{ "info": { "id": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "slug": "glowing-falcon", "projectID": "global", "directory": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PnyW7F", "path": "private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PnyW7F", "title": "Nudge content generation", "agent": "nudge-generation", "model": { "id": "gpt-6-astra", "providerID": "openai", "variant": "low" }, "version": "1.18.30", "summary": { "additions": 0, "deletions": 0, "files": 0 }, "cost": 0, "tokens": { "input": 26007, "output": 3016, "reasoning": 447, "cache": { "read": 33536, "write": 0 } }, "time": { "created": 1789698157391, "updated": 1789698275138 } }, "messages": [ { "info": { "role": "user", "time": { "created": 1789698157409 }, "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_0b252ab610017tt3PNtRSZADmJ", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa" }, "parts": [ { "type": "text", "text": "Generate Codeforces learning content: five progressive hints, an original editorial, and AC-quality C++.\n\nFollow the editorial-first research workflow in the system instructions. Use the supplied statement to verify that the source solves this exact problem.\n\nWrite clean Markdown with LaTeX as needed. Use quick and clever humor when appropriate. Tell it like it is (don't sugar-coat responses), and use very casual language. You are fully allowed to swear, just don't overdo it like a sailor (be natural). Deconstruct any false assumptions.\n\nGenerate content for Codeforces problem 2248E: \"Excuse for Breaks\" (rated 1900).\n\nKnown problem metadata:\n- Contest ID: 2248\n- Index: E\n- Name: Excuse for Breaks\n- Rating: 1900 (Codeforces difficulty rating)\n- Tags: binary search, brute force, greedy, math, two pointers\n- Problem URL: https://codeforces.com/contest/2248/problem/E\n\nUse the rating and tags as weak signals only. The supplied statement is the source of truth.\n\nProblem Statement:\n\n
E. Excuse for Breaks
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You are given three integers $$$n$$$, $$$m$$$, and $$$d$$$, and two arrays $$$p_1, p_2, \\ldots, p_m$$$ and $$$r_1, r_2, \\ldots, r_m$$$. The array $$$p$$$ is strictly increasing.

For a binary array $$$a$$$ of any positive finite length (so $$$|a|$$$ need not equal $$$n$$$), define its value $$$f(a)$$$ using the following pseudocode:

function f(a):
    v := 0
    c := 0
    for i from 1 to length(a):
        if a[i] is equal to 1:
            v := v + d
            c := c + 1
        else:
            c := 0
        for j from 1 to m:
            if c is equal to p[j]:
                v := v + r[j]
        if c is equal to n:
            c := 0
    return v
Here, \":=\" denotes the assignment operation.

Let $$$I(a)$$$ denote the array $$$[1,1,\\ldots,1]$$$ of length $$$|a|$$$. In other words, $$$I(a)$$$ consists of $$$|a|$$$ ones.

Determine whether there exists a binary (consisting only of zeros and ones) array $$$a$$$ such that $$$f(a) \\gt f(I(a))$$$.

Input

Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 2000$$$). The description of the test cases follows.

The first line of each test case contains three integers $$$n$$$, $$$m$$$, and $$$d$$$ ($$$1 \\le n \\le 10^9$$$, $$$0 \\le m \\le 2000$$$, $$$0 \\le d \\le 10^9$$$).

The $$$i$$$-th of the next $$$m$$$ lines contains two integers $$$p_i$$$ and $$$r_i$$$ ($$$1 \\le p_i \\le n$$$, $$$1 \\le r_i \\le 10^9$$$).

The array $$$p$$$ is strictly increasing.

It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$2000$$$.

Output

For each test case, output \"YES\" if such a binary array $$$a$$$ exists, and \"NO\" otherwise.

You can output the answer in any case (upper or lower). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be recognized as positive responses.

Example
Input
3
6 4 3
2 5
3 9
4 1
5 3
7 3 5
2 5
4 5
7 10
684492057 3 386217943
367971233 991739271
612599954 429216213
684492056 402931836
Output
YES
NO
YES
Note

In the first test case, you can choose $$$a = [1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1]$$$.

The array $$$a$$$ contains $$$16$$$ ones, so the total contribution of $$$d$$$ is $$$16 \\cdot 3 = 48$$$. Its runs of consecutive ones have lengths $$$9$$$, $$$4$$$, and $$$3$$$. The corresponding total rewards are $$$32$$$, $$$15$$$, and $$$14$$$, respectively. Therefore, $$$f(a) = 48 + 32 + 15 + 14 = 109$$$.

Moreover, $$$I(a)$$$ consists of $$$18$$$ ones. During the computation of $$$f(I(a))$$$, they form three complete blocks of $$$6$$$ ones, and each block contributes $$$6 \\cdot 3 + 5 + 9 + 1 + 3 = 36$$$. Thus, $$$f(I(a)) = 3 \\cdot 36 = 108$$$. Since $$$f(a) \\gt f(I(a))$$$, the answer is \"YES\".

In the second test case, no binary array $$$a$$$ satisfies $$$f(a) \\gt f(I(a))$$$, so the answer is \"NO\".

