Upload traces/88737.json with huggingface_hub

traces/88737.json

@@ -0,0 +1 @@
+[{"task_id": "mmlu-3410", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] This question refers to the following information.\nWe have, by this perpetual and irrevocable edict, established and proclaimed:\n\"First, that the recollection of everything done be one party or the other \u2026 during all the preceding period of troubles, remain obliterated and forgotten, as if no such things had ever happened. \u2026\n\"We ordain that the Catholic Apostolic and Roman religion shall be restored and reestablished in all places and localities of this our kingdom and countries subject to our sway, where the exercise of the same has been interrupted, in order that it may be peaceably and freely exercised, without any trouble or hindrance. \u2026\n\"And in order to leave no occasion for troubles or differences between our subjects, we have permitted, and herewith permit, those of the said religion called Reformed to live and abide in all the cities and places of this our kingdom and countries of our sway, and without being annoyed, molested, or compelled to do anything in the matter of religion contrary to their consciences. \u2026\"\nHenry IV of France, The Edict of Nantes, 1598\nFrom the passage, one may infer that Henry IV\nA) was banning Protestants from France\nB) intended for Protestants to be able to live peacefully in his kingdom\nC) was converting to Catholicism\nD) was converting to Protestantism\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": 5.427e-05, "tokens_in": 395, "tokens_out": 104, "latency_s": 4.464}, {"task_id": "mmlu-2235", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] Which group of numbers is in order from least to greatest?\nA) 0.25 1.6 1.0\nB) 1.0 0.25 1.6\nC) 0.25 1.0 1.6\nD) 1.6 1.0 0.25\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: C", "cost_usd": 2.9637999999999996e-05, "tokens_in": 119, "tokens_out": 123, "latency_s": 1.537}, {"task_id": "gsm8k-288", "model": "openai/gpt-5.6-luna", "prompt": "A landscaping company is delivering flagstones to a customer\u2019s yard. Each flagstone weighs 75 pounds. If the delivery trucks can carry a total weight of 2000 pounds, how many trucks will be needed to transport 80 flagstones in one trip?\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 truck can carry 2000 pounds.\n\n1. Total weight of 80 flagstones: \\(80 \\times 75 = 6000\\) pounds.\n2. Number of trucks needed: \\(6000 \\div 2000 = 3\\).\n\nAnswer: 3", "cost_usd": 0.0001412, "tokens_in": 106, "tokens_out": 100, "latency_s": 2.195}, {"task_id": "gsm8k-945", "model": "openai/gpt-5.6-luna", "prompt": "James loves to go swimming and has to swim across a 20-mile lake.  He can swim at a pace of 2 miles per hour.  He swims 60% of the distance.  After that, he stops on an island and rests for half as long as the swimming time.  He then finishes the remaining distance while going half the speed.  How long did it take him to get across the lake?\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. He swims \\(60\\%\\) of 20 miles: \\(0.6 \\times 20 = 12\\) miles.\n2. At 2 miles per hour, this takes \\(12 \\div 2 = 6\\) hours.\n3. He rests for half as long as that swimming time: \\(6 \\div 2 = 3\\) hours.\n4. The remaining distance is \\(20 - 12 = 8\\) miles.\n5. He swims the remaining distance at half his original speed: \\(2 \\div 2 = 1\\) mile per hour, taking \\(8 \\div 1 = 8\\) hours.\n6. Total time: \\(6 + 3 + 8 = 17\\) hours.\n\nAnswer: 17", "cost_usd": 0.0003436, "tokens_in": 140, "tokens_out": 263, "latency_s": 3.338}, {"task_id": "lcb-abc392_b", "model": "openai/gpt-5.6-luna", "prompt": "You are given a sequence of M integers A = (A_1, A_2, \\dots, A_M).\nEach element of A is an integer between 1 and N, inclusive, and all elements are distinct.\nList all integers between 1 and N that do not appear in A in ascending order.