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+[{"task_id": "mmlu-7069", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU]  Advertising is employed as a defence, to retain customers, and to increase brand usage. Advertising is used to reinforce existing attitudes, not necessarily to drastically change them. This idea is based on:\nA) Strong theory.\nB) Weak theory.\nC) ATR framework.\nD) DRIP model.\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": 6.636e-05, "tokens_in": 108, "tokens_out": 183, "latency_s": 15.692}, {"task_id": "mmlu-6851", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] In building a linear regression model for a particular data set, you observe the coefficient of one of the features having a relatively high negative value. This suggests that\nA) This feature has a strong effect on the model (should be retained)\nB) This feature does not have a strong effect on the model (should be ignored)\nC) It is not possible to comment on the importance of this feature without additional information\nD) Nothing can be determined.\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": "C", "cost_usd": 0.000295, "tokens_in": 138, "tokens_out": 614, "latency_s": 18.827}, {"task_id": "gsm8k-647", "model": "openai/gpt-5.6-luna", "prompt": "Bill is trying to dig a well in his backyard. He can dig 4 feet/hour through soil and half that fast through clay. If he has to dig through 24 feet of soil and 8 feet of clay, how long will it take him to dig the well?\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": "Soil: \\(24 \\div 4 = 6\\) hours.  \nClay speed: \\(4 \\div 2 = 2\\) feet/hour.  \nClay: \\(8 \\div 2 = 4\\) hours.  \nTotal: \\(6 + 4 = 10\\) hours.\n\nAnswer: 10", "cost_usd": 0.00017679999999999996, "tokens_in": 110, "tokens_out": 129, "latency_s": 2.537}, {"task_id": "gsm8k-151", "model": "openai/gpt-5.6-luna", "prompt": "Shawnda decides that her neighborhood kids could really use a bike inflation service. She decides the best way to charge is by the tire. Each tire costs 25 cents to inflate. On the first day, 5 people on bicycles came by to get both tires inflated. 3 people came by to get all their tricycle tires inflated. Finally, one person actually came by on a unicycle. How many dollars did she make that day?\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": "Each bicycle has 2 tires: \\(5 \\times 2 = 10\\) tires.  \nEach tricycle has 3 tires: \\(3 \\times 3 = 9\\) tires.  \nThe unicycle has 1 tire.  \n\nTotal tires: \\(10 + 9 + 1 = 20\\).  \nAt 25 cents per tire: \\(20 \\times \\$0.25 = \\$5\\).\n\nAnswer: 5", "cost_usd": 0.0002232, "tokens_in": 144, "tokens_out": 162, "latency_s": 3.536}, {"task_id": "lcb-abc400_a", "model": "openai/gpt-5.6-luna", "prompt": "In the ceremony commemorating ABC400, we want to arrange 400 people in a rectangular formation of A rows and B columns without any gaps.\nYou are given a positive integer A. Print the value of a positive integer B for which such an arrangement is possible. If there is no such positive integer B, print -1.\n\nInput\n\nThe input is given from Standard Input in the following format:\nA\n\nOutput\n\nPrint the value of B or -1 as specified by the problem statement.\n\nConstraints\n\n\n- A is an integer between 1 and 400, inclusive.\n\nSample Input 1\n\n10\n\nSample Output 1\n\n40\r\n\nWe can arrange 400 people in 10 rows and 40 columns.\n\nSample Input 2\n\n11\n\nSample Output 2\n\n-1\n\nSample Input 3\n\n400\n\nSample Output 3\n\n1\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()\na = int(data)\nanswer = 400 // a if 400 % a == 0 else -1\noutput = f\"{answer}\\n\"\nsys.stdout.write(output)\n", "cost_usd": 0.0002988000000000001, "tokens_in": 360, "tokens_out": 189, "latency_s": 3.322}, {"task_id": "lcb-abc400_a", "model": "openai/gpt-5.6-luna", "prompt": "In the ceremony commemorating ABC400, we want to arrange 400 people in a rectangular formation of A rows and B columns without any gaps.