\n\n\n\nSource lookup status:\n\n- Tutorial (en): https://codeforces.com/blog/entry/155640 loaded (200 OK; title: \"Codeforces Round 1113 (Div. 2) Editorial - Codeforces\"; mentions 2248E)\n\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 \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```", "id": "prt_0b252ab620019OdTWzkiY1wzX6", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ab610017tt3PNtRSZADmJ" } ] }, { "info": { "parentID": "msg_0b252ab610017tt3PNtRSZADmJ", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "low", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PnyW7F", "root": "/" }, "cost": 0, "tokens": { "total": 9354, "input": 9322, "output": 32, "reasoning": 0, "cache": { "write": 0, "read": 0 } }, "modelID": "gpt-6-astra", "providerID": "openai", "time": { "created": 1789698157546, "completed": 1789698165327 }, "finish": "tool-calls", "id": "msg_0b252abea001CggOKvv8d92GN8", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa" }, "parts": [ { "type": "step-start", "id": "prt_0b252b533001wb4BXGP6gH6ntZ", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252abea001CggOKvv8d92GN8" }, { "type": "tool", "tool": "webfetch", "callID": "call_GhvUmBftT8Q1Q7SVXVx0dqsz", "state": { "status": "completed", "input": { "url": "https://codeforces.com/blog/entry/155640", "format": "markdown" }, "output": "---\nimage: https://codeforces.com/codeforces.org/s/13758/images/codeforces-telegram-square-1024x1024.png\ntitle: Codeforces Round 1113 (Div. 2) Editorial - Codeforces\ndescription: Codeforces. Programming competitions and contests, programming community\n---\n\n \n\n \n[![Codeforces](//codeforces.com/codeforces.org/s/13758/images/codeforces-logo-with-telegram-555x130.png \"Codeforces\")](https://codeforces.com/) \n\n[![In English](//codeforces.com/codeforces.org/s/13758/images/flags/24/gb.png \"In English\")](https://codeforces.com/blog/entry/155640?locale=en) [![По-русски](//codeforces.com/codeforces.org/s/13758/images/flags/24/ru.png \"По-русски\")](https://codeforces.com/blog/entry/155640?locale=ru) \n\n[Enter](https://codeforces.com/enter?back=%2Fblog%2Fentry%2F155640)| [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* [Rating](https://codeforces.com/ratings)\n* 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![Necropost](//codeforces.com/codeforces.org/s/13758/images/icons/hourglass.png \"Necropost\")\n\n| | [Detailed →](https://codeforces.com/recent-actions) |\n| | --------------------------------------------------- |\n\n* [szdytom](https://codeforces.com/profile/szdytom)\n* [Blog](https://codeforces.com/blog/szdytom)\n* [Teams](https://codeforces.com/teams/with/szdytom)\n* [Submissions](https://codeforces.com/submissions/szdytom)\n* [Groups](https://codeforces.com/groups/with/szdytom)\n* [Contests](https://codeforces.com/contests/with/szdytom)\n* [Problemsetting](https://codeforces.com/contests/writer/szdytom)\n\n### [szdytom's blog](https://codeforces.com/blog/szdytom)\n\n[ Codeforces Round 1113 (Div. 2) Editorial ](https://codeforces.com/blog/entry/155640) \n\nBy [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\"), 7 weeks ago, ![In English](//codeforces.com/codeforces.org/s/13758/images/flags/24/gb.png \"In English\") \n\n[2248A - You Delete, I Delete](https://codeforces.com/contest/2248/problem/A \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nAlice deletes a `0`, Bob deletes a `1`. Which occurrence should each of them choose?\n\n**Hint2**\n\nDeleting a `0` makes the string lexicographically larger. Does the position of that `0` matter?\n\n**Hint3**\n\nAfter Alice's move, Bob sees one string and deletes a `1`. What is his best move?\n\n**Solution**\n\nDeleting a `0` at an earlier position is always better for Alice. Indeed, compare deleting the `0`s at positions $$$i \\\\lt j$$$: the two results agree before position $$$i$$$, and at position $$$i$$$ the first result has `1` while the second still has `0`. So Alice deletes the **first** `0`.\n\nAfter that, Bob wants to minimize the string. By the symmetric argument he deletes the **first** `1` of the remaining string, which is the first `1` of the original string, since deleting a `0` never changes the order of the `1`s.\n\nHence the answer is the original string with its first `0` and its first `1` removed.\n\n**Complexity:** $$$O(n)$$$ per test case.\n\n**Code**\n\n```\n#include \n#include \nconst int N = 105;\nchar str[N];\nvoid solve() {\n scanf(\"%s\", str + 1);\n int n = strlen(str + 1);\n bool c0 = 0, c1 = 0;\n for (int i = 1; i <= n; i++) {\n if (!c0 && str[i] == '0') {\n c0 = true;\n continue;\n }\n if (!c1 && str[i] == '1') {\n c1 = true;\n continue;\n }\n putchar(str[i]);\n }\n puts(\"\");\n}\nint main() {\n int t = 0;\n scanf(\"%d\", &t);\n while (t--) solve();\n return 0;\n}\n\n```\n\n**Code (Brute Force)**\n\n```\n#include\nusing namespace std;\n\n#define ll long long\n#define nline \"\\n\"\n#define sz(x) (int)x.size()\n\nvoid solve(){\n\tstring s; cin>>s;\n\tstring ans;\n\tfor(int i=0;i>test_cases;\n\twhile(test_cases--){\n\t\tsolve();\n\t}\n}\n\n```\n\n[2248B - Merge to Match](https://codeforces.com/contest/2248/problem/B \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nSort both arrays. Each operation reduces the number of elements by exactly one.\n\n**Hint2**\n\nAll values are distinct, so no final element can be an untouched original element. What lower bound on $$$n$$$ does this give?\n\n**Hint3**\n\nAfter sorting, in which interval of $$$a$$$ must the $$$i$$$-th smallest value $$$b\\_i$$$ lie?\n\n**Solution**\n\nSort both arrays. An operation merges two elements $$$x \\\\le y$$$ into any $$$z \\\\in \\[x,y\\]$$$, so think of each final element as the merge of a group of original elements whose min/max bracket its value.\n\n**Necessity.**\n\n* **$$$n \\\\ge 2m$$$.** The values in $$$b$$$ are distinct from those in $$$a$$$, so a final element consisting of a single original element is impossible. Every final element needs at least two originals.\n* **$$$a\\_i \\\\le b\\_i \\\\le a\\_{n-m+i}$$$.** The $$$i$$$ smallest final values are produced by $$$i$$$ groups containing at least $$$i$$$ original elements, all $$$\\\\le b\\_i$$$. Only $$$i-1$$$ original elements are smaller than $$$a\\_i$$$, so $$$b\\_i \\\\ge a\\_i$$$. The symmetric argument on the largest values gives $$$b\\_i \\\\le a\\_{n-m+i}$$$.\n\n**Sufficiency.** If the conditions hold, pair $$$a\\_i$$$ with $$$a\\_{n-m+i}$$$ for each $$$i$$$ (the pairs are disjoint since $$$n \\\\ge 2m$$$) and merge each pair into $$$b\\_i$$$. The remaining middle elements can be attached to any group one by one: merging the current group value $$$b\\_1$$$ with an extra element $$$e$$$ yields $$$b\\_1$$$ again, because $$$b\\_1$$$ lies between them.\n\n**Complexity:** $$$O((n+m)\\\\log(n+m))$$$ per test case.\n\n**Code**\n\n```\nusing namespace std;\n\nint main() {\n\tint T;\n\tscanf(\"%d\", &T);\n\tfor (int _ = 0; _ < T; ++_) {\n\t\tint n, m;\n\t\tscanf(\"%d %d\", &n, &m);\n\t\tvector a(n), b(m);\n\t\tfor (int &x : a) {\n\t\t scanf(\"%d\", &x);\n\t\t}\n\t\tfor (int &x : b) {\n\t\t\tscanf(\"%d\", &x);\n\t\t}\n\t\tsort(a.begin(), a.end());\n\t\tsort(b.begin(), b.end());\n\n\t\tif (n < 2 * m) {\n\t\t\tputs(\"NO\");\n\t\t\tcontinue;\n\t\t}\n\n\t\tint i = 0;\n\t\twhile (i < m && a[i] < b[i] && b[i] < a[n - m + i]) ++i;\n\t\tputs(i < m ? \"no\" : \"Yes\");\n\t}\n\treturn 0;\n}\n\n```\n\n[2248C - Maximize the Score](https://codeforces.com/contest/2248/problem/C \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nIf a value is chosen when some elements between its two occurrences were already deleted, move its deletion earlier. When is this never worse? Think about $$$(a+b)^2 \\\\ge a^2 + b^2$$$.\n\n**Hint2**\n\nConclude that the process is equivalent to partitioning the array into disjoint blocks: each block is either a single position, or the whole interval between two equal values at its ends.\n\n**Hint3**\n\nDP over the prefix. Consider the block containing the last position.\n\n**Solution**\n\nCall an operation **full** if it deletes the entire original interval between the two occurrences of the chosen value.\n\n**Lemma.** There is an optimal strategy in which every operation is full.\n\nSuppose a value is chosen with current span $$$p \\\\dots q$$$, while an interval $$$J$$$ strictly inside its original span was deleted earlier. Swap the two deletions: the value now covers the whole original interval. If the two original lengths are $$$A$$$ and $$$B$$$ ($$$A \\\\ge B$$$), the score changes from at most $$$(A-B)^2 + B^2$$$ to $$$A^2$$$, and $$$(A-B)^2 + B^2 \\\\le A^2$$$. So the total never decreases, and repeating this exchange makes every operation full.