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+[{"task_id": "mmlu-8406", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU]  Of the following social problems that could result from a genetic supermarket, which does Singer think is the least serious?\nA) loss of human diversity\nB) the effects of engineering children to have positional goods\nC) loss of equality of opportunity\nD) increased disparity between the rich and the poor\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: A", "cost_usd": 0.0001444212, "tokens_in": 108, "tokens_out": 816, "latency_s": 82.639}, {"task_id": "mmlu-10423", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] If a warrant carries a right to buy one share of common stock and is exercisable at $20 per common share while the market price of a share is $30, the theoretical value of the warrant is:\nA) $20.00\nB) $10.00\nC) $5.00\nD) $0.00\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": 5.6979999999999986e-05, "tokens_in": 123, "tokens_out": 142, "latency_s": 2.153}, {"task_id": "gsm8k-878", "model": "openai/gpt-5.6-luna", "prompt": "Julie, Micah, and Mitchell sold 32 glasses of lemonade at their lemonade stand. Julie sold 14 glasses and the boys sold an equal number of glasses. How many more glasses did Julie sell than Micah?\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": "Julie sold 14 glasses.\n\nThe boys sold 32 \u2212 14 = 18 glasses altogether.\n\nSince Micah and Mitchell sold equal amounts, Micah sold 18 \u00f7 2 = 9 glasses.\n\nJulie sold 14 \u2212 9 = 5 more glasses than Micah.\n\nAnswer: 5", "cost_usd": 0.00014800000000000002, "tokens_in": 98, "tokens_out": 107, "latency_s": 3.298}, {"task_id": "gsm8k-244", "model": "openai/gpt-5.6-luna", "prompt": "Griffin had 24 french fries, but Kyle took 5 of them.  Billy took twice as many as Kyle.  Ginger gave Griffin a handful of her fries, and then Colby took from Griffin 3 less than the number of fries that Kyle had taken.  If in the end Griffin had 27 fries, how many fries did Ginger give Griffin?\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": "1. Griffin started with 24 fries.\n2. Kyle took 5: \\(24 - 5 = 19\\).\n3. Billy took twice as many as Kyle: \\(2 \\times 5 = 10\\), leaving \\(19 - 10 = 9\\).\n4. Colby took 3 less than Kyle: \\(5 - 3 = 2\\), leaving \\(9 - 2 = 7\\).\n5. Griffin ended with 27 fries, so Ginger gave \\(27 - 7 = 20\\) fries.\n\nAnswer: 20", "cost_usd": 0.00026940000000000004, "tokens_in": 129, "tokens_out": 203, "latency_s": 3.091}, {"task_id": "lcb-abc388_a", "model": "openai/gpt-5.6-luna", "prompt": "You are given a string S. Here, the first character of S is an uppercase English letter, and the second and subsequent characters are lowercase English letters.\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nConstraints\n\n\n- S is a string of length between 1 and 100, inclusive.\n- The first character of S is an uppercase English letter.\n- The second and subsequent characters of S are lowercase English letters.\n\nSample Input 1\n\nKyoto\n\nSample Output 1\n\nKUPC\r\n\nThe first character of Kyoto is K, so concatenate K and UPC, and print KUPC.\n\nSample Input 2\n\nTohoku\n\nSample Output 2\n\nTUPC\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.", "response": "import sys\n\ndata = sys.stdin.buffer.read().split()\ns = data[0]\noutput = s[:1] + b\"UPC\"\nsys.stdout.buffer.write(output)\n", "cost_usd": 0.00030140000000000006, "tokens_in": 367, "tokens_out": 190, "latency_s": 2.6}, {"task_id": "lcb-abc388_a", "model": "openai/gpt-5.6-luna", "prompt": "You are given a string S. Here, the first character of S is an uppercase English letter, and the second and subsequent characters are lowercase English letters.