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[{"task_id": "mmlu-13604", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] Which of the following did Domhoff not identify as a process of decision-making in the USA?\nA) the ideology process\nB) the exploitation process\nC) the policy-formation process\nD) the candidate-selection process\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": "B", "cost_usd": 4.284e-05, "tokens_in": 94, "tokens_out": 191, "latency_s": 6.072}, {"task_id": "mmlu-9581", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] Lactose\n\nA) Is always the cause of milk intolerance\nB) Cannot be digested in the human gut\nC) Intolerance is present in up to 20% of the population\nD) Is excluded on a low FODMAP diet\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: D", "cost_usd": 0.00012824, "tokens_in": 98, "tokens_out": 409, "latency_s": 5.633}, {"task_id": "gsm8k-276", "model": "openai/gpt-5.6-luna", "prompt": "Carly had 80 cards, 2/5 of the cards had the letter A on them, 1/2 of the remaining had the letter B, 5/8 of the rest had the letter C on them, and the others had the letter D. How many of the cards had the letter D on them?\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": "Start with 80 cards.\n\n- Cards with A: \\( \\frac{2}{5}\\times 80=32\\)\n- Remaining: \\(80-32=48\\)\n- Cards with B: \\( \\frac{1}{2}\\times 48=24\\)\n- Remaining: \\(48-24=24\\)\n- Cards with C: \\( \\frac{5}{8}\\times 24=15\\)\n- Cards with D: \\(24-15=9\\)\n\nAnswer: 9", "cost_usd": 0.00022219999999999998, "tokens_in": 121, "tokens_out": 165, "latency_s": 3.258}, {"task_id": "gsm8k-513", "model": "openai/gpt-5.6-luna", "prompt": "Julie had $500. She spent 20% of it on clothes and then 25% of the remaining money on CDs. How much money did Julie have left?\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 spent 20% of $500 on clothes:\n\n20% of $500 = $100\n\nMoney remaining: $500 \u2212 $100 = $400\n\nShe spent 25% of $400 on CDs:\n\n25% of $400 = $100\n\nMoney left: $400 \u2212 $100 = $300\n\nAnswer: 300", "cost_usd": 0.00014959999999999995, "tokens_in": 88, "tokens_out": 110, "latency_s": 4.604}, {"task_id": "lcb-abc395_b", "model": "openai/gpt-5.6-luna", "prompt": "Overview: Create an N \\times N pattern as follows.\n\n###########\n#.........#\n#.#######.#\n#.#.....#.#\n#.#.###.#.#\n#.#.#.#.#.#\n#.#.###.#.#\n#.#.....#.#\n#.#######.#\n#.........#\n###########\n\n\nYou are given a positive integer N.\nConsider an N \\times N grid. Let (i,j) denote the cell at the i-th row from the top and the j-th column from the left. Initially, no cell is colored.\nThen, for i = 1,2,\\dots,N in order, perform the following operation:\n\n- Let j = N + 1 - i.\n- If i \\leq j, fill the rectangular region whose top-left cell is (i,i) and bottom-right cell is (j,j) with black if i is odd, or white if i is even. If some cells are already colored, overwrite their colors. \n- If i > j, do nothing.\n\nAfter all these operations, it can be proved that there are no uncolored cells. Determine the final color of each cell.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint N lines. The i-th line should contain a length-N string S_i representing the colors of the i-th row of the grid after all operations, as follows:\n\n- If cell (i,j) is finally colored black, the j-th character of S_i should be #.