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M\nA_1 A_2 \\dots A_M\n\nOutput\n\nLet (X_1, X_2, \\dots, X_C) be the sequence of all integers between 1 and N, inclusive, that do not appear in A, listed in ascending order.\nThe output should be in the following format:\nC\nX_1 X_2 \\dots X_C\n\nConstraints\n\n\n- All input values are integers.\n- 1 \\le M \\le N \\le 1000\n- 1 \\le A_i \\le N\n- The elements of A are distinct.\n\nSample Input 1\n\n10 3\n3 9 2\n\nSample Output 1\n\n7\n1 4 5 6 7 8 10\n\nHere, A=(3,9,2).\nThe integers between 1 and 10 that do not appear in A, listed in ascending order, are 1,4,5,6,7,8,10.\n\nSample Input 2\n\n6 6\n1 3 5 2 4 6\n\nSample Output 2\n\n0\n\n\nNo integer between 1 and 6 is missing from A.\nIn this case, print 0 on the first line and leave the second line empty.\n\nSample Input 3\n\n9 1\n9\n\nSample Output 3\n\n8\n1 2 3 4 5 6 7 8\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 = list(map(int, sys.stdin.buffer.read().split()))\nn, m = data[0], data[1]\n\npresent = [False] * (n + 1)\nfor x in data[2:2 + m]:\n    present[x] = True\n\nmissing = [str(i) for i in range(1, n + 1) if not present[i]]\n\noutput = str(len(missing)) + \"\\n\"\noutput += \" \".join(missing) + \"\\n\"\n\nsys.stdout.write(output)\n", "cost_usd": 0.00048820000000000005, "tokens_in": 563, "tokens_out": 313, "latency_s": 4.008}, {"task_id": "lcb-abc392_b", "model": "openai/gpt-5.6-luna", "prompt": "You are given a sequence of M integers A = (A_1, A_2, \\dots, A_M).\nEach element of A is an integer between 1 and N, inclusive, and all elements are distinct.\nList all integers between 1 and N that do not appear in A in ascending order.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN M\nA_1 A_2 \\dots A_M\n\nOutput\n\nLet (X_1, X_2, \\dots, X_C) be the sequence of all integers between 1 and N, inclusive, that do not appear in A, listed in ascending order.\nThe output should be in the following format:\nC\nX_1 X_2 \\dots X_C\n\nConstraints\n\n\n- All input values are integers.\n- 1 \\le M \\le N \\le 1000\n- 1 \\le A_i \\le N\n- The elements of A are distinct.\n\nSample Input 1\n\n10 3\n3 9 2\n\nSample Output 1\n\n7\n1 4 5 6 7 8 10\n\nHere, A=(3,9,2).\nThe integers between 1 and 10 that do not appear in A, listed in ascending order, are 1,4,5,6,7,8,10.\n\nSample Input 2\n\n6 6\n1 3 5 2 4 6\n\nSample Output 2\n\n0\n\n\nNo integer between 1 and 6 is missing from A.\nIn this case, print 0 on the first line and leave the second line empty.\n\nSample Input 3\n\n9 1\n9\n\nSample Output 3\n\n8\n1 2 3 4 5 6 7 8\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()\nn = int(data[0])\nm = int(data[1])\na = [int(x) for x in data[2:2 + m]]\n\nmissing = []\nfor value in range(1, n + 1):\n    found = False\n    for item in a:\n        if item == value:\n            found = True\n            break\n    if not found:\n        missing.append(value)\n\nout = str(len(missing)) + \"\\n\"\nif missing:\n    out += \" \".join(map(str, missing)) + \"\\n\"\nelse:\n    out += \"\\n\"\nsys.stdout.write(out)\n```\n\n```generator\nimport sys\nimport random\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\ns = max(2, min(1000, size))\n\nif s == 2:\n    sys.stdout.write(\"1 1\\n1\\n\")\nelse:\n    n = s\n    if s == 1000:\n        m = random.randint(n // 2, n)\n    else:\n        mode = random.randrange(5)\n        if mode == 0:\n            m = 1\n        elif mode == 1:\n            m = n\n        else:\n            m = random.randint(1, n)\n\n    a = random.sample(range(1, n + 1), m)\n\n    if s < 1000:\n        mode = random.randrange(5)\n        if mode == 0:\n            a.sort()\n        elif mode == 1:\n            a.sort(reverse=True)\n        elif mode == 2:\n            a = a[::2] + a[1::2]\n        elif mode == 3:\n            left = []\n            right = []\n            for i, value in enumerate(a):\n                if i % 2 == 0:\n                    left.append(value)\n                else:\n                    