\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nInput\n\nThe input is given from Standard Input in the following format:\nS\n\nOutput\n\nPrint the string formed by concatenating the first character of S and UPC in this order.\n\nConstraints\n\n\n- S is a string of length between 1 and 100, inclusive.\n- The first character of S is an uppercase English letter.\n- The second and subsequent characters of S are lowercase English letters.\n\nSample Input 1\n\nKyoto\n\nSample Output 1\n\nKUPC\r\n\nThe first character of Kyoto is K, so concatenate K and UPC, and print KUPC.\n\nSample Input 2\n\nTohoku\n\nSample Output 2\n\nTUPC\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\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\n\ndata = sys.stdin.buffer.read()\ntext = data.decode().split()[0]\n\nfirst = \"\"\nfor index, character in enumerate(text):\n    if index == 0:\n        first += character\n\nanswer_parts = []\nfor character in first:\n    answer_parts.append(character)\nfor character in \"UPC\":\n    answer_parts.append(character)\n\nsys.stdout.write(\"\".join(answer_parts))\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size <= 2:\n    s = \"A\"\nelse:\n    length = max(1, min(100, size))\n    first = random.choice(\"ABCDEFGHIJKLMNOPQRSTUVWXYZ\")\n    mode = seed % 5\n\n    if length == 1:\n        suffix = \"\"\n    elif mode == 0:\n        suffix = \"a\" * (length - 1)\n    elif mode == 1:\n        suffix = \"z\" * (length - 1)\n    elif mode == 2:\n        suffix = \"\".join(\"a\" if i % 2 == 0 else \"z\" for i in range(length - 1))\n    elif mode == 3:\n        suffix = \"\".join(random.choice(\"abcdefghijklmnopqrstuvwxyz\") for _ in range(length - 1))\n    else:\n        suffix = \"\".join(chr(ord(\"a\") + ((length - 2 - i) % 26)) for i in range(length - 1))\n\n    s = first + suffix\n\nsys.stdout.write(s + \"\\n\")\n```\n\n```boundary_generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size <= 2:\n    s = \"A\"\nelse:\n    length = max(1, min(100, size))\n    first = chr(ord(\"A\") + random.randrange(26))\n    suffix_chars = []\n\n    for i in range(length - 1):\n        value = random.randrange(26)\n        if i % 4 == 0:\n            value = (value + seed + i * 17) % 26\n        elif i % 4 == 1:\n            value = (25 - value + seed * 3 + i) % 26\n        elif i % 4 == 2:\n            value = (value ^ (seed + i)) % 26\n        suffix_chars.append(chr(ord(\"a\") + value))\n\n    if length > 8 and seed % 3 == 0:\n        pivot = length // 2\n        suffix_chars[pivot - 2:pivot + 2] = list(\"azaz\")\n    elif length > 12 and seed % 3 == 1:\n        for i in range(3, length - 1, 7):\n            suffix_chars[i] = \"z\"\n    elif length > 12:\n        for i in range(2, length - 1, 9):\n            suffix_chars[i] = \"a\"\n\n    s = first + \"\".join(suffix_chars)\n\nsys.stdout.write(s + \"\\n\")\n```", "cost_usd": 0.0015191999999999998, "tokens_in": 612, "tokens_out": 1164, "latency_s": 10.397}, {"task_id": "lcb-abc388_c", "model": "openai/gpt-5.6-luna", "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.\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\ndata = list(map(int, sys.stdin.buffer.read().split()))\nn = data[0]\na = data[1:]\n\ncount = 0\nanswer = 0\n\nfor j in range(n):\n    while count < j and 2 * a[count] <= a[j]:\n        count += 1\n    answer += count\n\nsys.stdout.write(str(answer) + \"\\n\")\n", "cost_usd": 0.0008695999999999999, "tokens_in": 730, "tokens_out": 603, "latency_s": 6.579}, {"task_id": "lcb-abc388_c", "model": "openai/gpt-5.6-luna", "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.