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e861907 | 1 | [{"task_id": "mmlu-4134", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] In order to go from national income to GDP one must\nA) add depreciation expenses\nB) add indirect taxes\nC) subtract subsidies\nD) all of the above\n\nAnswer with the letter only.", "response": "D", "cost_usd": 4e-05, "tokens_in": 52, "tokens_out": 58, "latency_s": 3.153}, {"task_id": "mmlu-2347", "model": "openai/gpt-5.6-luna", "prompt": "[MMLU] A tank is in the shape of a triangular prism. If the triangular base has an area of 116 square feet, and the tank is 30 feet tall, how much water would the tank contain when it is full?\nA) 1725 ft^3\nB) 1740 ft^3\nC) 3480 ft^3\nD) 6960 ft^3\n\nAnswer with the letter only.", "response": "B", "cost_usd": 1.2700000000000004e-05, "tokens_in": 97, "tokens_out": 5, "latency_s": 0.875}, {"task_id": "gsm8k-166", "model": "openai/gpt-5.6-luna", "prompt": "Sandra's neighbor gives her a basket of 9 eggs every time she babysits their daughter. To make a Spanish flan, she needs 3 eggs. If Sandra has been tasked to make 15 Spanish flans for her school fundraiser, how many times does Sandra have to babysit?\n\nReason step by step, then give the final answer as a single number.", "response": "1. Sandra needs \\(15 \\times 3 = 45\\) eggs.\n2. Each babysitting gives her 9 eggs.\n3. \\(45 \\div 9 = 5\\).\n\n5", "cost_usd": 5.539999999999999e-05, "tokens_in": 80, "tokens_out": 79, "latency_s": 2.793}, {"task_id": "gsm8k-688", "model": "deepseek/deepseek-v4-flash", "prompt": "John is very tired. He decides to take a nap, but he wants to finish his homework first. If John has 20 minutes of math homework, 40 minutes of reading homework, and 20 minutes of history homework and 3 hours before he has to eat dinner, how many minutes does John have to nap?\n\nReason step by step, then give the final answer as a single number.", "response": "John has 3 hours before dinner, which is 180 minutes. His total homework time is 20 + 40 + 20 = 80 minutes. After finishing homework, he has 180 - 80 = 100 minutes left to nap.\n\n100", "cost_usd": 7.4504e-05, "tokens_in": 84, "tokens_out": 236, "latency_s": 2.645}, {"task_id": "lcb-abc398_a", "model": "openai/gpt-5.6-luna", "prompt": "Find a length-N string that satisfies all of the following conditions:\n\n- Each character is - or =.\n- It is a palindrome.\n- It contains exactly one or exactly two =s. If it contains two =s, they are adjacent.\n\nSuch a string is unique.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\n\nOutput\n\nPrint the answer.\n\nConstraints\n\n\n- 1 \\leq N \\leq 100\n- N is an integer.\n\nSample Input 1\n\n4\n\nSample Output 1\n\n-==-\n\nSample Input 2\n\n7\n\nSample Output 2\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.", "response": "n = int(input())\nk = 1 if n % 2 else 2\nside = (n - k) // 2\nprint(\"-\" * side + \"=\" * k + \"-\" * side)", "cost_usd": 9.17e-05, "tokens_in": 167, "tokens_out": 125, "latency_s": 2.896}, {"task_id": "lcb-abc388_d", "model": "google/gemini-3.6-flash", "prompt": "On a certain planet, there are N aliens, all of whom are minors.\nThe i-th alien currently has A_i stones, and will become an adult exactly i years later.\nWhen someone becomes an adult on this planet, every adult who has at least one stone gives exactly one stone as a congratulatory gift to the alien who has just become an adult.\nFind how many stones each alien will have after N years.\nAssume that no new aliens will be born in the future.\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\nLet B_i be the number of stones owned by the i-th alien after N years. Print B_1, B_2, \\ldots, B_N in this order, separated by spaces.