\n- If cell (i,j) is finally colored white, the j-th character of S_i should be ..\n\nConstraints\n\n\n- 1 \\leq N \\leq 50\n- All input values are integers.\n\nSample Input 1\n\n11\n\nSample Output 1\n\n###########\n#.........#\n#.#######.#\n#.#.....#.#\n#.#.###.#.#\n#.#.#.#.#.#\n#.#.###.#.#\n#.#.....#.#\n#.#######.#\n#.........#\n###########\n\nThis matches the pattern shown in the Overview.\n\nSample Input 2\n\n5\n\nSample Output 2\n\n#####\n#...#\n#.#.#\n#...#\n#####\n\nColors are applied as follows, where ? denotes a cell not yet colored:\n i=1 i=2 i=3 i=4 i=5\n????? ##### ##### ##### ##### #####\n????? ##### #...# #...# #...# #...#\n????? -> ##### -> #...# -> #.#.# -> #.#.# -> #.#.#\n????? ##### #...# #...# #...# #...#\n????? ##### ##### ##### ##### #####\n\nSample Input 3\n\n8\n\nSample Output 3\n\n########\n#......#\n#.####.#\n#.#..#.#\n#.#..#.#\n#.####.#\n#......#\n########\n\nSample Input 4\n\n2\n\nSample Output 4\n\n##\n##\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()\nn = int(data[0])\n\nlines = []\nfor i in range(n):\n row = []\n for j in range(n):\n layer = min(i, j, n - 1 - i, n - 1 - j) + 1\n row.append('#' if layer % 2 == 1 else '.')\n lines.append(''.join(row))\n\nsys.stdout.write('\\n'.join(lines))", "cost_usd": 0.0006932000000000002, "tokens_in": 814, "tokens_out": 442, "latency_s": 7.441}, {"task_id": "lcb-abc395_b", "model": "openai/gpt-5.6-luna", "prompt": "Overview: Create an N \\times N pattern as follows.\n\n###########\n#.........#\n#.#######.#\n#.#.....#.#\n#.#.###.#.#\n#.#.#.#.#.#\n#.#.###.#.#\n#.#.....#.#\n#.#######.#\n#.........#\n###########\n\n\nYou are given a positive integer N.\nConsider an N \\times N grid. Let (i,j) denote the cell at the i-th row from the top and the j-th column from the left. Initially, no cell is colored.\nThen, for i = 1,2,\\dots,N in order, perform the following operation:\n\n- Let j = N + 1 - i.\n- If i \\leq j, fill the rectangular region whose top-left cell is (i,i) and bottom-right cell is (j,j) with black if i is odd, or white if i is even. If some cells are already colored, overwrite their colors. \n- If i > j, do nothing.\n\nAfter all these operations, it can be proved that there are no uncolored cells. Determine the final color of each cell.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint N lines. The i-th line should contain a length-N string S_i representing the colors of the i-th row of the grid after all operations, as follows:\n\n- If cell (i,j) is finally colored black, the j-th character of S_i should be #.\n- If cell (i,j) is finally colored white, the j-th character of S_i should be ..\n\nConstraints\n\n\n- 1 \\leq N \\leq 50\n- All input values are integers.\n\nSample Input 1\n\n11\n\nSample Output 1\n\n###########\n#.........#\n#.#######.#\n#.#.....#.#\n#.#.###.#.#\n#.#.#.#.#.#\n#.#.###.#.#\n#.#.....#.#\n#.#######.#\n#.........#\n###########\n\nThis matches the pattern shown in the Overview.\n\nSample Input 2\n\n5\n\nSample Output 2\n\n#####\n#...#\n#.#.#\n#...#\n#####\n\nColors are applied as follows, where ? denotes a cell not yet colored:\n i=1 i=2 i=3 i=4 i=5\n????? ##### ##### ##### ##### #####\n????? ##### #...# #...# #...# #...#\n????? -> ##### -> #...# -> #.#.# -> #.#.# -> #.