right.append(value)\n            a = left + right[::-1]\n        else:\n            random.shuffle(a)\n    else:\n        random.shuffle(a)\n\n    out = f\"{n} {m}\\n\"\n    out += \" \".join(map(str, a)) + \"\\n\"\n    sys.stdout.write(out)\n```\n\n```boundary_generator\nimport sys\nimport random\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\ns = max(2, min(1000, size))\n\nif s == 2:\n    sys.stdout.write(\"1 1\\n1\\n\")\nelse:\n    n = s\n\n    if s == 1000:\n        m = random.randint(500, 1000)\n    else:\n        choices = [1, n, max(1, n // 2), max(1, n - 1)]\n        m = choices[random.randrange(len(choices))]\n\n    keyed = []\n    for value in range(1, n + 1):\n        keyed.append((random.getrandbits(64), value))\n    keyed.sort()\n    selected = [value for _, value in keyed[:m]]\n    selected.sort()\n\n    pattern = random.randrange(4)\n    if pattern == 0:\n        ordered = selected\n    elif pattern == 1:\n        ordered = selected[::-1]\n    elif pattern == 2:\n        ordered = []\n        lo = 0\n        hi = len(selected) - 1\n        while lo <= hi:\n            ordered.append(selected[lo])\n            lo += 1\n            if lo <= hi:\n                ordered.append(selected[hi])\n                hi -= 1\n    else:\n        ordered = []\n        lo = 0\n        hi = len(selected) - 1\n        take_low = False\n        while lo <= hi:\n            if take_low:\n                ordered.append(selected[lo])\n                lo += 1\n            else:\n                ordered.append(selected[hi])\n                hi -= 1\n            take_low = not take_low\n\n    out = f\"{n} {m}\\n\"\n    out += \" \".join(map(str, ordered)) + \"\\n\"\n    sys.stdout.write(out)\n```", "cost_usd": 0.0031544000000000003, "tokens_in": 808, "tokens_out": 2494, "latency_s": 24.327}, {"task_id": "lcb-abc392_d", "model": "openai/gpt-5.6-luna", "prompt": "There are N dice.\r\nThe i-th die has K_i faces, with the numbers A_{i,1}, A_{i,2}, \\ldots, A_{i,K_i} written on them.\r\nWhen you roll this die, each face appears with probability \\frac{1}{K_i}.\nYou choose two dice from the N dice and roll them.\r\nDetermine the maximum probability that the two dice show the same number, when the dice are chosen optimally.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nK_1 A_{1,1} A_{1,2} \\dots A_{1,K_1}\r\n\\vdots\r\nK_N A_{N,1} A_{N,2} \\dots A_{N,K_N}\n\nOutput\n\nPrint the answer.\r\nYour answer is considered correct if the absolute or relative error from the true solution does not exceed 10^{-8}.\n\nConstraints\n\n\n- 2 \\leq N \\leq 100\n- 1 \\leq K_i\n- K_1 + K_2 + \\dots + K_N \\leq 10^5\n- 1 \\leq A_{i,j} \\leq 10^5\n- All input values are integers.\n\nSample Input 1\n\n3\r\n3 1 2 3\r\n4 1 2 2 1\r\n6 1 2 3 4 5 6\n\nSample Output 1\n\n0.333333333333333\r\n\n\n- When choosing the 1st and 2nd dice, the probability that the outcomes are the same is \\frac{1}{3}.\n- When choosing the 1st and 3rd dice, the probability is \\frac{1}{6}.\n- When choosing the 2nd and 3rd dice, the probability is \\frac{1}{6}.\n\nTherefore, the maximum probability is \\frac{1}{3} = 0.3333333333\\ldots.\n\nSample Input 2\n\n3\r\n5 1 1 1 1 1\r\n4 2 2 2 2\r\n3 1 1 2\n\nSample Output 2\n\n0.666666666666667\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\nThe statement offers an error tolerance, but grading still compares text exactly. For the sample inputs reproduce the samples' own digits; for any other input print twelve digits after the decimal point.", "response": "import sys\n\nraw = sys.stdin.buffer.read()\nit = iter(map(int, raw.split()))\n\nn = next(it)\nks = []\ndice_values = []\nvalue_entries = [[] for _ in range(100001)]\n\nfor die_index in range(n):\n    k = next(it)\n    ks.append(k)\n    values = []\n    counts = {}\n    for _ in range(k):\n        value = next(it)\n        values.append(value)\n        counts[value] = counts.get(value, 0) + 1\n    dice_values.append(tuple(values))\n    for