\n\nNow the deleted intervals are disjoint, and the whole process is exactly: partition the array into blocks, each being a single position or the interval between the two equal values at its ends, and collect $$$\\\\sum (\\\\text{block length})^2$$$.\n\nLet $$$dp\\[i\\]$$$ be the best score for the prefix $$$a\\_1 \\\\dots a\\_i$$$. The block containing position $$$i$$$ is either the singleton $$${i}$$$, or the interval $$$\\[l,i\\]$$$ where $$$l$$$ is the first occurrence of $$$a\\_i$$$:\n\n$$$ dp\\[i\\] = \\\\max\\\\big(dp\\[i-1\\] + 1,\\\\; dp\\[l-1\\] + (i-l+1)^2\\\\big). $$$\n\nThe singleton option never overestimates: if the mate of $$$a\\_i$$$ is still present in the optimal strategy of the prefix, deleting $$$\\[l,i\\]$$$ instead is at least as good, since $$$(i-l+1)^2 \\\\ge (i-l)^2 + 1$$$.\n\nThe answer is $$$dp\\[2n\\]$$$.\n\n**Complexity:** $$$O(n)$$$ per test case.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nint main() {\n\tint T;\n\tscanf(\"%d\", &T);\n\tfor (int _ = 0; _ < T; ++_) {\n\t\tint n;\n\t\tscanf(\"%d\", &n);\n\t\tvector a(2 * n);\n\t\tfor (int &x : a) {\n\t\t\tscanf(\"%d\", &x);\n\t\t}\n\n\t\tvector lp(2 * n, -1);\n\t\tvector occ(n, -1);\n\t\tfor (int i = 0; i < 2 * n; ++i) {\n\t\t\tif (occ[a[i] - 1] != -1) {\n\t\t\t\tlp[i] = occ[a[i] - 1];\n\t\t\t} else {\n\t\t\t\tocc[a[i] - 1] = i;\n\t\t\t}\n\t\t}\n\n\t\tvector dp(2 * n);\n\t\tdp[0] = 1;\n\t\tfor (int i = 1; i < 2 * n; ++i) {\n\t\t\tdp[i] = dp[i - 1] + 1;\n\t\t\tif (lp[i] != -1) {\n\t\t\t\tdp[i] = max(dp[i], (lp[i] == 0 ? 0 : dp[lp[i] - 1]) + 1LL * (i - lp[i] + 1) * (i - lp[i] + 1));\n\t\t\t}\n\t\t}\n\t\tprintf(\"%lld\\n\", dp[2 * n - 1]);\n\t}\n}\n\n```\n\n[2248D - Good Pair Queries](https://codeforces.com/contest/2248/problem/D \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nClassify each position by the pair $$$(s\\_i, t\\_i)$$$ and count the four types: $$$(0,0)$$$, $$$(0,1)$$$, $$$(1,0)$$$, $$$(1,1)$$$.\n\n**Hint2**\n\nFind a quantity that only decreases during the game, and equals $$$0$$$ at the end.\n\n**Hint3**\n\nWhich pairs of columns can be deleted together in one operation? When can all columns be paired up?\n\n**Solution**\n\nFor a substring pair let $$$x=N(0,1)$$$, $$$y=N(1,0)$$$, $$$u=N(0,0)$$$, $$$v=N(1,1)$$$, where $$$N(0,1)$$$ denotes the number occurrences of $$$(0,1)$$$.\n\n**Claim.** The pair is good iff $$$|x-y| \\\\le u+v$$$.\n\n**Necessity**: Take any allowed operation. Let the deleted columns have counts $$$u\\_0, v\\_0, x\\_0, y\\_0$$$ and total $$$k = u\\_0+v\\_0+x\\_0+y\\_0$$$. \n\n**Case $$$c=0$$$:** \nIn the deleted subsequence of $$$s$$$, character $$$0$$$ appears $$$u\\_0+x\\_0$$$ times; in $$$t$$$ it appears $$$u\\_0+y\\_0$$$ times. \nMode condition: \n\n$$$ u\\_0+x\\_0 \\\\ge \\\\lceil k/2 \\\\rceil,\\\\qquad u\\_0+y\\_0 \\\\ge \\\\lceil k/2 \\\\rceil . $$$\n\nAdding them: $$$2u\\_0 + x\\_0 + y\\_0 \\\\ge k = u\\_0+v\\_0+x\\_0+y\\_0$$$ $$$\\\\Rightarrow$$$ $$$u\\_0 \\\\ge v\\_0$$$. \nSubtracting one from the other gives $$$|x\\_0 - y\\_0| \\\\le u\\_0 - v\\_0 \\\\le u\\_0 + v\\_0$$$. \n\n**Case $$$c=1$$$:** similarly $$$v\\_0 \\\\ge u\\_0$$$ and $$$|x\\_0 - y\\_0| \\\\le v\\_0 - u\\_0 \\\\le u\\_0 + v\\_0$$$. \n\nThus **every operation** satisfies $$$|x\\_0 - y\\_0| \\\\le u\\_0+v\\_0$$$.\n\nNow define the potential $$$\\\\Phi = |x-y| - (u+v)$$$. \nAfter one operation the change is \n\n$$$ \\\\Phi' - \\\\Phi = \\\\bigl(|x-x\\_0 - (y-y\\_0)| - |x-y|\\\\bigr) + (u\\_0+v\\_0). $$$\n\nBy the triangle inequality, $$$|x-y| - |(x-y)-(x\\_0-y\\_0)| \\\\le |x\\_0-y\\_0|$$$. \nTherefore \n\n$$$ \\\\Phi' - \\\\Phi = (u\\_0+v\\_0) - \\\\bigl(|x-y| - |(x-y)-(x\\_0-y\\_0)|\\\\bigr) \\\\ge u\\_0+v\\_0 - |x\\_0-y\\_0| \\\\ge 0\\. $$$\n\nSo $$$\\\\Phi$$$ never decreases. \nThe empty pair has $$$\\\\Phi = 0$$$. Hence initially $$$\\\\Phi \\\\le 0$$$, i.e. $$$|x-y| \\\\le u+v$$$.\n\n**Sufficiency**: We can empty the strings by repeatedly deleting columns in groups of size 1 or 2.\n\nA single column can be deleted alone **iff** it is pure: $$$(0,0)$$$ (mode $$$0$$$) or $$$(1,1)$$$ (mode $$$1$$$). Two columns can be deleted together **iff** they are **not** two identical mixed columns:\n\n* $$$(0,0)$$$ pairs with anything.\n* $$$(1,1)$$$ pairs with anything.\n* $$$(0,1)$$$ pairs with $$$(1,0)$$$.\n\nSo we can pair $$$\\\\min(x,y)$$$ copies of $$$(0,1)$$$ with $$$(1,0)$$$, and the remaining $$$|x-y|$$$ mixed columns (all of the same type) are paired with pure columns $$$(0,0)$$$ or $$$(1,1)$$$. This is possible exactly because $$$u+v \\\\ge |x-y|$$$.\n\nThe leftover $$$u+v-|x-y|$$$ pure columns are paired arbitrarily among themselves. If the total number of columns is odd, exactly one pure column remains and can be deleted alone.\n\nAll these pairs / singletons are valid operations. After their deletion the strings become empty, so the pair is good.\n\nPrecompute prefix sums for the four column types. For a query $$$\\[l,r\\]$$$ obtain the four counts $$$u,v,x,y$$$ in constant time. Answer yes if $$$|x-y| \\\\le u+v$$$, otherwise no.\n\n**Complexity:** $$$O(n+q)$$$ per test case.\n\n**Code**\n\n```\n#include \n#include \n#include \n#include \n#include \n#include \ntypedef long long LL;\nusing namespace std;\ninline int read() {\n int num = 0 ,f = 1; char c = getchar();\n while (!isdigit(c)) f = c == '-' ? -1 : f ,c = getchar();\n while (isdigit(c)) num = (num << 1) + (num << 3) + (c ^ 48) ,c = getchar();\n return num * f;\n}\nconst int N = 2e5 + 5;\nchar s[N] ,t[N];\nint sum[4][N] ,n ,q;\ninline void fuling_solve() {\n n = read() ,q = read();\n scanf(\"%s%s\" ,s + 1 ,t + 1);\n for (int j = 0; j < 4; j++) sum[j][0] = 0;\n for (int i = 1; i <= n; i++) {\n int id = (s[i] ^ 48) * 2 + (t[i] ^ 48);\n for (int j = 0; j < 4; j++) sum[j][i] = sum[j][i - 1] + (id == j);\n }\n while (q--) {\n int l = read() ,r = read();\n int a = sum[0][r] - sum[0][l - 1];\n int b = sum[1][r] - sum[1][l - 1];\n int c = sum[2][r] - sum[2][l - 1];\n int d = sum[3][r] - sum[3][l - 1];\n puts(abs(b - c) <= a + d ? \"YES\" : \"NO\");\n }\n}\nsigned main() {\n int t = read();\n while (t--) fuling_solve();\n return 0;\n}\n\n```\n\n**Bonus Question**\n\nWe have consider to change this problem to a harder version: Find number of pairs $$$(l, r) (1 \\\\le l \\\\le r \\\\le n)$$$ such that $$$(s\\[l, r\\], t\\[l, r\\])$$$ is good.\n\nCan you solve this too?\n\n[2248E - Excuse for Breaks](https://codeforces.com/contest/2248/problem/E \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nLet $$$S\\_i = F(I(i))$$$. Since the counter resets at $$$n$$$, $$$S\\_{i+n} = S\\_i + S\\_n$$$.\n\n**Hint2**\n\nIf $$$S\\_x + S\\_y \\\\gt S\\_{x+y+1}$$$ for some $$$x,y$$$, the array $$$I(x),\\[0\\],I(y)$$$ already wins. Conversely, if no such pair exists, every array loses.\n\n**Hint3**\n\nFor fixed $$$y$$$, the function $$$S\\_x + S\\_y - S\\_{x+y+1}$$$ changes only when $$$x+1$$$ or $$$x+y+2$$$ is a reward point. Where is its maximum attained?\n\n**Solution**\n\nLet $$$S\\_i = F(I(i))$$$ ($$$S\\_0 = 0$$$). The counter resets exactly at $$$n$$$, so $$$S\\_{i+n} = S\\_i + S\\_n$$$; also $$$F$$$ is additive across a `0`: if a string ends with `0`, the next part starts from $$$c=0$$$.\n\nDefine\n\n$$$ M = \\\\max\\_{x,y \\\\ge 1} \\\\big(S\\_x + S\\_y - S\\_{x+y+1}\\\\big). $$$\n\n**Claim.** $$$F(s) \\\\le S\\_{|s|} + \\\\max(M,0)$$$ for every binary array $$$s$$$, with equality for $$$s = I(x),\\[0\\],I(y)$$$ at a maximizing pair. Hence the answer is `YES` iff $$$M \\\\gt 0$$$.\n\nInduct on the number of runs. If $$$s$$$ has no `0`, then $$$F(s) = S\\_{|s|}$$$. Otherwise write $$$s = I(x),\\[0\\],t$$$ with $$$x$$$ leading ones. Then\n\n$$$F(s) = S\\_x + F(t) \\\\le S\\_x + S\\_{|t|} + \\\\max(M,0) \\\\le S\\_{|s|} + \\\\max(M,0),$$$\n\nbecause $$$S\\_x + S\\_{|t|} - S\\_{x+|t|+1} \\\\le M$$$.\n\n**Only reward thresholds matter.** Let $$$g(x,y) = S\\_x + S\\_y - S\\_{x+y+1}$$$. By periodicity $$$g$$$ is $$$n$$$-periodic in each variable. Fix $$$y$$$:\n\n$$$g(x+1,y) - g(x,y) = (S\\_{x+1}-S\\_x) - (S\\_{x+y+2}-S\\_{x+y+1}),$$$\n\nwhich is positive only when $$$x+1$$$ is a reward point $$$p\\_i$$$. On the circle of residues modulo $$$n$$$, the maximum is attained on a plateau whose left end is congruent to some $$$p\\_i$$$ — so some $$$x = p\\_i$$$ attains it. The same argument in $$$y$$$ gives\n\n$$$ M = \\\\max\\_{1 \\\\le i,j \\\\le m} \\\\big(S\\_{p\\_i} + S\\_{p\\_j} - S\\_{p\\_i+p\\_j+1}\\\\big). $$$\n\n(If $$$m=0$$$, then $$$S\\_x = dx$$$ and $$$M = -d \\\\le 0$$$.)\n\n**Algorithm.** All needed values of $$$S$$$ follow from the periodic formula $$$S\\_i = \\\\lfloor i/n\\\\rfloor S\\_n + S\\_{i\\\\bmod n}$$$. For each pair $$$(i,j)$$$, locate $$$p\\_i+p\\_j+1$$$ among the sorted points $$${p\\_1,\\\\dots,p\\_m, n+p\\_1,\\\\dots,n+p\\_m}$$$ with a two-pointer scan; if $$$S\\_{p\\_i}+S\\_{p\\_j} \\\\gt S\\_{p\\_i+p\\_j+1}$$$ for some pair, print `YES`.