Upload traces/88715.json with huggingface_hub

traces/88715.json

@@ -0,0 +1 @@
+[{"task_id": "mmlu-189", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] Which of the following is a structural, fibrous protein found in the dermis?\nA) Collagen\nB) Heparin\nC) Lipocyte\nD) Melanin\n\nAnswer with the letter only.\n\nThink step by step about each option, then end with exactly: ANSWER: <letter>", "response": "ANSWER: A", "cost_usd": 0.0001154, "tokens_in": 73, "tokens_out": 84, "latency_s": 3.922}, {"task_id": "mmlu-8170", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU]  According to Mill, the value of a particular pleasure depends on\nA) its quantity alone.\nB) its quality alone.\nC) both its quantity and its quality.\nD) neither its quantity nor its quality.\n\nAnswer with the letter only.\n\nThink step by step about each option, then end with exactly: ANSWER: <letter>", "response": "ANSWER: C", "cost_usd": 0.0001226, "tokens_in": 79, "tokens_out": 89, "latency_s": 3.078}, {"task_id": "gsm8k-789", "model": "openai/gpt-5.6-luna", "prompt": "Carson lives in an old house where the pipes will freeze if the temperature inside the house falls below 32 degrees. Every log of wood Carson burns heats the house up by 5 degrees. If it was 45 degrees during the day, and it's 33 degrees colder during the night compared to the day, how many logs does Carson need to burn to prevent the pipes from freezing?\n\nReason step by step, then give the final answer as a single number.\n\nReason step by step, then end with the final answer as a single number.\n\n(ref 262777)", "response": "1. Night temperature: \\(45 - 33 = 12\\) degrees.\n2. Temperature needed: \\(32 - 12 = 20\\) degrees.\n3. Each log raises the temperature by 5 degrees: \\(20 \\div 5 = 4\\) logs.\n\n4", "cost_usd": 0.00017999999999999998, "tokens_in": 120, "tokens_out": 130, "latency_s": 3.242}, {"task_id": "gsm8k-535", "model": "openai/gpt-5.6-luna", "prompt": "Nick had twice as many candies as George. Then George ate 5 candies. Now George has 3 candies left. How many candies does Nick have?\n\nReason step by step, then give the final answer as a single number.\n\nReason step by step, then end with the final answer as a single number.\n\n(ref 723192)", "response": "George initially had \\(3 + 5 = 8\\) candies.\n\nNick had twice as many: \\(2 \\times 8 = 16\\).\n\n16", "cost_usd": 0.00013219999999999996, "tokens_in": 73, "tokens_out": 98, "latency_s": 1.465}, {"task_id": "lcb-abc392_a", "model": "moonshotai/kimi-k3", "prompt": "You are given a sequence of integers A = (A_1, A_2, A_3).\r\nLet B = (B_1, B_2, B_3) be any permutation of A.\r\nDetermine whether it is possible that B_1 \\times B_2 = B_3.\n\nInput\n\nThe input is given from Standard Input in the following format:\nA_1 A_2 A_3\n\nOutput\n\nIf it is possible that B_1 \\times B_2 = B_3, print Yes; otherwise, print No.\n\nConstraints\n\n\n- All input values are integers.\n- 1 \\le A_1, A_2, A_3 \\le 100\n\nSample Input 1\n\n3 15 5\n\nSample Output 1\n\nYes\r\n\nHere, A=(3,15,5).\r\nBy rearranging it as B=(3,5,15), we can satisfy B_1 \\times B_2 = B_3.\n\nSample Input 2\n\n5 3 2\n\nSample Output 2\n\nNo\r\n\nNo permutation of B satisfies B_1 \\times B_2 = B_3.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nSolve this competitive-programming problem in Python 3. Read all of stdin at once with sys.stdin.buffer.read().split() and index across tokens; build output in a list and emit it with one write at the end. Check the constraint block and pick an algorithm whose running time stays under ~1e8 steps at the largest allowed input. Return ONLY raw complete Python 3 source \u2014 no Markdown fences, no prose, no comments.\n\nWhen enumerating partitions into UNLABELED groups, generate each partition once: process items in order and place the next item into each existing group or one new group. Do not recursively choose arbitrary subsets as blocks \u2014 permuting block order repeats the same partition and can turn a feasible search into a timeout. Start from the first item, update aggregate values incrementally, put final values straight into a set, and skip existing groups with equal state: those branches are symmetric.\n\nOUTPUT FORMAT: this judge compares printed output EXACTLY, token for token, even when the statement promises a numeric tolerance, and the hidden tests may not use the same precision as the statement's worked examples. Apply BOTH: (1) if an input exactly matches one of the statement's worked examples, print that example's expected output byte-for-byte exactly as the statement shows it; (2) for every other input print floats at a fixed width with format(x, '.12f'). Never pin an input that is not a worked example; compute those normally.", "response": "import sys\n\ndef main():\n    data = sys.stdin.buffer.read().split()\n    a, b, c = int(data[0]), int(data[1]), int(data[2])\n    ok = (a * b == c) or (a * c == b) or (b * c == a)\n    out = [\"Yes\" if ok else \"No\"]\n    sys.stdout.write(\"\\n\".join(out) + \"\\n\")\n\nmain()", "cost_usd": 0.004251, "tokens_in": 690, "tokens_out": 189, "latency_s": 8.557}, {"task_id": "lcb-abc399_d", "model": "moonshotai/kimi-k3", "prompt": "N couples are seated in a line.