any possible starting number $x$ in that group is at least $b$, so $4x>n$. Such groups contribute zero and can be skipped.\n\n## 3. Turn one group into a grid\n\nFor a fixed $b$, put the number $b2^u3^v$ at row $v$, column $u$.\n\nUse a rectangular grid with\n$$\nW=1+\\left\\lfloor\\log_2\\frac nb\\right\\rfloor,\n\\qquad\nH=1+\\left\\lfloor\\log_3\\frac nb\\right\\rfloor.\n$$\nSome cells represent numbers greater than $n$. Treat those cells as permanently absent.\n\nFor $x=b2^u3^v$, the forbidden set occupies\n$$\n(v,u),\\quad(v,u+1),\\quad(v,u+2),\\quad(v+1,u).\n$$\nVisually:\n\n```text\nx    2x    4x\n3x\n```\n\nSo we must avoid keeping all four cells of this shape.\n\nThe grid width is at most $19$. That is small enough for a bitmask, even though the original set has half a million numbers.\n\n## 4. Scan cells with a profile DP\n\nProcess cells row by row, from left to right.\n\nLet $dp[mask]$ be the minimum cost after processing the current prefix, where the mask describes the last $W$ processed cells:\n\n- bit $0$ describes the most recently processed cell;\n- bit $W-1$ describes the oldest remembered cell;\n- a bit is $1$ if that cell was kept, and $0$ otherwise.\n\nInitially, $dp[0]=0$ and every other state is infinity. The initial zero bits are just padding before the grid starts.\n\n### Removing a cell\n\nFor a real cell containing number $z$, pay $m_z$ and shift in a zero:\n$$\nnewMask=(mask\\ll1)\\mathbin{\\&}(2^W-1).\n$$\n\nFor a nonexistent cell, this is the only permitted choice and costs zero.\n\n### Keeping a cell\n\nSuppose the current position is $(r,c)$. The only forbidden shape that can become complete now has this cell as its bottom cell. Its other three cells are\n$$\n(r-1,c),\\quad(r-1,c+1),\\quad(r-1,c+2).\n$$\nBefore processing the current cell, their statuses are stored in bits\n$$\nW-1,\\quad W-2,\\quad W-3.\n$$\nThis shape needs checking only when $r>0$ and $c+2<W$.\n\nIf all three bits are $1$, keeping the current cell is forbidden. Otherwise, keep it at no additional cost and shift in a one:\n$$\nnewMask=((mask\\ll1)\\mathbin{\\&}(2^W-1))\\mathbin{|}1.\n$$\n\nAfter the last cell, the group's answer is the minimum over all masks.\n\n### Why only one row of history is enough\n\nEvery forbidden shape spans two consecutive rows. When its bottom cell is reached, its three upper cells are still in the mask. Every shape is checked exactly at that moment, so older decisions can safely be forgotten.\n\n## 5. Implementation details\n\nUse two DP arrays and swap them after every cell. All costs and DP values need `long long`: the total removal cost can reach $5\\cdot10^{14}$.\n\nThe implementation groups the two predecessor masks that differ only in their oldest bit. Write them as\n$$\nlow\\quad\\text{and}\\quad low+2^{W-1},\n\\qquad 0\\le low<2^{W-1}.\n$$\nBoth shift to $2\\cdot low$ when removing the current cell, or to $2\\cdot low+1$ when keeping it.\n\nThe first predecessor has its oldest bit zero, so it can never complete the forbidden shape. Only the second predecessor needs the three-bit test. This fills each entry of the next array exactly once and avoids clearing it between cells.\n\nBecause we process only $b\\le n/4$, every processed grid has $W\\ge3$, so the three-bit mask is always well-defined. Generate powers using integer multiplication and 64-bit intermediates rather than floating-point logarithms.\n\n## Correctness proof\n\n**Lemma 1.** The initial beauty is $4$, and the progressions with characteristic $4$ are exactly $\\{x,2x,3x,4x\\}$.\n\nAs shown above, unless the first term equals the common difference, none of the last three terms divides another, leaving at most three contributing pairs. When the first term equals the difference, exactly four pairs contribute. The initial set contains $\\{1,2,3,4\\}$.