\n\n**Complexity:** $$$O(m^2)$$$ per test case ($$$\\\\sum m \\\\le 2000$$$).\n\n**Code**\n\n```\nimport sys\ninput = sys.stdin.readline\n\nclass Point:\n def __init__(self, t, v):\n self.t = t\n self.v = v\n\nsrw = [Point(0, 0) for _ in range(4100)]\n\nT = int(input())\nfor _ in range(T):\n n, m, d = map(int, input().split())\n for i in range(1, m+1):\n t, v = map(int, input().split())\n srw[i].t = t\n srw[i].v = v + srw[i-1].v\n for i in range(1, m+1):\n srw[m+i] = Point(n + srw[i].t, srw[m].v + srw[i].v)\n ok = False\n for i in range(1, m+1):\n k = i\n for j in range(1, m+1):\n st = srw[i].t + 1 + srw[j].t\n while k+1 <= 2*m and srw[k+1].t <= st:\n k += 1\n if srw[i].v + srw[i].t*d + srw[j].v + srw[j].t*d > srw[k].v + st*d:\n ok = True\n print(\"Yes\" if ok else \"No\")\n for i in range(1, 2*m+1):\n srw[i] = Point(0, 0)\n\n```\n\n[2248F - Matrix Elimination](https://codeforces.com/contest/2248/problem/F \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nDefine the deficit $$$g\\_{x,y} = \\\\text{rowSum}\\_x + \\\\text{colSum}\\_y - 3v\\_{x,y}$$$ of a cell. A cell is a peak iff its deficit is $$$\\\\le 0$$$. How much can one operation reduce a deficit?\n\n**Hint2**\n\nFor $$$n,m \\\\ge 2$$$, every operation reduces any deficit by at most $$$n+m-3$$$, and full-matrix operations achieve this bound for every cell at once. So the problem reduces to a $$$k$$$-th smallest value.\n\n**Hint3**\n\nIn one dimension a cell is a peak iff $$$2a\\_i \\\\ge S$$$ (total sum). Only \"full\" operations and operations that skip one endpoint matter, and the two endpoint types are never both useful. The rest is a couple of linear inequalities solved in $$$O(1)$$$.\n\n**Solution**\n\nLet $$$R\\_x = \\\\sum\\_j v\\_{x,j}$$$ and $$$C\\_y = \\\\sum\\_i v\\_{i,y}$$$. The peak condition is equivalent to\n\n$$$ g\\_{x,y} := R\\_x + C\\_y - 3v\\_{x,y} \\\\le 0\\. $$$\n\nIf an operation covers $$$a$$$ cells of row $$$x$$$ and $$$b$$$ cells of column $$$y$$$, it changes the deficit by $$$-(a+b-3)$$$ when it covers $$$(x,y)$$$, and by $$$-(a+b)$$$ otherwise. For $$$n,m \\\\ge 2$$$ this means the deficit is reduced by at most $$$n+m-3$$$ per operation: full coverage gives $$$a+b-3 \\\\le n+m-3$$$, while an operation avoiding $$$(x,y)$$$ can cover cells of only one of the two lines, contributing at most $$$\\\\max(n-1,m-1) \\\\le n+m-3$$$.\n\n**Case $$$n,m \\\\ge 2$$$.** After $$$T$$$ operations, cell $$$(x,y)$$$ can be a peak only if $$$g\\_{x,y} \\\\le T(n+m-3)$$$. Conversely, $$$T$$$ full-matrix operations make every cell with $$$\\\\lceil g\\_{x,y}/(n+m-3)\\\\rceil \\\\le T$$$ a peak simultaneously. Hence the answer is the $$$k$$$-th smallest value of\n\n$$$ \\\\text{need}\\_{x,y} = \\\\max\\\\left(0,\\\\ \\\\left\\\\lceil \\\\frac{R\\_x + C\\_y - 3v\\_{x,y}}{n+m-3}\\\\right\\\\rceil\\\\right). $$$\n\n**Case $$$n=1$$$ or $$$m=1$$$.** Rotate to a row of length $$$L=\\\\max(n,m)$$$, sum $$$S$$$. \nPeak condition: $$$a\\_i \\\\ge S-a\\_i \\\\iff h\\_i := S-2a\\_i \\\\le 0$$$.\n\nAn operation on a segment of length $$$\\\\ell$$$ subtracts 1 from each element inside.\n\n* If $$$i$$$ is inside: $$$h\\_i$$$ decreases by $$$\\\\ell-2$$$\n* If $$$i$$$ is outside: $$$h\\_i$$$ decreases by $$$\\\\ell$$$\n\nOnly three segment types are useful (others are dominated):\n\n1. **Full** $$$\\[1,L\\]$$$: every $$$h\\_i$$$ reduced by $$$L-2$$$.\n2. **Skip‑first** $$$\\[2,L\\]$$$: $$$h\\_1$$$ reduced by $$$L-1$$$, others by $$$L-3$$$.\n3. **Skip‑last** $$$\\[1,L-1\\]$$$: $$$h\\_L$$$ reduced by $$$L-1$$$, others by $$$L-3$$$.\n\n**Only full operations**: Same as the general formula with $$$n+m-3 = L-2$$$: \nAnswer = $$$k$$$-th smallest $$$\\\\left\\\\lceil \\\\frac{S-2a\\_i}{L-2} \\\\right\\\\rceil$$$ (with $$$\\\\max(0,\\\\cdot)$$$). (This is exactly the `xlim` part in the code).\n\n**Full + one skip type, with an endpoint among the peaks**: Assume the chosen endpoint is $$$p$$$ (either 1 or $$$L$$$), using $$$x$$$ full and $$$y$$$ skip‑$p$ operations ($$$T=x+y$$$). Take the other $$$k-1$$$ peaks as the cells with the **largest** $$$a\\_i$$$ (smallest $$$h\\_i$$$) among the remaining $$$L-1$$$ cells; let $$$v\\_k$$$ be the smallest of these. Reductions: - For $$$p$$$: $$$(L-2)x + (L-1)y = (L-2)T + y$$$ - For the other $$$k-1$$$ cells: $$$(L-2)x + (L-3)y = (L-2)T - y$$$\n\nRequirements:\n\n$$$ \\\\begin{aligned} (L-2)T + y &\\\\ge S - 2a\\_p \\\\quad &(1)\\\\\\\\ (L-2)T - y &\\\\ge S - 2v\\_k \\\\quad &(2)\\\\\\\\ 0 \\\\le y &\\\\le T . \\\\end{aligned} $$$\n\nFrom the constraints:\n\n* (2) with $$$y=0$$$ gives $$$T \\\\ge \\\\lceil (S-2v\\_k)/(L-2) \\\\rceil$$$\n* (1) with $$$y=T$$$ gives $$$T \\\\ge \\\\lceil (S-2a\\_p)/(L-1) \\\\rceil$$$\n* Adding (1) and (2): $$$2(L-2)T \\\\ge 2S-2a\\_p-2v\\_k \\\\iff T \\\\ge \\\\lceil (S-a\\_p-v\\_k)/(L-2) \\\\rceil$$$\n\nSet \n\n$$$ z = \\\\max\\\\left( \\\\lceil \\\\frac{S-a\\_p-v\\_k}{L-2} \\\\rceil,\\\\; \\\\lceil \\\\frac{S-2a\\_p}{L-1} \\\\rceil \\\\right) $$$\n\n(If the first bound $$$\\\\lceil (S-2v\\_k)/(L-2) \\\\rceil$$$ were strictly larger, then $$$y=0$$$ is optimal and Strategy of only full operations already covers it.)\n\nNow we must check whether there exists an integer $$$y$$$ satisfying the box constraints:\n\n$$$ \\\\max\\\\bigl(0,\\\\; S-2a\\_p - (L-2)z\\\\bigr) \\\\;\\\\le\\\\; y \\\\;\\\\le\\\\; \\\\min\\\\bigl(z,\\\\; (L-2)z - (S-2v\\_k)\\\\bigr). $$$\n\nIf this holds, $$$T=z$$$ is feasible and optimal for this endpoint; otherwise the endpoint strategy cannot beat the \"only full operations\" strategy.\n\nThe code’s `solveL` implements exactly this: it computes $$$z$$$, tests feasibility, and returns $$$z$$$ if valid or $$$\\\\infty$$$ otherwise. \nThe final answer is $$$\\\\min(\\\\text{Strategy full},\\\\; \\\\text{solveL for left endpoint},\\\\; \\\\text{solveL for right endpoint})$$$.\n\n**Complexity:** $$$O(nm)$$$ per test case.\n\n**Code**\n\n```\nimport math\n\ndef ceil_div(a, b):\n return (a + b - 1) // b\n\ndef solveL(v, v1, s, k):\n n = len(v) + 1\n if k == 1:\n return max(0, ceil_div(s - 2 * v1, n - 1))\n v.sort(reverse=True)\n vk = v[k - 2]\n z = max(ceil_div(s - v1 - vk, n - 2), ceil_div(s - 2 * v1, n - 1))\n if min(z, (n - 2) * z - s + 2 * vk) >= max(0, s - (n - 2) * z - 2 * v1):\n return z\n return math.inf\n\nT = int(input())\nfor _ in range(T):\n n, m, k = map(int, input().split())\n s = 0\n u, rs, cs = [], [0] * n, [0] * m\n v = []\n for _ in range(n):\n row = list(map(int, input().split()))\n v.append(row)\n s += sum(row)\n rs[_] = sum(row)\n for j in range(m):\n cs[j] += row[j]\n u.append(row[j])\n\n if n + m <= 3:\n u.sort()\n if k == 1:\n if n + m == 2 and u[0] < 0:\n print(-1)\n else:\n print(0)\n else:\n print(abs(u[0] - u[1]))\n continue\n\n xlim = []\n for i in range(n):\n for j in range(m):\n xlim.append(max(0, ceil_div(rs[i] + cs[j] - 3 * v[i][j], n + m - 3)))\n\n xlim.sort()\n ans = xlim[k - 1]\n\n if m == 1 or n == 1:\n ans = min(ans, solveL(u[:-1], u[-1], s, k))\n ans = min(ans, solveL(u[1:], u[0], s, k))\n print(ans)\n\n```\n\n[2248G - No Balance Left](https://codeforces.com/contest/2248/problem/G \"Codeforces Round 1113 (Div. 2)\")\n\n**Hint1**\n\nA purchase is a transaction: spend an achievable amount $$$x$$$, gain rebate $$$r(x)$$$, net change $$$r(x)-x$$$. The achievable spends form a knapsack.\n\n**Hint2**\n\nBelow the smallest spend that yields a net increase, the balance only decreases; reachability is a DP on a DAG.\n\n**Hint3**\n\nAbove that threshold the balance can grow without bound, and every transaction changes it by a multiple of a single gcd. What determines whether $$$0$$$ is reachable?\n\n**Hint4**\n\nIf the smallest threshold does not exceed the cheapest product, every purchase is discounted and no positive balance can ever become $$$0$$$.\n\n**Solution**\n\nLet $$$V = \\\\max(s, a\\_m)$$$ and compute the set $$$C$$$ of spend amounts achievable up to $$$V$$$ with an unbounded knapsack over the prices (a bitset, $$$O(V^2/64)$$$). The rebate for a spend $$$x$$$ is $$$r(x) = \\\\max{b\\_j : a\\_j \\\\le x}$$$, obtained by a prefix maximum.\n\nFor every achievable $$$x$$$:\n\n* if $$$r(x) \\\\lt x$$$, the net decrease is $$$d = x - r(x)$$$; keep $$$L\\[d\\] = \\\\min{x}$$$ — the cheapest way to lose exactly $$$d$$$;\n* if $$$r(x) \\\\gt x$$$, the purchase increases the balance; let $$$\\\\text{mnip}$$$ be the smallest such spend.\n\n**Below $$$\\\\text{mnip}$$$.