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+[{"task_id": "mmlu-8081", "model": "qwen/qwen3.7-flash", "prompt": "[MMLU] In the original 'Get Smart' TV series what is agent Maxwell Smart's code name?\nA) Agent 86\nB) Agent 99\nC) Larrabee\nD) The Chief\n\nAnswer with the letter only.", "response": "A", "cost_usd": 3.143e-05, "tokens_in": 64, "tokens_out": 227, "latency_s": 3.207}, {"task_id": "mmlu-992", "model": "qwen/qwen3.7-flash", "prompt": "[MMLU] Of the following ions, which has the smallest radius?\nA) K+\nB) Ca2+\nC) Sc3+\nD) Rb+\n\nAnswer with the letter only.", "response": "C", "cost_usd": 9.564000000000001e-05, "tokens_in": 55, "tokens_out": 723, "latency_s": 5.623}, {"task_id": "gsm8k-331", "model": "deepseek/deepseek-v4-flash", "prompt": "Nick is choosing between two jobs. Job A pays $15 an hour for 2000 hours a year, and is in a state with a 20% total tax rate. Job B pays $42,000 a year and is in a state that charges $6,000 in property tax and a 10% tax rate on net income after property tax. How much more money will Nick make at the job with a higher net pay rate, compared to the other job?\n\nReason step by step, then give the final answer as a single number.", "response": "Job A: Gross pay = $15/hour \u00d7 2000 hours = $30,000. After 20% tax, net = $30,000 \u00d7 0.80 = $24,000.\n\nJob B: Gross pay = $42,000. Property tax = $6,000. Net after property tax = $42,000 - $6,000 = $36,000. Income tax at 10% = $36,000 \u00d7 0.10 = $3,600. Total net = $42,000 - $6,000 - $3,600 = $32,400.\n\nDifference = $32,400 - $24,000 = $8,400.\n\n\\boxed{8400}", "cost_usd": 0.00027342, "tokens_in": 115, "tokens_out": 919, "latency_s": 21.522}, {"task_id": "gsm8k-722", "model": "deepseek/deepseek-v4-flash", "prompt": "Ron is fed up with the pothole in front of his house. If it doesn't get fixed, it's going to do $450 worth of damage to his car. Unfortunately, the city council refuses to fix it, and will fine Ron $120 for unauthorized road maintenance if he fixes it himself. Ron will also have to buy 3 buckets of asphalt that each cost $25. How much money does Ron save by fixing the pothole?\n\nReason step by step, then give the final answer as a single number.", "response": "Ron saves $255 by fixing the pothole himself.\n\nThe damage from not fixing it is $450. The cost to fix it includes a $120 fine and $75 for three buckets of asphalt, totaling $195. The savings are $450 - $195 = $255.\n\n\\boxed{255}", "cost_usd": 3.930219999999997e-05, "tokens_in": 111, "tokens_out": 154, "latency_s": 3.241}, {"task_id": "lcb-abc389_b", "model": "openai/gpt-5.6-luna", "prompt": "You are given an integer X not less than 2.\nFind the positive integer N such that N! = X.\nHere, N! denotes the factorial of N, and it is guaranteed that there is exactly one such N.\n\nInput\n\nThe input is given from Standard Input in the following format:\nX\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 2 \\leq X \\leq 3 \\times 10^{18}\n- There is exactly one positive integer N such that N!=X.\n- All input values are integers.\n\nSample Input 1\n\n6\n\nSample Output 1\n\n3\r\n\nFrom 3!=3\\times2\\times1=6, print 3.\n\nSample Input 2\n\n2432902008176640000\n\nSample Output 2\n\n20\r\n\nFrom 20!=2432902008176640000, print 20.\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.", "response": "import sys\n\nx = int(sys.stdin.readline())\nn = 1\nfactorial = 1\n\nwhile factorial < x:\n    n += 1\n    factorial *= n\n\nprint(n)", "cost_usd": 4.8099999999999977e-05, "tokens_in": 217, "tokens_out": 44, "latency_s": 1.134}, {"task_id": "lcb-abc390_d", "model": "openai/gpt-5.6-luna", "prompt": "There are N bags, labeled bag 1, bag 2, \\ldots, bag N.\r\nBag i (1 \\leq i \\leq N) contains A_i stones.\nTakahashi can perform the following operation any number of times, possibly zero:\n\nChoose two bags A and B, and move all stones from bag A into bag B.\n\nFind the number of different possible values for the following after repeating the operation.\n\n- B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N, where B_i is the final number of stones in bag i.\r\nHere, \\oplus denotes bitwise XOR.\n\n\nAbout bitwise XOR\r\nFor non-negative integers a and b, the bitwise XOR a \\oplus b is defined as follows:\r\n\r\n\r\nIn the binary representation of a \\oplus b, the digit in the 2^k place (k \\ge 0) is 1 if and only if exactly one of the digits in the 2^k place of a and b is 1; otherwise, it is 0.\r\n\r\n\r\nFor example, 3 \\oplus 5 = 6 (in binary, 011 \\oplus 101 = 110).\r\nIn general, for k non-negative integers x_1, x_2, \\ldots, x_k, their bitwise XOR x_1 \\oplus x_2 \\oplus \\cdots \\oplus x_k is defined as (\\cdots((x_1 \\oplus x_2) \\oplus x_3) \\oplus \\cdots) \\oplus x_k, which does not depend on the order of x_1, x_2, \\ldots, x_k.\r\n\nIt can be proved that under the constraints of this problem, the number of possible values is finite.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\ldots A_N\n\nOutput\n\nPrint the number of different possible values for B_1 \\oplus B_2 \\oplus \\cdots \\oplus B_N after repeating the operation.\n\nConstraints\n\n\n- 2 \\leq N \\leq 12\n- 1 \\leq A_i \\leq 10^{17}\n- All input values are integers.\n\nSample Input 1\n\n3\r\n2 5 7\n\nSample Output 1\n\n3\r\n\nFor example, if Takahashi chooses bags 1 and 3 for the operation, then the numbers of stones in bags 1, 2, 3 become 0, 5, 9.\r\nIf he stops at this point, the XOR is 0 \\oplus 5 \\oplus 9 = 12.\nThe other possible XOR values after repeating the operation are 0 and 14.\r\nTherefore, the possible values are 0, 12, 14; there are three values, so the output is 3.\n\nSample Input 2\n\n2\r\n100000000000000000 100000000000000000\n\nSample Output 2\n\n2\n\nSample Input 3\n\n6\r\n71 74 45 34 31 60\n\nSample Output 3\n\n84\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.", "response": "import sys\n\nsys.setrecursionlimit(100000)\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nN = data[0]\nA = data[1:]\n\nvalues = set()\nblocks = [0] * N\n\ndef dfs(i, block_count, current_xor):\n    if i == N:\n        values.add(current_xor)\n        return\n\n    a = A[i]\n\n    for j in range(block_count):\n        old = blocks[j]\n        new = old + a\n        blocks[j] = new\n        dfs(i + 1, block_count, current_xor ^ old ^ new)\n        blocks[j] = old\n\n    blocks[block_count] = a\n    dfs(i + 1, block_count + 1, current_xor ^ a)\n    blocks[block_count] = 0\n\ndfs(0, 0, 0)\nprint(len(values))", "cost_usd": 0.0009968, "tokens_in": 704, "tokens_out": 1544, "latency_s": 19.865}]