\n\n**Lemma 2.** Different groups indexed by $b$ can be optimized independently.\n\nAll numbers in any forbidden set have the same value after removing all factors of $2$ and $3$. Thus every constraint belongs to one group, and the groups contain disjoint numbers. Their minimum costs therefore add.\n\n**Lemma 3.** The profile DP computes the minimum valid removal cost within one group.\n\nInduct on the number of processed cells. The initial state correctly describes an empty prefix. At a real cell, the transitions consider both possible decisions: remove it at its cost or keep it for free. Removing cannot complete a forbidden shape. Keeping is rejected exactly when the three already processed cells of its forbidden shape were all kept. No other shape can become complete at this step. Nonexistent cells are correctly forced absent.\n\nThe mask retains every earlier decision needed by a future check. Therefore, for each mask, retaining only the cheapest valid prefix loses no useful solution. The invariant follows, and the minimum final state is optimal for the group.\n\n**Theorem.** The algorithm returns the minimum cost required to decrease the beauty.\n\nBy Lemma 1, decreasing the beauty is equivalent to breaking every forbidden set. Lemma 3 optimally breaks those sets inside each group, and Lemma 2 allows their costs to be summed. Groups with $b>n/4$ contain no constraints and require no removals. Hence the returned total is optimal.\n\n## Complexity\n\nFor a group $b$, the DP takes\n$$\nO(H_bW_b2^{W_b})\n$$\ntime and $O(2^{W_b})$ auxiliary space.\n\nSince $2^{W_b}\\le2n/b$ and both dimensions are $O(\\log(n/b))$, summing over groups gives\n$$\nO\\!\\left(n\\sum_{b=1}^{\\lfloor n/4\\rfloor}\n\\frac{\\log^2(n/b)}b\\right)\n=O(n\\log^3 n)\n$$\ntime.\n\nThe largest DP has $2^{19}$ states, and groups are processed one at a time. Including the input costs, total space is $O(n)$.",
              "solution": "#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    int n;\n    cin >> n;\n    vector<ll> m(n + 1);\n    for (int i = 1; i <= n; ++i) cin >> m[i];\n\n    const ll INF = 1LL << 60;\n    ll answer = 0;\n\n    for (int b = 1; b <= n / 4; ++b) {\n        if (b % 2 == 0 || b % 3 == 0) continue;\n\n        int w = 0;\n        for (ll x = b; x <= n; x *= 2) ++w;\n        int states = 1 << w, half = states / 2;\n        int triple = 7 << (w - 3);\n\n        vector<ll> dp(states, INF), next(states);\n        dp[0] = 0;\n\n        int row = 0;\n        for (ll first = b; first <= n; first *= 3, ++row) {\n            ll value = first;\n            for (int col = 0; col < w; ++col, value *= 2) {\n                bool exists = value <= n;\n                ll cost = exists ? m[value] : 0;\n                bool check = row > 0 && col + 2 < w;\n\n                // Pair predecessors differing only in the oldest bit.\n                for (int low = 0; low < half; ++low) {\n                    ll a = dp[low], c = dp[low + half];\n                    next[2 * low] = min(a, c) + cost;\n\n                    if (check && ((low + half) & triple) == triple)\n                        c = INF;\n                    next[2 * low + 1] = exists ? min(a, c) : INF;\n                }\n                dp.swap(next);\n            }\n        }\n        answer += *min_element(dp.begin(), dp.end());\n    }\n\n    cout << answer << '\\n';\n}\n"
            },
            "output": "Structured output captured successfully.",
            "metadata": {
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            "title": "Structured Output",
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