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+[{"task_id": "mmlu-4724", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] A student decides that, having already spent three hours studying for an exam, she should spend one more hour studying for the same exam. Which of the following is most likely true?\nA) The marginal benefit of the fourth hour is certainly less than the marginal cost of the fourth hour.\nB) The marginal benefit of the fourth hour is at least as great as the marginal cost of the fourth hour.\nC) Without knowing the student\u2019s opportunity cost of studying, we have no way of knowing whether or not her marginal benefits outweigh her marginal costs.\nD) The marginal cost of the third hour was likely greater than the marginal cost of the fourth hour.\n\nAnswer with the letter only.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "Answer: B", "cost_usd": 0.00012502, "tokens_in": 257, "tokens_out": 318, "latency_s": 4.161}, {"task_id": "mmlu-9164", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] For which of these two scenarios does the main character (who uses I/me/my) do something clearly morally wrong, according to ordinary moral standards in the US as of 2020? Scenario 1 | I added a red sock to my roommates white laundry. Scenario 2 | I asked the homeless shelter for food even though I make plenty of money.\nA) Wrong, Wrong\nB) Wrong, Not wrong\nC) Not wrong, Wrong\nD) Not wrong, Not wrong\n\nAnswer with the letter only.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "A", "cost_usd": 4.104e-05, "tokens_in": 148, "tokens_out": 154, "latency_s": 3.279}, {"task_id": "gsm8k-269", "model": "openai/gpt-5.6-luna", "prompt": "A juggler can juggle 16 balls.  Half of the balls are golf balls, and half of the golf balls are blue.  How many blue golf balls are there?\n\nReason step by step, then give the final answer as a single number.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "There are 16 balls total.\n\nHalf are golf balls: 16 \u00f7 2 = 8.\n\nHalf of the golf balls are blue: 8 \u00f7 2 = 4.\n\nAnswer: 4", "cost_usd": 0.0001228, "tokens_in": 92, "tokens_out": 87, "latency_s": 2.274}, {"task_id": "gsm8k-680", "model": "openai/gpt-5.6-luna", "prompt": "The elevator in Jack's building supports a maximum load of 700 kg. An adult weighs an average of 80 kg. If Jack rides the elevator with 8 other adults, by how much will they have exceeded the maximum load of the elevator?\n\nReason step by step, then give the final answer as a single number.\n\nEnd with one final line of exactly the form 'Answer: X' and nothing afterwards \u2014 X is the option letter when options are listed, otherwise the final value.", "response": "There are 9 adults total: Jack plus 8 others.\n\nTheir total weight is:\n9 \u00d7 80 kg = 720 kg\n\nThe excess load is:\n720 kg \u2212 700 kg = 20 kg\n\nAnswer: 20 kg", "cost_usd": 0.0001588, "tokens_in": 104, "tokens_out": 115, "latency_s": 1.499}, {"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.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nTake the largest stated constraints and derive the required complexity before writing code. Long inputs demand a near-linear pass without repeated slicing or rescanning; tiny object counts may instead allow exhaustive enumeration over canonical states.", "response": "import sys\n\nx = int(sys.stdin.buffer.read())\nn = 1\nfactorial = 1\n\nwhile factorial < x:\n    n += 1\n    factorial *= n\n\nsys.stdout.write(str(n) + \"\\n\")", "cost_usd": 0.00037620000000000004, "tokens_in": 405, "tokens_out": 246, "latency_s": 3.316}, {"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.\n\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nTake the largest stated constraints and derive the required complexity before writing code. Long inputs demand a near-linear pass without repeated slicing or rescanning; tiny object counts may instead allow exhaustive enumeration over canonical states.\n\nNow build test tooling for that solution and output nothing else. First a REFERENCE program: obviously correct, allowed to be very slow \u2014 simulate directly or enumerate every possibility, and deliberately avoid the algorithmic idea used above so the two can disagree. It reads the same stdin format and prints the same output format; handling small inputs only is fine. Second a GENERATOR program: it takes two command line integers, seed and size, calls random.seed(seed), and prints one random valid input in exactly the statement's input format, size being a rough magnitude clamped to the constraints. At size 2 print the smallest legal input. At maximum size, every seed must produce genuinely independent high-entropy values, while smaller sizes should also cover repeats, range extremes and adversarial orderings. Never draw most maximum-size