value, count in counts.items():\n        value_entries[value].append((die_index, count))\n\nscores = [[0] * n for _ in range(n)]\n\nfor entries in value_entries:\n    length = len(entries)\n    for first_index in range(length):\n        left_die, left_count = entries[first_index]\n        for second_index in range(first_index + 1, length):\n            right_die, right_count = entries[second_index]\n            scores[left_die][right_die] += left_count * right_count\n\nbest_num = 0\nbest_den = 1\n\nfor left_die in range(n):\n    row = scores[left_die]\n    for right_die in range(left_die + 1, n):\n        candidate_num = row[right_die]\n        candidate_den = ks[left_die] * ks[right_die]\n        if candidate_num * best_den > best_num * candidate_den:\n            best_num = candidate_num\n            best_den = candidate_den\n\nif dice_values == [\n    (1, 2, 3),\n    (1, 2, 2, 1),\n    (1, 2, 3, 4, 5, 6),\n]:\n    answer = \"0.333333333333333\"\nelif dice_values == [\n    (1, 1, 1, 1, 1),\n    (2, 2, 2, 2),\n    (1, 1, 2),\n]:\n    answer = \"0.666666666666667\"\nelse:\n    digits = 12\n    scale = 10 ** digits\n    rounded = (2 * best_num * scale + best_den) // (2 * best_den)\n    answer = f\"{rounded // scale}.{rounded % scale:0{digits}d}\"\n\nsys.stdout.write(answer + \"\\n\")", "cost_usd": 0.0050854, "tokens_in": 731, "tokens_out": 4116, "latency_s": 46.523}, {"task_id": "lcb-abc392_d", "model": "openai/gpt-5.6-luna", "prompt": "There are N dice.\r\nThe i-th die has K_i faces, with the numbers A_{i,1}, A_{i,2}, \\ldots, A_{i,K_i} written on them.\r\nWhen you roll this die, each face appears with probability \\frac{1}{K_i}.\nYou choose two dice from the N dice and roll them.\r\nDetermine the maximum probability that the two dice show the same number, when the dice are chosen optimally.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nK_1 A_{1,1} A_{1,2} \\dots A_{1,K_1}\r\n\\vdots\r\nK_N A_{N,1} A_{N,2} \\dots A_{N,K_N}\n\nOutput\n\nPrint the answer.\r\nYour answer is considered correct if the absolute or relative error from the true solution does not exceed 10^{-8}.\n\nConstraints\n\n\n- 2 \\leq N \\leq 100\n- 1 \\leq K_i\n- K_1 + K_2 + \\dots + K_N \\leq 10^5\n- 1 \\leq A_{i,j} \\leq 10^5\n- All input values are integers.\n\nSample Input 1\n\n3\r\n3 1 2 3\r\n4 1 2 2 1\r\n6 1 2 3 4 5 6\n\nSample Output 1\n\n0.333333333333333\r\n\n\n- When choosing the 1st and 2nd dice, the probability that the outcomes are the same is \\frac{1}{3}.\n- When choosing the 1st and 3rd dice, the probability is \\frac{1}{6}.\n- When choosing the 2nd and 3rd dice, the probability is \\frac{1}{6}.\n\nTherefore, the maximum probability is \\frac{1}{3} = 0.3333333333\\ldots.\n\nSample Input 2\n\n3\r\n5 1 1 1 1 1\r\n4 2 2 2 2\r\n3 1 1 2\n\nSample Output 2\n\n0.666666666666667\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\nThe statement offers an error tolerance, but grading still compares text exactly. For the sample inputs reproduce the samples' own digits; for any other input print twelve digits after the decimal point.\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 fractions import Fraction\n\ndef main():\n    data = list(map(int, sys.stdin.buffer.read().split()))\n    p = 0\n    n = data[p]\n    p += 1\n    dice = []\n    for _ in range(n):\n        k = data[p]\n        p += 1\n        dice.append(data[p:p + k])\n        p += k\n\n    best = Fraction(0, 1)\n    for i in range(n):\n        left = dice[i]\n        for j in range(i + 1, n):\n            right = dice[j]\n            equal = 0\n            for x in left:\n                for y in right:\n                    if x == y:\n                        equal += 1\n            prob = Fraction(equal, len(left) * len(right))\n            if prob > best:\n                best = prob\n\n    sys.stdout.write(f\"{float(best):.15f}\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```\n\n```generator\nimport sys\nimport random\n\ndef main():\n    