** Every affordable transaction has $$$r(x) \\\\le x$$$, so the balance only decreases. Define $$$f\\[0\\] = 1$$$ and\n\n$$$ f\\[i\\] = \\\\bigvee\\_{d \\\\le i} \\\\big(L\\[d\\] \\\\le i \\\\text{ and } f\\[i-d\\]\\\\big). $$$\n\nThis is a DAG (all transitions go down); using the minimum spend for each decrease $$$d$$$ is optimal because a cheaper transaction is only more flexible. A shifted bitset computes all $$$f\\[i\\]$$$ in $$$O(V^2/64)$$$.\n\n**At or above $$$\\\\text{mnip}$$$.** A net-increasing purchase is affordable and can be repeated, so the balance can be made arbitrarily large. Every transaction changes the balance by a multiple of\n\n$$$ g = \\\\gcd(I,\\\\ D,\\\\ c\\_1, \\\\dots, c\\_n,\\\\ b\\_m), $$$\n\nso the residue of the balance modulo $$$g$$$ is invariant, and from a sufficiently large balance every value in the same residue class is reachable (Frobenius). Any path to $$$0$$$ must enter the decreasing region: for the last purchase, $$$b - x + r(x) = 0$$$ with $$$x \\\\le b$$$ forces $$$r(x) = 0$$$ and $$$x = b \\\\lt a\\_1$$$. Therefore a balance $$$i \\\\ge \\\\text{mnip}$$$ can reach $$$0$$$ iff $$$i \\\\equiv 0 \\\\pmod g$$$ and some positive balance below $$$\\\\text{mnip}$$$ is reachable.\n\nThe second condition is automatic when $$$a\\_1 \\\\gt c\\_{\\\\min}$$$: the spend $$$c\\_{\\\\min}$$$ (the cheapest product) has rebate $$$0$$$, so $$$f\\[c\\_{\\\\min}\\] = 1$$$ and $$$c\\_{\\\\min} \\\\lt \\\\text{mnip}$$$. Conversely, if $$$a\\_1 \\\\le c\\_{\\\\min}$$$, every purchase has $$$r(x) \\\\ge b\\_1 \\\\ge 1$$$, so a purchase of cost $$$x \\\\le h$$$ leaves balance $$$h - x + r(x) \\\\gt 0$$$ — no positive balance can ever become $$$0$$$.\n\nHere $$$I$$$ and $$$D$$$ are the net increases/decreases found above, and the costs and $$$b\\_m$$$ stand in for the spends beyond $$$V$$$: for $$$x \\\\ge a\\_m$$$ the net change is $$$x - b\\_m$$$, and since all achievable spends are multiples of $$$\\\\gcd(c\\_1,\\\\dots,c\\_n)$$$, the whole tail contributes exactly $$$\\\\gcd(c\\_1,\\\\dots,c\\_n, b\\_m)$$$.\n\n**Complexity:** $$$O(V^2/64)$$$ time and $$$O(V)$$$ memory, with $$$V = \\\\max(s, a\\_m) \\\\le 125\\\\,000$$$.\n\n**Code**\n\n```\n#include \nusing namespace std;\n\nconst int maxV = 125000 + 5;\n\nint Lgcd(const vector &v) {\n\treturn accumulate(v.begin(), v.end(), 0, [](int a, int b) { return gcd(a, b); });\n}\n\nint main() {\n\tint n, m, s;\n\tscanf(\"%d %d %d\", &n, &m, &s);\n\tvector c(n);\n\tvector> ab(m);\n\tfor (auto &x : c)\n\t\tscanf(\"%d\", &x);\n\tfor (auto &[a, b] : ab)\n\t\tscanf(\"%d %d\", &a, &b);\n\tab.insert(ab.begin(), make_pair(0, 0));\n\tm += 1;\n\n\tbitset p;\n\tp.set(0);\n\tfor (int x : c)\n\t\tp.set(x);\n\n\tconst int V = max(s, ab.back().first);\n\tfor (int i = 1; i <= V; ++i) {\n\t\tif (p[i])\n\t\t\tp |= (p << i);\n\t}\n\n\tvector l(V + 1, s + 1);\n\tl[0] = 0;\n\tint mnip = s + 1;\n\tvector I, D;\n\tfor (int i = 1, abp = 0; i <= V; ++i) {\n\t\tif (!p[i])\n\t\t\tcontinue;\n\t\twhile (abp + 1 < m && ab[abp + 1].first <= i)\n\t\t\tabp += 1;\n\n\t\tauto [a, b] = ab[abp];\n\t\tif (b <= i) {\n\t\t\tl[i - b] = min(l[i - b], i);\n\t\t\tD.push_back(i - b);\n\t\t} else {\n\t\t\tI.push_back(b - i);\n\t\t\tmnip = min(mnip, i);\n\t\t}\n\t}\n\n\tvector> lv;\n\tfor (int i = 1; i <= V; ++i) {\n\t\tif (l[i] <= s)\n\t\t\tlv.emplace_back(l[i], i);\n\t}\n\tconst int ln = lv.size();\n\n\tsort(lv.begin(), lv.end());\n\tsort(c.begin(), c.end());\n\tsort(I.begin(), I.end());\n\tsort(D.begin(), D.end());\n\n\tbitset f, L;\n\tf.set(0);\n\tfor (int i = 1, lp = 0; i < mnip; ++i) {\n\t\twhile (lp < ln && lv[lp].first <= i) {\n\t\t\tL.set(maxV - lv[lp].second);\n\t\t\tlp += 1;\n\t\t}\n\n\t\tf[i] = ((f & (L >> (maxV - i)))._Find_first() < V);\n\t}\n\n\tint Ig = Lgcd(I), Dg = Lgcd(D), cg = Lgcd(c);\n\tint ugcd = gcd(gcd(Ig, Dg), gcd(cg, ab.back().second));\n\n\tfor (int i = 1; i < mnip; ++i)\n\t\tputs(f[i] ? \"YES\" : \"NO\");\n\tfor (int i = mnip; i <= s; ++i)\n\t\tputs((i % ugcd == 0 && ab[1].first > c[0]) ? \"YES\" : \"NO\");\n\treturn 0;\n}\n\n```\n\nI promise I'll improve this soon. I know its unclear and skipped some steps for now.\n\nAlso check out:\n\n* \n* \n\n![](//codeforces.com/codeforces.org/s/13758/images/icons/paperclip-16x16.png) Tutorial of [Codeforces Round 1113 (Div. 2)](https://codeforces.com/contest/2248) \n\n* [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/voteup-gray.png \"Vote: I like it\")](#)\n* +154\n* [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/votedown-gray.png \"Vote: I do not like it\")](#)\n\n* [![Author](//codeforces.com/codeforces.org/s/13758/images/blog/user_16x16.png \"Author\")](https://codeforces.com/profile/szdytom) [szdytom ](https://codeforces.com/profile/szdytom)\n* ![Publication date](//codeforces.com/codeforces.org/s/13758/images/blog/date_16x16.png \"Publication date\") 7 weeks ago\n* [![Comments](//codeforces.com/codeforces.org/s/13758/images/blog/comments_16x16.png \"Comments\")](https://codeforces.com/blog/entry/155640#comments) [83 ](https://codeforces.com/blog/entry/155640#comments)\n\n \n![Comments](//codeforces.com/codeforces.org/s/13758/images/icons/comments-48x48.png \"Comments\") Comments (83) \n\n[Write comment?](#) \n\n| » [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") | 7 weeks ago, [show](#) # |\n| -------------------------------------------------------------------- | ------------------------ |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/1198980/avatar/905c3a9d5029cd8e.jpg) ](https://codeforces.com/profile/szdytom) [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382276 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) _Auto comment: topic has been updated by [szdytom](https://codeforces.com/profile/szdytom \"Master szdytom\") ([previous revision](https://codeforces.com/topic/156282/en6), [new revision](https://codeforces.com/topic/156282/en7), [compare](https://codeforces.com/topic/156282/diff/en6/en7))._ [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n\n| » [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [show (+1)](#) # |\n| ---------------------------------------------------------- | ----------------------------- |\n\n| » [ ![](https://codeforces.com/userpic.codeforces.org/4921754/avatar/6939fd62eb19593f.jpg) ](https://codeforces.com/profile/rdor) [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382280 \"Link to comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +56 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) is it just me or B felt harder than C and D? [→](#) [Reply](#) |\n| ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n* | » » [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 7 weeks ago, [show](#) # |\n| ---------------------------------------------------------------- | ------------------------ |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/overi) [overi](https://codeforces.com/profile/overi \"Newbie overi\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382291 \"Link to comment\") [^](#comment-1382280 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) agreed [→](#) [Reply](#) |\n| --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n* | » » [petryshynserhii361](https://codeforces.com/profile/petryshynserhii361 \"Specialist petryshynserhii361\") | 7 weeks ago, [show (+1)](#) # |\n| ----------------------------------------------------------------------------------------------------------- | ----------------------------- |\n\n| » » [ ![](https://codeforces.com/userpic.codeforces.org/no-avatar.jpg) ](https://codeforces.com/profile/petryshynserhii361) [petryshynserhii361](https://codeforces.com/profile/petryshynserhii361 \"Specialist petryshynserhii361\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382301 \"Link to comment\") [^](#comment-1382280 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) 0 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) Not really for me, i actually struggled more with C because my dumb ahh couldnt realise it was just a dp for what felt like eternity. Also congrats on becoming specialist! [→](#) [Reply](#) |\n| ----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n * | » » » [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [show](#) # |\n| -------------------------------------------------------------- | ------------------------ |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4921754/avatar/6939fd62eb19593f.jpg) ](https://codeforces.com/profile/rdor) [rdor](https://codeforces.com/profile/rdor \"Pupil rdor\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382306 \"Link to comment\") [^](#comment-1382301 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) +1 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) thanks!! :D [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |\n * | » » » [JeffreyLee](https://codeforces.com/profile/JeffreyLee \"Expert JeffreyLee\") | 7 weeks ago, [show (+1)](#) # |\n| --------------------------------------------------------------------------------- | ----------------------------- |\n\n| » » » [ ![](https://codeforces.com/userpic.codeforces.org/4029260/avatar/63a4474c60b159c.jpg) ](https://codeforces.com/profile/JeffreyLee) [JeffreyLee](https://codeforces.com/profile/JeffreyLee \"Expert JeffreyLee\") | 7 weeks ago, [hide](#) [#](https://codeforces.com/blog/entry/155640?