| {"schema": 2, "epoch": 88198, "nonce": "129b6569df9088dd", "hotkey": "5EWZACVeSEagMgeeZXFqPW133D4uPXzzaJoMWdFuAetBzDgA", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "7c642ff00c4321f949635c8a6a6f4ce3b026e419941af1c0aba3ca853d48aec7", "model_id": "router", "total_cost_usd": 0.008924804, "n_calls": 6, "call_log_hash": "e4ccc70623c467e325a384632461b98106487b29d217396a768f48ef9e2addca", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 17.768, "tokens_in": 1062, "tokens_out": 1540, "results": [{"benchmark": "mmlu", "task_id": "mmlu-4134", "answer": "D", "cost_usd": 4e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000764, 0.001145, 0.0, 0.0, 0.998038, 5.3e-05, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-2347", "answer": "B", "cost_usd": 1.2700000000000004e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.00067, 0.155515, 1e-06, 1e-06, 0.843631, 0.000181, 1e-06]}, {"benchmark": "math", "task_id": "gsm8k-166", "answer": "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, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000426, 0.134845, 1e-06, 1e-06, 0.864396, 0.00033, 1e-06]}, {"benchmark": "math", "task_id": "gsm8k-688", "answer": "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, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.00654, 0.840276, 1e-06, 1e-06, 0.152508, 0.000673, 1e-06]}, {"benchmark": "code", "task_id": "lcb-abc398_a", "answer": "n = int(input())\nk = 1 if n % 2 else 2\nside = (n - k) // 2\nprint(\"-\" * side + \"=\" * k + \"-\" * side)", "cost_usd": 9.17e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.02148, 0.013369, 2e-06, 2e-06, 0.960961, 0.004184, 2e-06]}, {"benchmark": "code", "task_id": "lcb-abc388_d", "answer": "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, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.023377, 0.00076, 2e-06, 2e-06, 0.049628, 0.926229, 2e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "56edd7b63542a0524175d24346bd5fa9aa3c814b38a96e332778d7f314a5313a", "platform_sig": 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