values from a tiny pool. Also write a second BOUNDARY GENERATOR with a different construction strategy. It has the same seed and size interface, but concentrates on valid maximum-size, high-diversity and worst-case-shaped inputs rather than ordinary random cases. Respond with exactly three fenced blocks:\n```reference\n<program>\n```\n```generator\n<program>\n```\n```boundary_generator\n<program>\n```", "response": "```reference\nimport sys\nimport math\n\nx = int(sys.stdin.buffer.read())\nfor n in range(1, 100):\n    if math.factorial(n) == x:\n        sys.stdout.write(str(n))\n        break\n```\n\n```generator\nimport sys\nimport random\nimport math\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size <= 2:\n    n = 2\nelse:\n    limit = min(max(size, 2), 20)\n    r = random.random()\n    if r < 0.1:\n        n = 2\n    elif r < 0.2:\n        n = limit\n    else:\n        n = random.randint(2, limit)\n\nsys.stdout.write(str(math.factorial(n)) + \"\\n\")\n```\n\n```boundary_generator\nimport sys\nimport random\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size <= 2:\n    n = 2\nelse:\n    limit = min(max(size, 2), 20)\n    bits = random.getrandbits(256)\n    folded = bits ^ (bits >> 37) ^ (bits >> 91) ^ (bits >> 173)\n    if folded % 11 == 0:\n        n = 2\n    elif folded % 11 == 1:\n        n = limit\n    else:\n        n = 2 + (folded % (limit - 1))\n\nvalue = 1\nfor k in range(2, n + 1):\n    value *= k\n\nsys.stdout.write(str(value) + \"\\n\")\n```", "cost_usd": 0.0017932, "tokens_in": 650, "tokens_out": 1386, "latency_s": 15.392}, {"task_id": "lcb-abc400_c", "model": "openai/gpt-5.6-luna", "prompt": "A positive integer X is called a good integer if and only if it satisfies the following condition:\n\n- There exists a pair of positive integers (a,b) such that X = 2^a \\times b^2.\n\nFor example, 400 is a good integer because 400 = 2^2 \\times 10^2.\nGiven a positive integer N, find the number of good integers between 1 and N, inclusive.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the number of good integers between 1 and N, inclusive.\n\nConstraints\n\n\n- 1 \\leq N \\leq 10^{18}\n- N is an integer.\n\nSample Input 1\n\n20\n\nSample Output 1\n\n5\r\n\nThere are five good integers between 1 and 20: 2, 4, 8, 16, and 18.\r\nThus, print 5.\n\nSample Input 2\n\n400\n\nSample Output 2\n\n24\n\nSample Input 3\n\n1234567890\n\nSample Output 3\n\n42413\r\n\nNote that the input might not fit in a 32-bit integer type.\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\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nTake the largest stated constraints and derive the required complexity before writing code. Long inputs demand a near-linear pass without repeated slicing or rescanning; tiny object counts may instead allow exhaustive enumeration over canonical states.", "response": "import sys\nimport math\n\ndata = sys.stdin.buffer.read().split()\nn = int(data[0])\n\nanswer = 0\npower = 2\nwhile power <= n:\n    limit = math.isqrt(n // power)\n    answer += (limit + 1) // 2\n    power <<= 1\n\nsys.stdout.write(str(answer))", "cost_usd": 0.0008059999999999999, "tokens_in": 466, "tokens_out": 594, "latency_s": 6.656}, {"task_id": "lcb-abc400_c", "model": "openai/gpt-5.6-luna", "prompt": "A positive integer X is called a good integer if and only if it satisfies the following condition:\n\n- There exists a pair of positive integers (a,b) such that X = 2^a \\times b^2.\n\nFor example, 400 is a good integer because 400 = 2^2 \\times 10^2.\nGiven a positive integer N, find the number of good integers between 1 and N, inclusive.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the number of good integers between 1 and N, inclusive.\n\nConstraints\n\n\n- 1 \\leq N \\leq 10^{18}\n- N is an integer.\n\nSample Input 1\n\n20\n\nSample Output 1\n\n5\r\n\nThere are five good integers between 1 and 20: 2, 4, 8, 16, and 18.\r\nThus, print 5.\n\nSample Input 2\n\n400\n\nSample Output 2\n\n24\n\nSample Input 3\n\n1234567890\n\nSample Output 3\n\n42413\r\n\nNote that the input might not fit in a 32-bit integer type.\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\nThe checker compares stdout token by token; anything but an exact match scores zero. Mirror the published sample outputs literally \u2014 identical lines, and identical decimal digit counts wherever the samples print decimals.\n\nUse Python 3. Read the whole of stdin with one buffered read before parsing; assemble the entire output first and emit it with a single write. Prefer exact integer or Fraction arithmetic over floats for counting and comparison. Reply with the program alone: no commentary, no code fences.