#comment-1382388 \"Link to comment\") [^](#comment-1382301 \"Parent comment\") \\| [![Vote: I like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-voteup-blue.png \"Vote: I like it\")](#) \\-8 [![Vote: I do not like it](//codeforces.com/codeforces.org/s/13758/images/actions/comment-votedown-blue.png \"Vote: I do not like it\")](#) I think C and F are fraud problems,if you guess the conclusion and succeed,you will solve them immediately. [→](#) [Reply](#) |\n| ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- |\n\n * | » » » » [silverfish](https://codeforces.com/profile/silverfish \"Specialist silverfish\") | 7 weeks ago, [show](#) # |\n| --------------------------------------------------------------------------------------- | ------------------------ |\n\n\n...157878 bytes truncated...\n\nThe tool call succeeded but the output was truncated. Full output saved to: /Users/zayd/.local/share/opencode/tool-output/tool_0b252ca42001xCJTSbFgjOMFCK\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.", "metadata": { "truncated": true, "outputPath": "/Users/zayd/.local/share/opencode/tool-output/tool_0b252ca42001xCJTSbFgjOMFCK" }, "title": "https://codeforces.com/blog/entry/155640 (text/markdown; charset=utf-8)", "time": { "start": 1789698160673, "end": 1789698165317 } }, "metadata": { "openai": { "itemId": "fc_0a152624c1a27375016aaca06fddc087d0b24fd4ed979bc13b" } }, "id": "prt_0b252b536001v729f1mzbQDT0L", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252abea001CggOKvv8d92GN8" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 9354, "input": 9322, "output": 32, "reasoning": 0, "cache": { "write": 0, "read": 0 } }, "cost": 0, "id": "prt_0b252ca4c001xHb25iAu0iM9oO", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252abea001CggOKvv8d92GN8" } ] }, { "info": { "parentID": "msg_0b252ab610017tt3PNtRSZADmJ", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "low", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PnyW7F", "root": "/" }, "cost": 0, "tokens": { "total": 25485, "input": 15618, "output": 436, "reasoning": 343, "cache": { "write": 0, "read": 9088 } }, "modelID": "gpt-6-astra", "providerID": "openai", "time": { "created": 1789698165329, "completed": 1789698192599 }, "finish": "tool-calls", "id": "msg_0b252ca51001ebfeT1pZZUX270", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa" }, "parts": [ { "type": "step-start", "id": "prt_0b252d3e9001AFGAcDoyoBt5HI", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ca51001ebfeT1pZZUX270" }, { "type": "reasoning", "text": "**Fixing the threshold proof**", "time": { "start": 1789698167787, "end": 1789698180488 }, "metadata": { "openai": { "itemId": "rs_0e61573989a5cfe1016aaca077be4c87d0b12454d0a2951e08" } }, "id": "prt_0b252d3eb001MsipfK5eQitVY7", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ca51001ebfeT1pZZUX270" }, { "type": "reasoning", "text": "**Verifying expanded candidates**", "time": { "start": 1789698180490, "end": 1789698180600 }, "metadata": { "openai": { "itemId": "rs_0e61573989a5cfe1016aaca077be4c87d0b12454d0a2951e08", "reasoningEncryptedContent": 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} }, "id": "prt_0b253058a0013G07w9HujfHgw7", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ca51001ebfeT1pZZUX270" }, { "type": "tool", "tool": "bash", "callID": "call_EnLO3UR0QyEFa2vzlMYixjHo", "state": { "status": "completed", "input": { "command": "python3 -c 'import random\nrandom.seed(2248)\nfor _ in range(10000):\n n=random.randint(1,10); d=random.randint(0,10)\n p=sorted(random.sample(range(1,n+1),random.randint(0,n))); r=[random.randint(1,12) for x in p]\n pref=[0]\n for v in r: pref.append(pref[-1]+v)\n def S(x): return x*d+(x//n)*sum(r)+sum(v for q,v in zip(p,r) if q<=x%n)\n expected=any(S(x)+S(y)>S(x+y+1) for x in range(n) for y in range(n))\n points=[q+b*n for b in range(3) for q in p]; sums=[0]\n for v in r*3: sums.append(sums[-1]+v)\n ans=False\n for i,x in enumerate(p):\n k=0\n for j,y in enumerate(p):\n while ksums[k]+d\n assert ans==expected,(n,d,p,r)\n dp=[-10**9]*n; dp[0]=0\n for length in range(1,4*n+2):\n ndp=[-10**9]*n; ndp[0]=max(dp)\n for c,v in enumerate(dp):\n nc=(c+1)%n; reward=d+sum(rv for q,rv in zip(p,r) if q==c+1)\n ndp[nc]=max(ndp[nc],v+reward)\n dp=ndp\n assert expected or max(dp)<=S(length)\nprint(\"10000 randomized cases passed: candidate scan versus all residue pairs, plus short-length DP checks.\")'" }, "output": "10000 randomized cases passed: candidate scan versus all residue pairs, plus short-length DP checks.\n", "metadata": { "output": "10000 randomized cases passed: candidate scan versus all residue pairs, plus short-length DP checks.\n", "exit": 0, "truncated": false }, "title": "python3 -c 'import random\nrandom.seed(2248)\nfor _ in range(10000):\n n=random.randint(1,10); d=random.randint(0,10)\n p=sorted(random.sample(range(1,n+1),random.randint(0,n))); r=[random.randint(1,12) for x in p]\n pref=[0]\n for v in r: pref.append(pref[-1]+v)\n def S(x): return x*d+(x//n)*sum(r)+sum(v for q,v in zip(p,r) if q<=x%n)\n expected=any(S(x)+S(y)>S(x+y+1) for x in range(n) for y in range(n))\n points=[q+b*n for b in range(3) for q in p]; sums=[0]\n for v in r*3: sums.append(sums[-1]+v)\n ans=False\n for i,x in enumerate(p):\n k=0\n for j,y in enumerate(p):\n while ksums[k]+d\n assert ans==expected,(n,d,p,r)\n dp=[-10**9]*n; dp[0]=0\n for length in range(1,4*n+2):\n ndp=[-10**9]*n; ndp[0]=max(dp)\n for c,v in enumerate(dp):\n nc=(c+1)%n; reward=d+sum(rv for q,rv in zip(p,r) if q==c+1)\n ndp[nc]=max(ndp[nc],v+reward)\n dp=ndp\n assert expected or max(dp)<=S(length)\nprint(\"10000 randomized cases passed: candidate scan versus all residue pairs, plus short-length DP checks.\")'", "time": { "start": 1789698191277, "end": 1789698192596 } }, "metadata": { "openai": { "itemId": "fc_0e61573989a5cfe1016aaca084703c87d086ea5054c0d6dc17" } }, "id": "prt_0b2530625001XVnOjOvl1ZnRFk", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ca51001ebfeT1pZZUX270" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 25485, "input": 15618, "output": 436, "reasoning": 343, "cache": { "write": 0, "read": 9088 } }, "cost": 0, "id": "prt_0b25334d5001jeLmY7KWI89oSw", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b252ca51001ebfeT1pZZUX270" } ] }, { "info": { "parentID": "msg_0b252ab610017tt3PNtRSZADmJ", "role": "assistant", "mode": "nudge-generation", "agent": "nudge-generation", "variant": "low", "path": { "cwd": "/private/var/folders/t7/b_w5m78j2mbb1ct051p9sv500000gn/T/nudge-opencode-generation-PnyW7F", "root": "/" }, "cost": 0, "tokens": { "total": 28167, "input": 1067, "output": 2548, "reasoning": 104, "cache": { "write": 0, "read": 24448 } }, "modelID": "gpt-6-astra", "providerID": "openai", "time": { "created": 1789698192601, "completed": 1789698275133 }, "finish": "tool-calls", "structured": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "Let $S(k)$ be the score of $k$ consecutive ones, with $S(0)=0$. A zero resets the counter, so an array’s score is the sum of $S(k)$ over its runs of ones. Also, $S(k+n)=S(k)+S(n)$." }, { "order": 2, "content": "Try an array with just one zero: $1^x0 1^y$. It wins exactly when $S(x)+S(y)>S(x+y+1)$. Why is checking one zero enough? If this inequality never holds, you can repeatedly merge two runs and their separating zero into one all-ones run without decreasing the score." }, { "order": 3, "content": "Separate the ordinary earnings from the bonuses: write $S(k)=kd+B(k)$, where $B(k)$ is the total bonus earned by $k$ consecutive ones. The winning condition becomes $B(x)+B(y)>B(x+y+1)+d$. The zero costs exactly $d$; everything else is about bonuses." }, { "order": 4, "content": "Suppose a pair $(x,y)$ wins. Shorten $x$ to the most recent position at which its run earned a bonus. Its own bonus total stays unchanged, while the all-ones benchmark cannot gain bonuses from becoming shorter. Then use periodicity to replace that position by some $p_i$. Do the same for $y$. Thus, only pairs $(p_i,p_j)$ matter." }, { "order": 5, "content": "Build the sorted bonus positions for three consecutive cycles: $p_i$, $n+p_i$, and $2n+p_i$, together with prefix sums of their rewards. For each fixed $i$, scan $j$ in increasing order and maintain a pointer counting positions at most $p_i+p_j+1$. If their total reward is $C$ and the original reward prefix sums are $R_i,R_j$, check $R_i+R_j>C+d$. The pointer only moves forward, giving $O(m^2)$ time." } ], "editorial": "The array can have any finite length, so trying to construct it directly is a trap. Instead, we’ll show that **if any array wins, an array with exactly one zero wins—and both runs can end at reward thresholds.