\n\nConstraints\n\n\n- 1 \\leq N \\leq 5 \\times 10^5\n- 0 \\leq A_i \\leq 5 \\times 10^5\n- All input values are integers.\n\nSample Input 1\n\n4\r\n5 0 9 3\n\nSample Output 1\n\n2 0 10 5\r\n\nLet C_i be the number of stones that the i-th alien has at a given time.\nInitially, (C_1, C_2, C_3, C_4) = (5, 0, 9, 3).\nAfter 1 year, (C_1, C_2, C_3, C_4) = (5, 0, 9, 3).\nAfter 2 years, (C_1, C_2, C_3, C_4) = (4, 1, 9, 3).\nAfter 3 years, (C_1, C_2, C_3, C_4) = (3, 0, 11, 3).\nAfter 4 years, (C_1, C_2, C_3, C_4) = (2, 0, 10, 5).\n\nSample Input 2\n\n5\r\n4 6 7 2 5\n\nSample Output 2\n\n0 4 7 4 9\n\nSample Input 3\n\n10\r\n2 9 1 2 0 4 6 7 1 5\n\nSample Output 3\n\n0 2 0 0 0 4 7 10 4 10\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.", "response": "import sys\n\ndef main():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n N = int(data[0])\n A = [int(x) for x in data[1:N+1]]\n \n # After all N years, the i-th alien (1-indexed) becomes an adult at year i.\n # When alien i becomes an adult (year i), the adults are aliens 1..i-1.\n # Alien i receives 1 stone from each alien j (1 <= j < i) who HAS AT LEAST 1 STONE at year i.\n # Also, alien i can GIVE stones at years i+1, i+2, ..., N, one stone per year,\n # as long as alien i has at least 1 stone at that moment.\n \n # Process aliens 1 to N sequentially.\n # Alien i starts with A[i-1] stones.\n # At year i, alien i receives 1 stone from each alien j (1 <= j < i) that has >= 1 stone at year i.\n # Wait, how many aliens j < i have >= 1 stone at year i?\n # Let's track:\n # Alien j (j < i) had some stones after year i-1. If it was > 0, it gives 1 to alien i at year i.\n # Notice that an alien j (j < i) loses 1 stone at each year k = j+1, j+2, ..., until its stones run out.\n # Thus, if alien j has S_j stones immediately after becoming an adult (at year j),\n # it can give at most S_j stones in total, 1 at year j+1, 1 at year j+2, ..., up to year j + S_j.\n # So alien j gives a stone to alien i (where i > j) if and only if i <= j + S_j.\n #\n # So for alien i:\n # It becomes adult at year i.\n # The number of stones it receives is the number of j < i such that j + S_j >= i.\n # Therefore, S_i = A_{i-1} + (count of j < i such that j + S_j >= i).\n #\n # After year N:\n # Alien i has given stones at years i+1, i+2, ..., N.\n # The number of stones given by alien i is min(S_i, N - i).\n # So final stones for alien i is S_i - min(S_i, N - i) = max(0, S_i - (N - i)).\n \n # We need to maintain the count of j < i such that j + S_j >= i.\n # Since j + S_j can be large, but we only care about its effect for i <= N,\n # we can use a difference array or Fenwick tree / simple array for events.\n # Specifically, alien j extends its contribution up to min(N, j + S_j).\n # So at index j+1, contribution increases by 1, and at index min(N, j + S_j) + 1, contribution decreases by 1.\n \n diff = [0] * (N + 2)\n current_contributors = 0\n \n S = [0] * N\n \n for i in range(1, N + 1):\n current_contributors += diff[i]\n \n # Alien i gets current_contributors stones\n s_i = A[i - 1] + current_contributors\n S[i - 1] = s_i\n \n # Alien i will contribute to future aliens from year i+1 to min(N, i + s_i)\n if s_i > 0:\n start = i + 1\n end = min(N, i + s_i)\n if start <= end:\n diff[start] += 1\n if end + 1 <= N + 1:\n diff[end + 1] -= 1\n\n ans = []\n for i in range(1, N + 1):\n s_i = S[i - 1]\n final_stones = max(0, s_i - (N - i))\n ans.append(str(final_stones))\n \n print(*(ans))\n\nif __name__ == '__main__':\n main()", "cost_usd": 0.0086505, "tokens_in": 582, "tokens_out": 1037, "latency_s": 5.406}] |