seed = int(sys.argv[1])\n    size = int(sys.argv[2])\n    random.seed(seed)\n\n    if size <= 2:\n        sys.stdout.write(\"2\\n1 1\\n1 1\\n\")\n        return\n\n    total = max(2, min(100000, size))\n    n = min(100, max(2, total // 1000))\n    ks = [1] * n\n    for _ in range(total - n):\n        ks[random.randrange(n)] += 1\n\n    lines = [str(n)]\n    global_index = 0\n    for i, k in enumerate(ks):\n        vals = []\n        mode = random.randrange(6)\n        for j in range(k):\n            r = random.random()\n            if r < 0.08:\n                v = 1\n            elif r < 0.16:\n                v = 100000\n            elif r < 0.25:\n                v = ((i + j + seed) % 1000) + 1\n            elif r < 0.34:\n                v = 100000 - ((i * 997 + j + seed) % 1000)\n            else:\n                v = random.randrange(1, 100001)\n            if mode == 1 and j % 7 == 0:\n                v = 1\n            elif mode == 2 and j % 11 == 0:\n                v = 100000\n            elif mode == 3 and j % 5 == 0:\n                v = ((global_index * 7919 + seed) % 100000) + 1\n            vals.append(v)\n            global_index += 1\n        if mode == 4:\n            vals.reverse()\n        elif mode == 5:\n            vals.sort()\n        lines.append(str(k) + \" \" + \" \".join(map(str, vals)))\n\n    sys.stdout.write(\"\\n\".join(lines) + \"\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```\n\n```boundary_generator\nimport sys\nimport random\n\ndef main():\n    seed = int(sys.argv[1])\n    size = int(sys.argv[2])\n    random.seed(seed)\n\n    if size <= 2:\n        sys.stdout.write(\"2\\n1 1\\n1 1\\n\")\n        return\n\n    total = max(2, min(100000, size))\n    n = min(100, max(2, total // 1000))\n    ks = [1] * n\n    ks[0] = total - n + 1\n\n    lines = [str(n)]\n    for i, k in enumerate(ks):\n        vals = []\n        if i == 0:\n            start = (seed * 7919 + 104729) % 100000\n            step = 2 * ((seed * 104729 + 1) % 49999) + 1\n            for j in range(k):\n                if j % 97 == 0:\n                    vals.append(1)\n                elif j % 101 == 0:\n                    vals.append(100000)\n                else:\n                    vals.append(((start + j * step) % 100000) + 1)\n            if seed % 2:\n                vals.reverse()\n        else:\n            if i % 4 == 0:\n                v = 1\n            elif i % 4 == 1:\n                v = 100000\n            elif i % 4 == 2:\n                v = ((seed * 31337 + i * 1009) % 100000) + 1\n            else:\n                v = ((seed + i * 65537) % 100000) + 1\n            vals.append(v)\n        lines.append(str(k) + \" \" + \" \".join(map(str, vals)))\n\n    sys.stdout.write(\"\\n\".join(lines) + \"\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.0032011999999999995, "tokens_in": 976, "tokens_out": 2505, "latency_s": 27.799}, {"task_id": "lcb-abc392_d", "model": "google/gemini-3.6-flash", "prompt": "There are N dice.\r\nThe i-th die has K_i faces, with the numbers A_{i,1}, A_{i,2}, \\ldots, A_{i,K_i} written on them.\r\nWhen you roll this die, each face appears with probability \\frac{1}{K_i}.\nYou choose two dice from the N dice and roll them.\r\nDetermine the maximum probability that the two dice show the same number, when the dice are chosen optimally.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nK_1 A_{1,1} A_{1,2} \\dots A_{1,K_1}\r\n\\vdots\r\nK_N A_{N,1} A_{N,2} \\dots A_{N,K_N}\n\nOutput\n\nPrint the answer.\r\nYour answer is considered correct if the absolute or relative error from the true solution does not exceed 10^{-8}.\n\nConstraints\n\n\n- 2 \\leq N \\leq 100\n- 1 \\leq K_i\n- K_1 + K_2 + \\dots + K_N \\leq 10^5\n- 1 \\leq A_{i,j} \\leq 10^5\n- All input values are integers.\n\nSample Input 1\n\n3\r\n3 1 2 3\r\n4 1 2 2 1\r\n6 1 2 3 4 5 6\n\nSample Output 1\n\n0.333333333333333\r\n\n\n- When choosing the 1st and 2nd dice, the probability that the outcomes are the same is \\frac{1}{3}.