\nYou are given a positive integer A. Print the value of a positive integer B for which such an arrangement is possible. If there is no such positive integer B, print -1.\n\nInput\n\nThe input is given from Standard Input in the following format:\nA\n\nOutput\n\nPrint the value of B or -1 as specified by the problem statement.\n\nConstraints\n\n\n- A is an integer between 1 and 400, inclusive.\n\nSample Input 1\n\n10\n\nSample Output 1\n\n40\r\n\nWe can arrange 400 people in 10 rows and 40 columns.\n\nSample Input 2\n\n11\n\nSample Output 2\n\n-1\n\nSample Input 3\n\n400\n\nSample Output 3\n\n1\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().split()\na = int(data[0])\n\nanswer = -1\nfor b in range(1, 401):\n    if a * b == 400:\n        answer = b\n        break\n\nsys.stdout.write(str(answer) + \"\\n\")\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nmagnitude = max(1, min(400, size))\n\nif size == 2:\n    value = 1\nelif magnitude == 400:\n    value = random.randint(1, 400)\nelse:\n    choice = random.randrange(6)\n    if choice == 0:\n        value = 1\n    elif choice == 1:\n        value = 400\n    elif choice == 2:\n        value = random.choice([1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400])\n    elif choice == 3:\n        value = random.randint(1, magnitude)\n    elif choice == 4:\n        value = max(1, min(400, magnitude + random.randint(-2, 2)))\n    else:\n        value = random.randint(1, 400)\n\nsys.stdout.write(str(value) + \"\\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\nmagnitude = max(1, min(400, size))\n\nif size == 2:\n    value = 1\nelif magnitude == 400:\n    raw = random.getrandbits(64)\n    value = raw % 400 + 1\n    if value in (1, 2, 4, 5, 8, 10, 16, 20, 25, 40, 50, 80, 100, 200, 400):\n        value = value % 399 + 1\nelse:\n    mode = random.getrandbits(3)\n    if mode == 0:\n        value = 1\n    elif mode == 1:\n        value = 400\n    elif mode == 2:\n        value = max(1, min(400, magnitude))\n    elif mode == 3:\n        value = max(1, min(400, magnitude - 1))\n    elif mode == 4:\n        value = random.choice([3, 6, 7, 9, 11, 13, 17, 19, 21, 23, 27, 31])\n    else:\n        value = random.randint(1, 400)\n\nsys.stdout.write(str(value) + \"\\n\")\n```", "cost_usd": 0.0018658, "tokens_in": 605, "tokens_out": 1454, "latency_s": 12.798}, {"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.\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\nfrom array import array\n\ndef solve():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    n = data[0]\n    values = data[1:]\n\n    bells = [\n        1, 1, 2, 5, 15, 52, 203,\n        877, 4140, 21147, 115975, 678570, 4213597\n    ]\n    bound = bells[n]\n\n    capacity = 1\n    while capacity * 3 < bound * 4:\n        capacity <<= 1\n\n    table = array(\"Q\", [0]) * capacity\n    mask = capacity - 1\n    hash_constant = 0x9E3779B97F4A7C15\n\n    blocks = [values[0]]\n    unique_count = 0\n    zero_seen = False\n\n    def dfs(pos, current_xor):\n        nonlocal unique_count, zero_seen\n\n        if pos == n:\n            if current_xor == 0:\n                if not zero_seen:\n                    zero_seen = True\n                    unique_count += 1\n                return\n\n            h = current_xor * hash_constant\n            h ^= h >> 32\n            h ^= h >> 64\n            h ^= h >> 96\n            h &= mask\n\n            while True:\n                stored = table[h]\n                if stored == 0:\n                    table[h] = current_xor\n                    unique_count += 1\n                    return\n                if stored == current_xor:\n                    return\n                h = (h + 1) & mask\n\n        value = values[pos]\n        block_count = len(blocks)\n\n        for j in range(block_count):\n            old_sum = blocks[j]\n            merged_sum = old_sum + value\n            blocks[j] = merged_sum\n            dfs(pos + 1, current_xor ^ old_sum ^ merged_sum)\n            blocks[j] = old_sum\n\n        blocks.append(value)\n        dfs(pos + 1, current_xor ^ value)\n        blocks.pop()\n\n    dfs(1, values[0])\n\n    output = str(unique_count) + \"\\n\"\n    sys.stdout.write(output)\n\nif __name__ == \"__main__\":\n    solve()\n", "cost_usd": 0.022657200000000002, "tokens_in": 852, "tokens_out": 18739, "latency_s": 174.844}, {"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.