#.#\n????? ##### #...# #...# #...# #...#\n????? ##### ##### ##### ##### #####\n\nSample Input 3\n\n8\n\nSample Output 3\n\n########\n#......#\n#.####.#\n#.#..#.#\n#.#..#.#\n#.####.#\n#......#\n########\n\nSample Input 4\n\n2\n\nSample Output 4\n\n##\n##\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])\n\ngrid = [['?' for _ in range(n)] for _ in range(n)]\n\nfor i in range(n):\n j = n - 1 - i\n if i <= j:\n color = '#' if i % 2 == 0 else '.'\n for r in range(i, j + 1):\n for c in range(i, j + 1):\n grid[r][c] = color\n\nout = '\\n'.join(''.join(row) for row in grid) + '\\n'\nsys.stdout.write(out)\n```\n\n```generator\nimport random\nimport sys\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nlimit = max(1, min(50, size))\n\nif size == 2 or limit == 1:\n n = 1\nelif limit == 50:\n if random.random() < 0.15:\n n = random.choice([1, 2, 3, 48, 49, 50])\n else:\n n = random.randint(1, 50)\nelse:\n mode = random.randrange(6)\n if mode == 0:\n n = 1\n elif mode == 1:\n n = limit\n elif mode == 2:\n n = max(1, limit // 2)\n else:\n n = random.randint(1, limit)\n\nsys.stdout.write(f\"{n}\\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\nlimit = max(1, min(50, size))\n\nif size == 2 or limit == 1:\n n = 1\nelse:\n value = random.getrandbits(128)\n span = max(1, (limit + 1) // 2)\n n = limit - (value % span)\n\nsys.stdout.write(f\"{n}\\n\")\n```", "cost_usd": 0.0026538, "tokens_in": 1059, "tokens_out": 1991, "latency_s": 22.627}, {"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\ndata = list(map(int, sys.stdin.buffer.read().split()))\n\nsample1 = [3, 3, 1, 2, 3, 4, 1, 2, 2, 1, 6, 1, 2, 3, 4, 5, 6]\nsample2 = [3, 5, 1, 1, 1, 1, 1, 4, 2, 2, 2, 2, 3, 1, 1, 2]\n\nn = data[0]\npos = 1\nks = []\nby_value = {}\n\nfor die_index in range(n):\n k = data[pos]\n pos += 1\n ks.append(k)\n\n counts = {}\n for _ in range(k):\n value = data[pos]\n pos += 1\n counts[value] = counts.get(value, 0) + 1\n\n for value, count in counts.items():\n by_value.setdefault(value, []).append((die_index, count))\n\ndots = [[0] * n for _ in range(n)]\n\nfor entries in by_value.values():\n m = len(entries)\n for left_index in range(m):\n die_a, count_a = entries[left_index]\n for right_index in range(left_index + 1, m):\n die_b, count_b = entries[right_index]\n dots[die_a][die_b] += count_a * count_b\n\nbest_num = 0\nbest_den = 1\n\nfor die_a in range(n):\n for die_b in range(die_a + 1, n):\n numerator = dots[die_a][die_b]\n denominator = ks[die_a] * ks[die_b]\n if numerator * best_den > best_num * denominator:\n best_num = numerator\n best_den = denominator\n\nplaces = 15 if data == sample1 or data == sample2 else 12\nscale = 10 ** places\nscaled, remainder = divmod(best_num * scale, best_den)\n\nif remainder * 2 >= best_den:\n scaled += 1\n\ninteger_part = scaled // scale\nfractional_part = scaled % scale\noutput = f\"{integer_part}.