\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\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nif not data:\n    sys.exit()\n\nn = data[0]\na = data[1:1 + n]\nanswer = 0\n\nfor i in range(n):\n    top = a[i]\n    for j in range(i + 1, n):\n        bottom = a[j]\n        if top * 2 <= bottom:\n            answer += 1\n\nsys.stdout.write(str(answer))\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nn = max(2, min(500000, size))\n\nif n == 2:\n    sys.stdout.write(\"2\\n1 1\\n\")\n    sys.exit()\n\nif n == 500000:\n    values = [random.getrandbits(30) + 1 for _ in range(n)]\nelse:\n    mode = seed % 5\n\n    if mode == 0:\n        values = [random.randint(1, 10**9) for _ in range(n)]\n    elif mode == 1:\n        pool = [\n            1, 2, 3, 4, 5, 10, 16, 32, 64, 100,\n            10**3, 10**4, 10**5, 10**6, 10**7,\n            10**8, 5 * 10**8, 10**9\n        ]\n        values = [random.choice(pool) for _ in range(n)]\n    elif mode == 2:\n        values = []\n        for _ in range(n):\n            kind = random.randrange(6)\n            if kind == 0:\n                values.append(1)\n            elif kind == 1:\n                values.append(10**9)\n            elif kind == 2:\n                values.append(random.randint(1, 20))\n            elif kind == 3:\n                values.append(random.randint(10**9 - 20, 10**9))\n            else:\n                values.append(random.randint(1, 10**9))\n    elif mode == 3:\n        center = random.randint(1, 10**9)\n        width = random.randint(1, 10**6)\n        lo = max(1, center - width)\n        hi = min(10**9, center + width)\n        values = [random.randint(lo, hi) for _ in range(n)]\n    else:\n        values = []\n        for _ in range(n):\n            if random.randrange(4) == 0:\n                values.append(random.choice([1, 2, 10**9 - 1, 10**9]))\n            else:\n                values.append(random.randint(1, 10**9))\n\nvalues.sort()\nout = str(n) + \"\\n\" + \" \".join(map(str, values)) + \"\\n\"\nsys.stdout.write(out)\n```\n\n```boundary_generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nn = max(2, min(500000, size))\n\nif n == 2:\n    sys.stdout.write(\"2\\n1 1\\n\")\n    sys.exit()\n\nif n == 500000:\n    low_count = n * 3 // 5\n    high_count = n - low_count\n    low = random.sample(range(1, 500000001), low_count)\n    high = random.sample(range(500000001, 1000000001), high_count)\n    values = low + high\nelse:\n    mode = seed % 4\n\n    if mode == 0:\n        values = []\n        for i in range(n):\n            if i < n // 3:\n                values.append(random.randint(1, 5))\n            elif i < 2 * n // 3:\n                values.append(random.randint(10**8, 2 * 10**8))\n            else:\n                values.append(random.randint(8 * 10**8, 10**9))\n    elif mode == 1:\n        values = []\n        for i in range(n):\n            if i % 7 == 0:\n                values.append(1)\n            elif i % 7 == 1:\n                values.append(10**9)\n            else:\n                block = i % 30\n                base = min(10**9, 1 << min(block, 29))\n                values.append(min(10**9, base + random.randint(0, max(1, base // 3))))\n    elif mode == 2:\n        values = []\n        for _ in range(n):\n            branch = random.randrange(3)\n            if branch == 0:\n                values.append(random.randint(1, 100))\n            elif branch == 1:\n                values.append(random.randint(499999900, 500000100))\n            else:\n                values.append(random.randint(999999900, 10**9))\n    else:\n        values = []\n        left = max(1, n // 2)\n        for _ in range(left):\n            values.append(random.randint(1, 250000000))\n        for _ in range(n - left):\n            values.append(random.randint(500000000, 10**9))\n\nvalues.sort()\nout = str(n) + \"\\n\" + \" \".join(map(str, values)) + \"\\n\"\nsys.stdout.write(out)\n```", "cost_usd": 0.0034218, "tokens_in": 975, "tokens_out": 2689, "latency_s": 26.379}]