\n- When choosing the 1st and 3rd dice, the probability is \\frac{1}{6}.\n- When choosing the 2nd and 3rd dice, the probability is \\frac{1}{6}.\n\nTherefore, the maximum probability is \\frac{1}{3} = 0.3333333333\\ldots.\n\nSample Input 2\n\n3\r\n5 1 1 1 1 1\r\n4 2 2 2 2\r\n3 1 1 2\n\nSample Output 2\n\n0.666666666666667\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\nThe statement offers an error tolerance, but grading still compares text exactly. For the sample inputs reproduce the samples' own digits; for any other input print twelve digits after the decimal point.\n\nWrite ONE more REFERENCE solution for this problem and nothing else. Correct by construction, arbitrarily slow, and taking a different route from every solution above \u2014 if one counted, enumerate; if one was greedy, search exhaustively. Same stdin format, same output format. Respond with exactly one fenced block:\n```reference\n<program>\n```", "response": "import sys\nfrom collections import Counter\nfrom fractions import Fraction\n\ndef main():\n    raw_input = sys.stdin.read()\n    tokens = raw_input.split()\n    if not tokens:\n        return\n    \n    N = int(tokens[0])\n    idx = 1\n    dice_counts = []\n    K_list = []\n    \n    for _ in range(N):\n        K = int(tokens[idx])\n        idx += 1\n        faces = [int(x) for x in tokens[idx : idx + K]]\n        idx += K\n        K_list.append(K)\n        dice_counts.append(Counter(faces))\n    \n    max_prob = Fraction(0, 1)\n    \n    for i in range(N):\n        c1 = dice_counts[i]\n        k1 = K_list[i]\n        for j in range(i + 1, N):\n            c2 = dice_counts[j]\n            k2 = K_list[j]\n            \n            small, big = (c1, c2) if len(c1) < len(c2) else (c2, c1)\n            dot = sum(cnt * big[val] for val, cnt in small.items() if val in big)\n            prob = Fraction(dot, k1 * k2)\n            if prob > max_prob:\n                max_prob = prob\n\n    if tokens == [\"3\", \"3\", \"1\", \"2\", \"3\", \"4\", \"1\", \"2\", \"2\", \"1\", \"6\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\"]:\n        sys.stdout.write(\"0.333333333333333\\n\")\n    elif tokens == [\"3\", \"5\", \"1\", \"1\", \"1\", \"1\", \"1\", \"4\", \"2\", \"2\", \"2\", \"2\", \"3\", \"1\", \"1\", \"2\"]:\n        sys.stdout.write(\"0.666666666666667\\n\")\n    else:\n        sys.stdout.write(f\"{float(max_prob):.12f}\\n\")\n\nif __name__ == \"__main__\":\n    main()\n\n```reference\nimport sys\n\ndef main():\n    raw_input = sys.stdin.read()\n    tokens = raw_input.split()\n    if not tokens:\n        return\n\n    N = int(tokens[0])\n    idx = 1\n    dice_faces = []\n\n    for _ in range(N):\n        K = int(tokens[idx])\n        idx += 1\n        faces = [int(x) for x in tokens[idx : idx + K]]\n        idx += K\n        dice_faces.append(faces)\n\n    best_prob = 0.0\n\n    for i in range(N):\n        faces1 = dice_faces[i]\n        len1 = len(faces1)\n        for j in range(i + 1, N):\n            faces2 = dice_faces[j]\n            len2 = len(faces2)\n            \n            matches = 0\n            for f1 in faces1:\n                for f2 in faces2:\n                    if f1 == f2:\n                        matches += 1\n            \n            p = matches / (len1 * len2)\n            if p > best_prob:\n                best_prob = p\n\n    if tokens == [\"3\", \"3\", \"1\", \"2\", \"3\", \"4\", \"1\", \"2\", \"2\", \"1\", \"6\", \"1\", \"2\", \"3\", \"4\", \"5\", \"6\"]:\n        sys.stdout.write(\"0.333333333333333\\n\")\n    elif tokens == [\"3\", \"5\", \"1\", \"1\", \"1\", \"1\", \"1\", \"4\", \"2\", \"2\", \"2\", \"2\", \"3\", \"1\", \"1\", \"2\"]:\n        sys.stdout.write(\"0.666666666666667\\n\")\n    else:\n        sys.stdout.write(f\"{best_prob:.12f}\\n\")\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.023890499999999995, "tokens_in": 844, "tokens_out": 6202, "latency_s": 31.722}]