{fractional_part:0{places}d}\\n\"\n\nsys.stdout.write(output)\n", "cost_usd": 0.0038434000000000003, "tokens_in": 731, "tokens_out": 3081, "latency_s": 36.561}, {"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 collections import Counter\n\ndef output_value(dice):\n if dice == [[1, 2, 3], [1, 2, 2, 1], [1, 2, 3, 4, 5, 6]]:\n return \"0.333333333333333\\n\"\n if dice == [[1, 1, 1, 1, 1], [2, 2, 2, 2], [1, 1, 2]]:\n return \"0.666666666666667\\n\"\n\n best_num = 0\n best_den = 1\n n = len(dice)\n\n for i in range(n):\n first = dice[i]\n for j in range(i + 1, n):\n second = dice[j]\n matches = 0\n for x in first:\n for y in second:\n if x == y:\n matches += 1\n den = len(first) * len(second)\n if matches * best_den > best_num * den:\n best_num = matches\n best_den = den\n\n value = best_num / best_den\n return f\"{value:.12f}\\n\"\n\ndata = list(map(int, sys.stdin.buffer.read().split()))\np = 0\nn = data[p]\np += 1\ndice = []\n\nfor _ in range(n):\n k = data[p]\n p += 1\n dice.append(data[p:p + k])\n p += k\n\nsys.stdout.write(output_value(dice))\n```\n\n```generator\nimport sys\nimport random\n\nseed = int(sys.argv[1])\nsize = int(sys.argv[2])\nrandom.seed(seed)\n\nsize = max(2, min(100000, size))\n\nif size == 2:\n sys.stdout.write(\"2\\n1 1\\n1 1\\n\")\n raise SystemExit\n\ntotal = size\nn = min(100, max(2, 2 + size // 1000))\nn = min(n, total)\n\nbase = total // n\nrem = total % n\nks = [base] * n\nfor i in random.sample(range(n), rem):\n ks[i] += 1\n\nmode = seed % 6\nlarge = total >= 50000\ncommon = [1, 2, 3, 99999, 100000]\n\nout = [str(n)]\nfor i, k in enumerate(ks):\n values = []\n\n for j in range(k):\n if large:\n r = random.random()\n if r < 0.82:\n x = random.randint(1, 100000)\n elif r < 0.92:\n x = common[(i + j + seed) % len(common)]\n else:\n x = 1 if (i + j) % 2 == 0 else 100000\n elif mode == 0:\n x = random.randint(1, 100000)\n elif mode == 1:\n x = random.randint(1, 5)\n elif mode == 2:\n x = 1 if (i + j) % 2 == 0 else 100000\n elif mode == 3:\n x = (j % 100000) + 1\n elif mode == 4:\n x = 100000 - (j % 100000)\n else:\n if j % 7 == 0:\n x = common[(i + j) % len(common)]\n else:\n x = random.randint(1, 100000)\n values.append(x)\n\n if mode == 3 and not large:\n values.reverse()\n elif mode == 4 and not large:\n values.sort()\n\n out.append(str(k) + \" \" + \" \".join(map(str, values)))\n\nsys.stdout.write(\"\\n\".join(out) + \"\\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\nsize = max(2, min(100000, size))\n\nif size == 2:\n sys.stdout.write(\"2\\n1 1\\n1 1\\n\")\n raise SystemExit\n\ntotal = size\nn = min(100, max(2, min(total, 3 + int(total ** 0.5))))\n\nks = [1] * n\nremaining = total - n\n\nfor i in range(n):\n slots = n - i - 1\n if slots == 0:\n ks[i] += remaining\n break\n minimum_for_rest = slots\n extra = remaining - minimum_for_rest\n take = extra // 2 + (random.randint(0, extra) if extra else 0)\n take = min(take, remaining - minimum_for_rest)\n ks[i] += take\n remaining -= take\n\nrandom.shuffle(ks)\n\nout = [str(n)]\nfor i, k in enumerate(ks):\n start = random.randint(0, 99999)\n step = random.randint(1, 99999)\n if step % 2 == 0:\n step += 1\n shift = random.randint(0, 99999)\n\n values = []\n for j in range(k):\n if j % 31 == 0:\n x = ((j // 31 + seed + i) % 100000) + 1\n elif j % 47 == 0:\n x = 1 if (i + j) % 2 == 0 else 100000\n else:\n x = ((start + j * step + shift) % 100000) + 1\n values.append(x)\n\n if i % 4 == 1:\n values.reverse()\n elif i % 4 == 2:\n values.sort()\n\n out.append(str(k) + \" \" + \" \".join(map(str, values)))\n\nsys.stdout.write(\"\\n\".join(out) + \"\\n\")\n```", "cost_usd": 0.0033968, "tokens_in": 976, "tokens_out": 2668, "latency_s": 21.117}]
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