\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\nfrom collections import deque\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\nn = data[0]\na = tuple(data[1:])\n\nseen = {a}\nq = deque([a])\nvalues = set()\n\nwhile q:\n    state = q.popleft()\n    x = 0\n    for v in state:\n        x ^= v\n    values.add(x)\n\n    for i in range(n):\n        if state[i] == 0:\n            continue\n        for j in range(n):\n            if i == j:\n                continue\n            nxt = list(state)\n            nxt[j] += nxt[i]\n            nxt[i] = 0\n            nxt = tuple(nxt)\n            if nxt not in seen:\n                seen.add(nxt)\n                q.append(nxt)\n\nsys.stdout.write(str(len(values)) + \"\\n\")\n```\n\n```generator\nimport sys\nimport random\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nif size <= 2:\n    sys.stdout.write(\"2\\n1 1\\n\")\n    raise SystemExit\n\nn = max(2, min(12, size))\n\nif n == 12:\n    a = [random.randint(1, 10**17) for _ in range(n)]\n    if seed % 13 == 0:\n        a[1] = a[0]\n    if seed % 7 == 0:\n        a[0] = 1\n        a[-1] = 10**17\n    mode = seed % 4\n    if mode == 1:\n        a.sort()\n    elif mode == 2:\n        a.sort(reverse=True)\n    elif mode == 3:\n        a = a[::2] + a[1::2]\nelse:\n    limit = 1 << min(35, n + 7)\n    mode = seed % 6\n    if mode == 0:\n        a = [random.randint(1, limit) for _ in range(n)]\n    elif mode == 1:\n        x = random.randint(1, limit)\n        a = [x if i % 2 == 0 else random.randint(1, limit) for i in range(n)]\n    elif mode == 2:\n        a = [1 if i % 2 == 0 else limit for i in range(n)]\n    elif mode == 3:\n        a = [limit - i for i in range(n)]\n    elif mode == 4:\n        a = [1 << (i % min(30, n + 5)) for i in range(n)]\n        random.shuffle(a)\n    else:\n        a = [random.randint(1, limit) for _ in range(n)]\n        a[-1] = a[0]\n\n    if seed % 5 == 0:\n        a[0] = 1\n    if seed % 7 == 0:\n        a[-1] = limit\n    if seed % 3 == 0:\n        a.reverse()\n\nsys.stdout.write(str(n) + \"\\n\")\nsys.stdout.write(\" \".join(map(str, a)) + \"\\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    sys.stdout.write(\"2\\n1 1\\n\")\n    raise SystemExit\n\nn = max(2, min(12, size))\n\nif n == 12:\n    a = []\n    a.append(1 + random.randrange(1 << 20))\n    a.append(10**17 - random.randrange(1 << 20))\n    a.append((1 << 56) + random.randrange(1 << 40))\n    a.append((1 << 55) - random.randrange(1 << 35))\n    for _ in range(n - 4):\n        x = random.getrandbits(57)\n        x = 1 + (x % 10**17)\n        a.append(x)\n\n    if seed % 3 == 0:\n        a = a[::-1]\n    elif seed % 3 == 1:\n        a = sorted(a, key=lambda x: (x.bit_count(), x))\n    else:\n        a = a[::2] + a[1::2]\nelse:\n    limit = 1 << min(35, n + 7)\n    a = []\n    for i in range(n):\n        r = random.getrandbits(32)\n        if i % 5 == 0:\n            v = 1 + (r % min(limit, 1 << 8))\n        elif i % 5 == 1:\n            v = limit - (r % min(limit - 1, 1 << 8))\n        elif i % 5 == 2:\n            v = 1 << (i % min(30, n + 5))\n        elif i % 5 == 3:\n            v = (1 << min(30, n + 5)) - 1 - (r % min(limit - 1, 1 << 10))\n            v = max(1, v)\n        else:\n            v = 1 + (r % limit)\n        a.append(v)\n\n    if seed % 2 == 0:\n        a.sort(reverse=True)\n    else:\n        a = a[1::2] + a[::2]\n\nsys.stdout.write(str(n) + \"\\n\")\nsys.stdout.write(\" \".join(map(str, a)) + \"\\n\")\n```", "cost_usd": 0.0039953, "tokens_in": 1097, "tokens_out": 3101, "latency_s": 35.688}]