** That leaves only $m^2$ candidates.\n\n### 1. Describe an all-ones run\n\nLet $S(k)$ be the score of $k$ consecutive ones, and define $S(0)=0$.\n\nThe counter resets after every $n$ ones, so\n\n$$S(k+n)=S(k)+S(n).$$\n\nLet $B(k)$ count only the bonus rewards earned during those $k$ ones. Then\n\n$$S(k)=kd+B(k).$$\n\nIf $R=\\sum_{i=1}^{m}r_i$, we have\n\n$$B(k)=\\left\\lfloor\\frac{k}{n}\\right\\rfloor R+\n\\sum_{p_i\\le k\\bmod n}r_i.$$\n\nIn particular, a reward at $p_i=n$ is collected before the counter resets. It belongs to the completed cycle.\n\nBecause rewards are positive, $B(k)$ is nondecreasing.\n\n### 2. One zero is enough\n\nConsider the array $1^x0 1^y$, where $1^x$ means $x$ consecutive ones. The zero separates two independent runs, so its advantage over all ones is\n\n$$G(x,y)=S(x)+S(y)-S(x+y+1).$$\n\nIf $G(x,y)>0$, we immediately have a valid answer.\n\nConversely, suppose $G(x,y)\\le 0$ for every pair of nonnegative integers $x,y$. Take any binary array and write it as runs of ones separated by zeros, allowing empty runs for leading, trailing, or consecutive zeros.\n\nWe can replace two neighboring runs and their separating zero by $x+y+1$ ones. This keeps the length unchanged and cannot decrease the score, because\n\n$$S(x)+S(y)\\le S(x+y+1).$$\n\nOther zeros still separate the remaining runs, so their scores are unaffected. Repeat until no zeros remain. The result is the all-ones array of the same length, with at least the original score.\n\nTherefore,\n\n$$\\boxed{\\text{Answer is YES iff some }G(x,y)>0.}$$\n\nNotice that if either run is empty, the pair cannot win: for example, $G(0,y)=S(y)-S(y+1)\\le 0$.\n\n### 3. Reduce infinitely many lengths to reward positions\n\nSubstituting $S(k)=kd+B(k)$ gives\n\n$$G(x,y)=B(x)+B(y)-B(x+y+1)-d.$$\n\nThe ordinary earnings mostly cancel. A break costs exactly $d$—the bonuses have to pay for it.\n\nSuppose $(x,y)$ is a winning pair. If $B(x)=0$, then monotonicity gives\n\n$$G(x,y)=B(y)-B(x+y+1)-d\\le 0,$$\n\nwhich is impossible. Thus, the first run has earned a bonus.\n\nLet $t\\le x$ be the last position where that run earned one. Such positions have the form\n\n$$t=qn+p_i.$$\n\nThere are no bonuses between $t$ and $x$, so $B(t)=B(x)$. Meanwhile,\n\n$$B(t+y+1)\\le B(x+y+1).$$\n\nConsequently,\n\n$$G(t,y)\\ge G(x,y)>0.$$\n\nShortening the run to its last reward does not hurt.\n\nNext, periodicity gives\n\n$$G(x+n,y)=G(x,y),$$\n\nsince adding $n$ to the first run adds the same $S(n)$ to both sides of the comparison. Hence we can replace $t=qn+p_i$ by $p_i$ without changing the advantage.\n\nApply the same argument to the second run. We obtain a winning pair $(p_i,p_j)$.\n\nSo the entire problem becomes checking\n\n$$\\boxed{B(p_i)+B(p_j)>B(p_i+p_j+1)+d}$$\n\nfor all $1\\le i,j\\le m$.\n\nIf $m=0$, there are no bonuses to exploit, and the answer is `NO`.\n\n### 4. Check every pair with two pointers\n\nDefine the original reward prefix sums\n\n$$R_0=0,\\qquad R_i=\\sum_{h=1}^{i}r_h.$$\n\nThen $B(p_i)=R_i$.\n\nTo evaluate the right-hand side efficiently, build the chronological list of reward events for three cycles:\n\n- positions $p_1,\\ldots,p_m$;\n- positions $n+p_1,\\ldots,n+p_m$;\n- positions $2n+p_1,\\ldots,2n+p_m$.\n\nEach cycle repeats the rewards $r_1,\\ldots,r_m$. The combined list is already sorted. Build its reward prefix sums $C$, with $C_0=0$.\n\nWhy three cycles? Our queried length is\n\n$$L=p_i+p_j+1\\le 2n+1\\le 3n.$$\n\nThree cycles cover every query, including the tiny edge case $n=1$.\n\nFor each fixed $i$:\n\n1. Start an event pointer $k=0$.\n2. Process $j=1,2,\\ldots,m$.\n3. Advance $k$ while the next event position is at most $L=p_i+p_j+1$.\n4. Now $B(L)=C_k$. Check whether\n $$R_i+R_j>C_k+d.$$\n5. If so, output `YES`.\n\nFor fixed $i$, $L$ increases with $j$, so the pointer never moves backward. If no pair wins, output `NO`.\n\n### Correctness argument\n\nIf the algorithm outputs `YES`, it found indices $i,j$ satisfying\n\n$$B(p_i)+B(p_j)>B(p_i+p_j+1)+d.$$\n\nEquivalently, $S(p_i)+S(p_j)>S(p_i+p_j+1)$, so the array $1^{p_i}0 1^{p_j}$ wins.\n\nIf any binary array wins, the run-merging argument guarantees some winning one-zero pair. Shortening each run to its last reward and removing complete cycles produces a winning pair $(p_i,p_j)$. The algorithm checks every such pair, and its event prefix sum computes the benchmark’s bonuses exactly. It must therefore output `YES`.\n\nThus the algorithm outputs `YES` exactly when a valid array exists.\n\n### Complexity and implementation details\n\nFor each of the $m$ choices of $i$, the inner loop visits $m$ candidates and the pointer crosses at most $3m$ events. The running time is $O(m^2)$, with $O(m)$ extra space. Since $\\sum m\\le 2000$, the total quadratic work is at most $2000^2$ up to constant factors.\n\nUse `long long` for event positions and reward sums. Use `<=` when including events at the queried length, because rewards are collected on that position. The final comparison must be strict: tying the all-ones score does not count.", "solution": "#include \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 int t;\n cin >> t;\n while (t--) {\n ll n, d;\n int m;\n cin >> n >> m >> d;\n\n vector p(m), r(m), pref(m + 1);\n for (int i = 0; i < m; ++i) {\n cin >> p[i] >> r[i];\n pref[i + 1] = pref[i] + r[i];\n }\n\n vector pos(3 * m), sum(3 * m + 1);\n for (int cycle = 0; cycle < 3; ++cycle) {\n for (int i = 0; i < m; ++i) {\n int k = cycle * m + i;\n pos[k] = cycle * n + p[i];\n sum[k + 1] = sum[k] + r[i];\n }\n }\n\n bool ok = false;\n for (int i = 0; i < m && !ok; ++i) {\n int k = 0;\n for (int j = 0; j < m; ++j) {\n ll len = p[i] + p[j] + 1;\n while (k < 3 * m && pos[k] <= len) ++k;\n if (pref[i + 1] + pref[j + 1] > sum[k] + d) {\n ok = true;\n break;\n }\n }\n }\n\n cout << (ok ? \"YES\\n\" : \"NO\\n\");\n }\n}\n" }, "id": "msg_0b25334d9001tiZMj8IUupKJA7", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa" }, "parts": [ { "type": "step-start", "id": "prt_0b2533fe4001SE7K8BaZGoBAqW", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b25334d9001tiZMj8IUupKJA7" }, { "type": "reasoning", "text": "**Refining the induction proof**", "time": { "start": 1789698195431, "end": 1789698200548 }, "metadata": { "openai": { "itemId": "rs_003f88e5d85d5632016aaca093608087d09a8ff712c91858d2", "reasoningEncryptedContent": "gAAAAABqrKCYqKmG1Jymjq4jYEyA5wrhvKKwVxwisvkzz9b_eBRR55-lAt_UJi7zrIvF4iKR48N10nd7_3wlXcBa07YX7rBqEsrOCzlX_WTO5nA_INmkvZVQH3BWTZLTqdxeUEeChOfoE055udRyMJDVeFT8lTtPHycSnnIYBFqWHyx8mqMl-GHVqVPYl-NkbNEOFdvX94k1SXPj58GK9Hgiop5TYC5LSYp50MCvLeat6cdyLMsKVYyEkb8I_J-hjs6pWxBXZNU9lnDhAGX3844OVnahJu4RSC6GjHIrVgX1Wxy0Q_mSfkkg5QHGa_cv9PctD2fb8lzhAtyBnezwxZLkdIrU79HbhjXDOzWJciwCEkx-kH8ll6GtQe-dGsgtQZKCdYP1-Vh0u4th_svDYW8XVp_6KDpWZETAwZqb1L25-96ckcgTjage3fO0KqP3KIHbYs1lDNaM7_jfZqAB0ee8Tw-_v20eYrlrcoNmmRv1ZHfaqAOrctZjioht1aJrw_ktGVVNW3zBI5JZrjzdXtUrHFTYb_lig1aJCDdI7VpSlXMLtGCd2gi6dBoCV7SiTt0rwMEORL3kAuR71yYZ8TuMaJg-H52GjA1DhdNWBWXJtbdIJu2uwzUjmm8ZVFQFi6jlRWtpEFFmM7eZYgeT3H0RPggvipgcwu7SFjszGX4ay19Ul7iirUKFSquRu3V1QY8sav-YspouCuzL1_qSd5RiZtaGJASO1N8pb_nOnioczpcN_diJ2370eL1R9-XVs9HhJKkaIr3pYIJQDagM3TYOJqVAojoEL99Z7gxA3vbw2Ew0dyWSHQapRrmb6PFd8y_ReGCtlXz_urgSIKwuZ-faarSB_hPy1pxvAVl7OLfzpWZ7n_nwCZQTEce4MrVzXpxwCiUbnvTvOR2tTIVGIo1Ytbu3kaxOXY66KNiDMj0WOaU5AO1b3dfumcWPKSPQUOa-qimcvF7JhblMS8rGx4m7nHWULlBF_3mNGf5bBIc3J5q3P5FCwzJzMdkdOCxH-tsVwkNr86LXGA3ElhqrR4otWPoEh2P4YqqSxFytF5-xAYneQFvs0UqQuQmCzAblay938QAH31pQgkUddBQHvzirXCcGHpIIj1SjecQmEUXvKwrAMAJjHSvxlslZT5_7Hvf89YnqVp3UVcQhio9EAUEf_2i9VvJZZcxEiG23iIpMyDeiB2qrOpMOHKqF8zJ2sevlpxYcsPyQ52AxpQSHah5RV_5KBSzE-CZK3Q48-CoMV6AMZuq-zEB-B5aiD3bg6T2qlYKxENYqzEKEoGYaGDvb2VouAngKAfMK8_X0aMiVh0wvFqqf50AzYrZwKakzi7fOoG8Vckx6ejcMPN52Y5hpiyHS7fIAUCeezW-rsQ9nTMv9HevlKu2jlJQWPnpN_K8z-4UgykZDKxyPOptXNGiiYJcLiovrWw4_9ZQ6VIr06bhrvx8Wh5YP4qxy4Xk_5oke3MA7NgQE38rI97LcDkLiXNmgSWCDSvTa6E9RjccVGOIQbFZzlYLCkHVBg72BarQjpAwnP2y8Pj0WZcIZ_k9CcIHirCoWIATAhthN6-mDGkrdN4OeDnIe-mPwkJxMj5lyTAiznzmB-TtQKghmbHUwd_Pcygc5BQha3uh0RDSVWmoOMOT1oZj3fXpmAWovMGIZFVXiz5-pavtUX7WEOaD1DbE0fafPdE-o_uWWKkG3M3FlFzwcSq-0vNknjVjAIiZsFixT6VjMtB7PpCrM7m3Cme5oaK6T_nvxttMcKYuhnhVrFKlHLkd8ceJugKZL6RDoCFMFqHGf-I8OVUYcZOOdyQ4X1vWRZBCuXbmarkbCBRt-kgct2Vo=" } }, "id": "prt_0b2533fe7001A60dUAUYNTRJGs", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b25334d9001tiZMj8IUupKJA7" }, { "type": "tool", "tool": "StructuredOutput", "callID": "call_olgAc7cnr0TI8wpoii9QQYGj", "state": { "status": "completed", "input": { "status": "success", "reason": null, "hints": [ { "order": 1, "content": "Let $S(k)$ be the score of $k$ consecutive ones, with $S(0)=0$. A zero resets the counter, so an array’s score is the sum of $S(k)$ over its runs of ones. Also, $S(k+n)=S(k)+S(n)$." }, { "order": 2, "content": "Try an array with just one zero: $1^x0 1^y$. It wins exactly when $S(x)+S(y)>S(x+y+1)$. Why is checking one zero enough? If this inequality never holds, you can repeatedly merge two runs and their separating zero into one all-ones run without decreasing the score." }, { "order": 3, "content": "Separate the ordinary earnings from the bonuses: write $S(k)=kd+B(k)$, where $B(k)$ is the total bonus earned by $k$ consecutive ones. The winning condition becomes $B(x)+B(y)>B(x+y+1)+d$. The zero costs exactly $d$; everything else is about bonuses." }, { "order": 4, "content": "Suppose a pair $(x,y)$ wins. Shorten $x$ to the most recent position at which its run earned a bonus. Its own bonus total stays unchanged, while the all-ones benchmark cannot gain bonuses from becoming shorter. Then use periodicity to replace that position by some $p_i$. Do the same for $y$. Thus, only pairs $(p_i,p_j)$ matter." }, { "order": 5, "content": "Build the sorted bonus positions for three consecutive cycles: $p_i$, $n+p_i$, and $2n+p_i$, together with prefix sums of their rewards. For each fixed $i$, scan $j$ in increasing order and maintain a pointer counting positions at most $p_i+p_j+1$. If their total reward is $C$ and the original reward prefix sums are $R_i,R_j$, check $R_i+R_j>C+d$. The pointer only moves forward, giving $O(m^2)$ time." } ], "editorial": "The array can have any finite length, so trying to construct it directly is a trap. Instead, we’ll show that **if any array wins, an array with exactly one zero wins—and both runs can end at reward thresholds.** That leaves only $m^2$ candidates.\n\n### 1. Describe an all-ones run\n\nLet $S(k)$ be the score of $k$ consecutive ones, and define $S(0)=0$.\n\nThe counter resets after every $n$ ones, so\n\n$$S(k+n)=S(k)+S(n).$$\n\nLet $B(k)$ count only the bonus rewards earned during those $k$ ones. Then\n\n$$S(k)=kd+B(k).$$\n\nIf $R=\\sum_{i=1}^{m}r_i$, we have\n\n$$B(k)=\\left\\lfloor\\frac{k}{n}\\right\\rfloor R+\n\\sum_{p_i\\le k\\bmod n}r_i.$$\n\nIn particular, a reward at $p_i=n$ is collected before the counter resets. It belongs to the completed cycle.\n\nBecause rewards are positive, $B(k)$ is nondecreasing.\n\n### 2. One zero is enough\n\nConsider the array $1^x0 1^y$, where $1^x$ means $x$ consecutive ones. The zero separates two independent runs, so its advantage over all ones is\n\n$$G(x,y)=S(x)+S(y)-S(x+y+1).$$\n\nIf $G(x,y)>0$, we immediately have a valid answer.\n\nConversely, suppose $G(x,y)\\le 0$ for every pair of nonnegative integers $x,y$. Take any binary array and write it as runs of ones separated by zeros, allowing empty runs for leading, trailing, or consecutive zeros.\n\nWe can replace two neighboring runs and their separating zero by $x+y+1$ ones. This keeps the length unchanged and cannot decrease the score, because\n\n$$S(x)+S(y)\\le S(x+y+1).$$\n\nOther zeros still separate the remaining runs, so their scores are unaffected. Repeat until no zeros remain. The result is the all-ones array of the same length, with at least the original score.\n\nTherefore,\n\n$$\\boxed{\\text{Answer is YES iff some }G(x,y)>0.}$$\n\nNotice that if either run is empty, the pair cannot win: for example, $G(0,y)=S(y)-S(y+1)\\le 0$.\n\n### 3. Reduce infinitely many lengths to reward positions\n\nSubstituting $S(k)=kd+B(k)$ gives\n\n$$G(x,y)=B(x)+B(y)-B(x+y+1)-d.$$\n\nThe ordinary earnings mostly cancel. A break costs exactly $d$—the bonuses have to pay for it.\n\nSuppose $(x,y)$ is a winning pair. If $B(x)=0$, then monotonicity gives\n\n$$G(x,y)=B(y)-B(x+y+1)-d\\le 0,$$\n\nwhich is impossible. Thus, the first run has earned a bonus.\n\nLet $t\\le x$ be the last position where that run earned one. Such positions have the form\n\n$$t=qn+p_i.$$\n\nThere are no bonuses between $t$ and $x$, so $B(t)=B(x)$. Meanwhile,\n\n$$B(t+y+1)\\le B(x+y+1).$$\n\nConsequently,\n\n$$G(t,y)\\ge G(x,y)>0.$$\n\nShortening the run to its last reward does not hurt.\n\nNext, periodicity gives\n\n$$G(x+n,y)=G(x,y),$$\n\nsince adding $n$ to the first run adds the same $S(n)$ to both sides of the comparison. Hence we can replace $t=qn+p_i$ by $p_i$ without changing the advantage.\n\nApply the same argument to the second run. We obtain a winning pair $(p_i,p_j)$.\n\nSo the entire problem becomes checking\n\n$$\\boxed{B(p_i)+B(p_j)>B(p_i+p_j+1)+d}$$\n\nfor all $1\\le i,j\\le m$.\n\nIf $m=0$, there are no bonuses to exploit, and the answer is `NO`.\n\n### 4. Check every pair with two pointers\n\nDefine the original reward prefix sums\n\n$$R_0=0,\\qquad R_i=\\sum_{h=1}^{i}r_h.$$\n\nThen $B(p_i)=R_i$.\n\nTo evaluate the right-hand side efficiently, build the chronological list of reward events for three cycles:\n\n- positions $p_1,\\ldots,p_m$;\n- positions $n+p_1,\\ldots,n+p_m$;\n- positions $2n+p_1,\\ldots,2n+p_m$.\n\nEach cycle repeats the rewards $r_1,\\ldots,r_m$. The combined list is already sorted. Build its reward prefix sums $C$, with $C_0=0$.\n\nWhy three cycles? Our queried length is\n\n$$L=p_i+p_j+1\\le 2n+1\\le 3n.$$\n\nThree cycles cover every query, including the tiny edge case $n=1$.\n\nFor each fixed $i$:\n\n1. Start an event pointer $k=0$.\n2. Process $j=1,2,\\ldots,m$.\n3. Advance $k$ while the next event position is at most $L=p_i+p_j+1$.\n4. Now $B(L)=C_k$. Check whether\n $$R_i+R_j>C_k+d.$$\n5. If so, output `YES`.\n\nFor fixed $i$, $L$ increases with $j$, so the pointer never moves backward. If no pair wins, output `NO`.\n\n### Correctness argument\n\nIf the algorithm outputs `YES`, it found indices $i,j$ satisfying\n\n$$B(p_i)+B(p_j)>B(p_i+p_j+1)+d.$$\n\nEquivalently, $S(p_i)+S(p_j)>S(p_i+p_j+1)$, so the array $1^{p_i}0 1^{p_j}$ wins.\n\nIf any binary array wins, the run-merging argument guarantees some winning one-zero pair. Shortening each run to its last reward and removing complete cycles produces a winning pair $(p_i,p_j)$. The algorithm checks every such pair, and its event prefix sum computes the benchmark’s bonuses exactly. It must therefore output `YES`.\n\nThus the algorithm outputs `YES` exactly when a valid array exists.\n\n### Complexity and implementation details\n\nFor each of the $m$ choices of $i$, the inner loop visits $m$ candidates and the pointer crosses at most $3m$ events. The running time is $O(m^2)$, with $O(m)$ extra space. Since $\\sum m\\le 2000$, the total quadratic work is at most $2000^2$ up to constant factors.\n\nUse `long long` for event positions and reward sums. Use `<=` when including events at the queried length, because rewards are collected on that position. The final comparison must be strict: tying the all-ones score does not count.", "solution": "#include \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 int t;\n cin >> t;\n while (t--) {\n ll n, d;\n int m;\n cin >> n >> m >> d;\n\n vector p(m), r(m), pref(m + 1);\n for (int i = 0; i < m; ++i) {\n cin >> p[i] >> r[i];\n pref[i + 1] = pref[i] + r[i];\n }\n\n vector pos(3 * m), sum(3 * m + 1);\n for (int cycle = 0; cycle < 3; ++cycle) {\n for (int i = 0; i < m; ++i) {\n int k = cycle * m + i;\n pos[k] = cycle * n + p[i];\n sum[k + 1] = sum[k] + r[i];\n }\n }\n\n bool ok = false;\n for (int i = 0; i < m && !ok; ++i) {\n int k = 0;\n for (int j = 0; j < m; ++j) {\n ll len = p[i] + p[j] + 1;\n while (k < 3 * m && pos[k] <= len) ++k;\n if (pref[i + 1] + pref[j + 1] > sum[k] + d) {\n ok = true;\n break;\n }\n }\n }\n\n cout << (ok ? \"YES\\n\" : \"NO\\n\");\n }\n}\n" }, "output": "Structured output captured successfully.", "metadata": { "valid": true }, "title": "Structured Output", "time": { "start": 1789698275102, "end": 1789698275105 } }, "metadata": { "openai": { "itemId": "fc_003f88e5d85d5632016aaca098766487d089ab7e757c230304" } }, "id": "prt_0b25353e80010konRJMbrh4HjU", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b25334d9001tiZMj8IUupKJA7" }, { "reason": "tool-calls", "type": "step-finish", "tokens": { "total": 28167, "input": 1067, "output": 2548, "reasoning": 104, "cache": { "write": 0, "read": 24448 } }, "cost": 0, "id": "prt_0b254773a001lHDkAovfSsphXK", "sessionID": "ses_f4dad54b0ffeOMw11gc2kxPsfa", "messageID": "msg_0b25334d9001tiZMj8IUupKJA7" } ] } ] }