\nCount the number of pairs of couples such that neither couple was originally sitting next to each other, and both couples can end up sitting next to each other by swapping seats among those four people.\n\nThere is a sequence A = (A_1, A_2, \\dots, A_{2N}) of length 2N. Each of the integers 1, 2, \\dots, N appears exactly twice in A.\nFind the number of integer pairs (a, b) satisfying 1 \\leq a < b \\leq N and all of the following conditions:\n\n- The two occurrences of a in A are not adjacent.\n- The two occurrences of b in A are not adjacent.\n- By performing the following operation one or more times in any order, it is possible to reach a state where the two occurrences of a in A are adjacent and the two occurrences of b in A are also adjacent.\n- Choose an integer pair (i, j) (1 \\leq i \\leq 2N, 1 \\leq j \\leq 2N) such that A_i = a and A_j = b, and swap A_i with A_j.\n\n\n\nYou are given T test cases; solve each of them.\n\nInput\n\nThe input is given from Standard Input in the following format, where \\mathrm{case}_i denotes the i-th test case:\nT\n\\mathrm{case}_1\n\\mathrm{case}_2\n\\vdots\n\\mathrm{case}_T\n\nEach test case is given in the following format:\nN\nA_1 A_2 \\dots A_{2N}\n\nOutput\n\nPrint T lines. The i-th line should contain the answer for the i-th test case.\n\nConstraints\n\n\n- 1 \\leq T \\leq 2 \\times 10^5\n- 1 \\leq N \\leq 2 \\times 10^5\n- 1 \\leq A_i \\leq N\n- Each of 1, 2, \\dots, N appears exactly twice in A.\n- The sum of N over all test cases is at most 2 \\times 10^5.\n- All input values are integers.\n\nSample Input 1\n\n3\n3\n1 2 3 3 1 2\n4\n1 1 2 2 3 3 4 4\n5\n1 2 3 4 5 1 2 3 4 5\n\nSample Output 1\n\n1\n0\n4\n\nConsider the first test case.\n(a, b) = (1, 2) satisfies the conditions in the problem statement, for the following reasons:\n\n- The two occurrences of 1 in A are not adjacent.\n- The two occurrences of 2 in A are not adjacent.\n- By performing the operation where (i, j) = (1, 6) and swapping A_1 with A_6, you can reach a state where the two occurrences of 1 are adjacent and the two occurrences of 2 are also adjacent.\n\n(1, 2) is the only pair (a, b) that satisfies the conditions.\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.\n\nSolve this competitive-programming problem in Python 3. Read all of stdin at once with sys.stdin.buffer.read().split() and index across tokens; build output in a list and emit it with one write at the end. Check the constraint block and pick an algorithm whose running time stays under ~1e8 steps at the largest allowed input. Return ONLY raw complete Python 3 source \u2014 no Markdown fences, no prose, no comments.\n\nWhen enumerating partitions into UNLABELED groups, generate each partition once: process items in order and place the next item into each existing group or one new group. Do not recursively choose arbitrary subsets as blocks \u2014 permuting block order repeats the same partition and can turn a feasible search into a timeout. Start from the first item, update aggregate values incrementally, put final values straight into a set, and skip existing groups with equal state: those branches are symmetric.\n\nOUTPUT FORMAT: this judge compares printed output EXACTLY, token for token, even when the statement promises a numeric tolerance, and the hidden tests may not use the same precision as the statement's worked examples. Apply BOTH: (1) if an input exactly matches one of the statement's worked examples, print that example's expected output byte-for-byte exactly as the statement shows it; (2) for every other input print floats at a fixed width with format(x, '.12f'). Never pin an input that is not a worked example; compute those normally.", "response": "", "cost_usd": 0.2323664, "tokens_in": 1068, "tokens_out": 16384, "latency_s": 486.671}, {"task_id": "lcb-abc399_d", "model": "openai/gpt-5.6-luna", "prompt": "That fails a worked example from the statement. Input:\n3\n3\n1 2 3 3 1 2\n4\n1 1 2 2 3 3 4 4\n5\n1 2 3 4 5 1 2 3 4 5\nExpected:\n1\n0\n4\nYour output:\n<empty>\nFix the GENERAL logic \u2014 do not special-case this input. Return ONLY raw complete Python 3 source \u2014 no Markdown fences, no prose, no comments.", "response": "import sys\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nt = data[0]\nidx = 1\nout = []\n\nfor _ in range(t):\n    n = data[idx]\n    idx += 1\n    a = data[idx:idx + 2 * n]\n    idx += 2 * n\n\n    edges = {}\n    for i in range(2 * n - 1):\n        x = a[i]\n        y = a[i + 1]\n        if x == y:\n            continue\n        if x < y:\n            key = (x, y)\n            orient = 0\n        else:\n            key = (y, x)\n            orient = 1\n        edges.setdefault(key, []).append((i, orient))\n\n    ans = 0\n    for items in edges.values():\n        m = len(items)\n        found = False\n        for i in range(m):\n            p, o = items[i]\n            for j in range(i + 1, m):\n                q, r = items[j]\n                if o == r and abs(p - q) > 1:\n                    found = True\n                    break\n            if found:\n                break\n        if found:\n            ans += 1\n\n    out.append(str(ans))\n\nsys.stdout.write(\"\\n\".join(out))", "cost_usd": 0.0021947999999999968, "tokens_in": 1119, "tokens_out": 1596, "latency_s": 18.019}]