\n\nBefore finishing: when a loop compares or swaps two picked items, store them in fresh local names instead of rebinding the loop variables, or later iterations read corrupted state. Duplicated elements count once per occurrence \u2014 never deduplicate through a set unless the statement says distinct.\n\nTake the largest stated constraints and derive the required complexity before writing code. Long inputs demand a near-linear pass without repeated slicing or rescanning; tiny object counts may instead allow exhaustive enumeration over canonical states.\n\nNow build test tooling for that solution and output nothing else. First a REFERENCE program: obviously correct, allowed to be very slow \u2014 simulate directly or enumerate every possibility, and deliberately avoid the algorithmic idea used above so the two can disagree. It reads the same stdin format and prints the same output format; handling small inputs only is fine. Second a GENERATOR program: it takes two command line integers, seed and size, calls random.seed(seed), and prints one random valid input in exactly the statement's input format, size being a rough magnitude clamped to the constraints. At size 2 print the smallest legal input. At maximum size, every seed must produce genuinely independent high-entropy values, while smaller sizes should also cover repeats, range extremes and adversarial orderings. Never draw most maximum-size values from a tiny pool. Also write a second BOUNDARY GENERATOR with a different construction strategy. It has the same seed and size interface, but concentrates on valid maximum-size, high-diversity and worst-case-shaped inputs rather than ordinary random cases. Respond with exactly three fenced blocks:\n```reference\n<program>\n```\n```generator\n<program>\n```\n```boundary_generator\n<program>\n```", "response": "```reference\nimport sys\nimport math\n\ndata = sys.stdin.buffer.read()\nn = int(data)\n\nanswer = 0\nfor x in range(2, n + 1):\n    y = x\n    while y % 2 == 0:\n        y //= 2\n    r = math.isqrt(y)\n    if r * r == y:\n        answer += 1\n\nsys.stdout.write(str(answer) + \"\\n\")\n```\n\n```generator\nimport sys\nimport random\nimport math\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nMAX_N = 10**18\n\nif size == 2:\n    n = 1\nelse:\n    limit = max(1, min(MAX_N, size))\n    if limit == MAX_N:\n        n = random.randrange(1, MAX_N + 1)\n    elif limit <= 10:\n        n = random.randrange(1, limit + 1)\n    else:\n        mode = random.randrange(8)\n        if mode == 0:\n            n = 1\n        elif mode == 1:\n            n = limit\n        elif mode == 2:\n            n = max(1, limit - 1)\n        elif mode == 3:\n            p = 1 << random.randrange(limit.bit_length())\n            n = min(limit, p)\n        elif mode == 4:\n            p = 1 << random.randrange(limit.bit_length())\n            n = min(limit, max(1, p + random.randrange(-3, 4)))\n        elif mode == 5:\n            c = random.randrange(1, math.isqrt(limit) + 1)\n            n = c * c\n        elif mode == 6:\n            k = random.randrange(1, max(2, limit.bit_length()))\n            power = 1 << min(k, limit.bit_length() - 1)\n            cmax = math.isqrt(limit // power)\n            if cmax:\n                c = random.randrange(1, cmax + 1)\n                n = power * c * c\n            else:\n                n = random.randrange(1, limit + 1)\n        else:\n            n = random.randrange(1, limit + 1)\n\nsys.stdout.write(str(n) + \"\\n\")\n```\n\n```boundary_generator\nimport sys\nimport random\nimport math\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nMAX_N = 10**18\n\nif size == 2:\n    n = 1\nelse:\n    limit = max(1, min(MAX_N, size))\n    if limit == MAX_N:\n        k = random.randrange(0, 31)\n        cap = math.isqrt(MAX_N >> k)\n        low = max(1, cap // 2)\n        c = random.randrange(low, cap + 1)\n        n = (1 << k) * c * c\n    elif limit <= 16:\n        n = random.choice([1, limit, max(1, limit - 1)])\n    else:\n        candidates = [1, limit, max(1, limit - 1)]\n        bit = 1\n        while bit <= limit:\n            candidates.append(bit)\n            if bit > 1:\n                candidates.append(max(1, bit - 1))\n                candidates.append(min(limit, bit + 1))\n            bit <<= 1\n\n        root = math.isqrt(limit)\n        for c in [1, max(1, root - 1), root]:\n            if c * c <= limit:\n                candidates.append(c * c)\n\n        k = random.randrange(0, max(1, limit.bit_length()))\n        power = 1 << min(k, limit.bit_length() - 1)\n        cmax = math.isqrt(limit // power)\n        if cmax >= 1:\n            c = random.choice([1, cmax, max(1, cmax - 1)])\n            candidates.append(power * c * c)\n\n        n = random.choice([x for x in candidates if 1 <= x <= limit])\n\nsys.stdout.write(str(n) + \"\\n\")\n```", "cost_usd": 0.0033486000000000006, "tokens_in": 711, "tokens_out": 2672, "latency_s": 24.662}]