| [{"task_id": "mmlu-5254", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] An \"A+\" course grade is which kind of reinforcer?\nA) primary\nB) secondary\nC) continuous\nD) partial\n\nAnswer with the letter only.", "response": "B", "cost_usd": 7.5e-06, "tokens_in": 45, "tokens_out": 5, "latency_s": 1.151}, {"task_id": "mmlu-12856", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] h former therapy clients.\nA) Psychologists are prohibited from having sexual intimacies with former therapy clients under any circumstances.\nB) Psychologists are prohibited from having sexual intimacies with former therapy clients for at least one year following the termination of therapy.\nC) Psychologists are prohibited from having sexual intimacies with former therapy clients in certain circumstances (e.g., when there is a risk of exploitation).\nD) Psychologists are not prohibited from having sexual intimacies with former therapy clients.\n\nAnswer with the letter only.", "response": "C", "cost_usd": 0.0002063, "tokens_in": 113, "tokens_out": 325, "latency_s": 3.835}, {"task_id": "gsm8k-810", "model": "openai/gpt-5.6-luna", "prompt": "In one year, the number of students on campus doubles at the end of every month. If there are 10 students on campus at the beginning of the year, how many additional students would have joined by the end of May, above and beyond the number of students already on campus at the beginning of the year?\n\nReason step by step, then give the final answer as a single number.", "response": "Starting with 10 students:\n\n- End of January: 20\n- End of February: 40\n- End of March: 80\n- End of April: 160\n- End of May: 320\n\nAdditional students: \\(320 - 10 = 310\\)\n\n310", "cost_usd": 0.00011039999999999997, "tokens_in": 84, "tokens_out": 170, "latency_s": 2.902}, {"task_id": "gsm8k-277", "model": "openai/gpt-5.6-luna", "prompt": "It takes Matthew 3 minutes to dig a small hole for shrubs and 10 minutes to dig a large hole for trees. How many hours will it take him to dig 30 small holes and 15 large holes?\n\nReason step by step, then give the final answer as a single number.", "response": "30 small holes: \\(30 \\times 3 = 90\\) minutes \n15 large holes: \\(15 \\times 10 = 150\\) minutes \nTotal: \\(90 + 150 = 240\\) minutes \nConvert to hours: \\(240 \\div 60 = 4\\)\n\n4", "cost_usd": 0.0001002, "tokens_in": 66, "tokens_out": 156, "latency_s": 3.136}, {"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())\nfactorial = 1\nn = 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": 1.313}, {"task_id": "lcb-abc388_c", "model": "qwen/qwen3.7-flash", "prompt": "There are N mochi (rice cakes) arranged in ascending order of size.\r\nThe size of the i-th mochi (1 \\leq i \\leq N) is A_i.\nGiven two mochi A and B, with sizes a and b respectively, you can make one kagamimochi (a stacked rice cake) by placing mochi A on top of mochi B if and only if a is at most half of b.\nYou choose two mochi out of the N mochi, and place one on top of the other to form one kagamimochi.\nFind how many different kinds of kagamimochi can be made.\nTwo kagamimochi are distinguished if at least one of the mochi is different, even if the sizes of the mochi are the same.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\cdots A_N\n\nOutput\n\nPrint the number of different kinds of kagamimochi that can be made.\n\nConstraints\n\n\n- 2 \\leq N \\leq 5 \\times 10^5\n- 1 \\leq A_i \\leq 10^9 \\ (1 \\leq i \\leq N)\n- A_i \\leq A_{i+1} \\ (1 \\leq i < N)\n- All input values are integers.\n\nSample Input 1\n\n6\r\n2 3 4 4 7 10\n\nSample Output 1\n\n8\r\n\nThe sizes of the given mochi are as follows:\n\nIn this case, you can make the following eight kinds of kagamimochi:\n\nNote that there are two kinds of kagamimochi where a mochi of size 4 is topped by a mochi of size 2, and two kinds where a mochi of size 10 is topped by a mochi of size 4.\n\nSample Input 2\n\n3\r\n387 388 389\n\nSample Output 2\n\n0\r\n\nIt is possible that you cannot make any kagamimochi.\n\nSample Input 3\n\n32\r\n1 2 4 5 8 10 12 16 19 25 33 40 50 64 87 101 149 175 202 211 278 314 355 405 412 420 442 481 512 582 600 641\n\nSample Output 3\n\n388\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": "```python\nimport sys\n\ndef solve():\n # Read all input from standard input efficiently\n # split() splits by any whitespace including newlines\n input_data = sys.stdin.read().split()\n \n if not input_data:\n return\n\n iterator = iter(input_data)\n \n try:\n n = int(next(iterator))\n except StopIteration:\n return\n\n # Convert the rest of the tokens to integers\n # This creates a list of integers for the mochi sizes\n a = list(map(int, iterator))\n \n # Defensive check: if input has fewer elements than N, adjust N\n if len(a) < n:\n n = len(a)\n \n count = 0\n ptr = 0\n \n # Iterate through each mochi considering it as the bottom mochi.\n # The array 'a' is sorted in non-decreasing order.\n # For a bottom mochi with size 'bottom_val', we want to count how many\n # mochi with size 'top_val' satisfy 2 * top_val <= bottom_val.\n # Since 'a' is sorted, the valid top mochi will always be a prefix of the array.\n # We maintain a pointer 'ptr' which indicates the number of valid top mochi found so far.\n # As we iterate through larger bottom mochi (increasing 'bottom_val'), the threshold\n # increases, so the set of valid top mochi can only grow. Thus, 'ptr' only moves forward.\n \n for j in range(n):\n bottom_val = a[j]\n \n # Advance ptr while the condition 2 * a[ptr] <= bottom_val holds.\n # ptr represents the count of elements a[0]...a[ptr-1] that satisfy the condition.\n # We check a[ptr] (the next candidate) against the current bottom_val.\n while ptr < n and 2 * a[ptr] <= bottom_val:\n ptr += 1\n \n # Add the number of valid tops for the current bottom mochi to the total count.\n count += ptr\n \n print(count)\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.00223814, "tokens_in": 626, "tokens_out": 17072, "latency_s": 124.837}] |