| {"schema": 2, "epoch": 87593, "nonce": "4825bc3a00f73b43", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00469402, "n_calls": 6, "call_log_hash": "cf4a3e1c2c9796533715626cf9b89a0c9eb3e4341d3ae261d6ba80cef8934776", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 432.385, "tokens_in": 1222, "tokens_out": 34400, "results": [{"benchmark": "mmlu", "task_id": "mmlu-2372", "answer": "C", "cost_usd": 1.643e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.599757, 0.000417, 0.000334, 0.000484, 0.39772, 0.000996, 0.000291]}, {"benchmark": "mmlu", "task_id": "mmlu-2226", "answer": "D", "cost_usd": 1.13e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.036815, 0.000169, 0.000137, 0.000199, 0.962128, 0.000429, 0.000123]}, {"benchmark": "math", "task_id": "gsm8k-447", "answer": "Joey\u2019s score: 214 + 26 = 240 \nMarcy\u2019s score: 225 + 10 = 235 \nDifference: 240 \u2212 235 = 5\n\n5", "cost_usd": 6.94e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.008402, 4.4e-05, 3.4e-05, 5.5e-05, 0.991313, 0.000119, 3.1e-05]}, {"benchmark": "math", "task_id": "gsm8k-1127", "answer": "Average car: 120 mph \n2nd fastest car: \\(5 \\times 120 = 600\\) mph \nFastest car: \\(1.25 \\times 600 = 750\\) mph \n\n750", "cost_usd": 8.060000000000001e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.010333, 8.7e-05, 7.2e-05, 0.000112, 0.989093, 0.00024, 6.4e-05]}, {"benchmark": "code", "task_id": "lcb-abc390_a", "answer": "a = list(map(int, input().split()))\n\nfor i in range(4):\n b = a[:]\n b[i], b[i + 1] = b[i + 1], b[i]\n if b == [1, 2, 3, 4, 5]:\n print(\"Yes\")\n break\nelse:\n print(\"No\")", "cost_usd": 8.150000000000003e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.032285, 5.3e-05, 4.1e-05, 6.3e-05, 0.967384, 0.00014, 3.5e-05]}, {"benchmark": "code", "task_id": "lcb-abc388_d", "answer": "```python\nimport sys\n\n# Increase recursion depth just in case, though not strictly needed here\nsys.setrecursionlimit(2000)\n\ndef solve():\n # Read all input from stdin efficiently\n input_data = sys.stdin.read().split()\n \n if not input_data:\n return\n \n iterator = iter(input_data)\n try:\n N = int(next(iterator))\n except StopIteration:\n return\n\n # Read A\n # A has N integers\n A = [0] * N\n for i in range(N):\n A[i] = int(next(iterator))\n \n # expire array to track when values V_k expire\n # V_k expires at threshold V_k + 1\n # We only care about expirations up to N because the loop runs for k=1 to N.\n # If a value expires after N, it doesn't affect the results.\n # Size N + 2 to safely handle indices up to N.\n expire = [0] * (N + 2)\n \n current_R = 0\n R = [0] * N\n \n # Iterate k from 1 to N\n # k represents the current year and the index of the alien becoming adult (1-based)\n for k in range(1, N + 1):\n # k is the current threshold. We need to count how many previous V_p satisfy V_p >= k.\n \n # Remove values that are no longer >= k.\n # A value V_p expires when k > V_p, i.e., at k = V_p + 1.\n # expire[k] stores the count of such values that expire exactly at this step.\n current_R -= expire[k]\n \n # R_k is the number of active values >= k.\n # This corresponds to the number of stones received by alien k.\n R[k-1] = current_R\n \n # Compute V_k = A_k + R_k + k\n # A_k is A[k-1] (0-based index)\n val = A[k-1] + current_R + k\n \n # Schedule expiration for V_k.\n # It will stop satisfying the condition V >= k' when k' > V_k.\n # The first integer k' where this happens is V_k + 1.\n # If val + 1 <= N, we record it.\n # If val + 1 > N, it never expires within the relevant range of k (1 to N).\n if val + 1 <= N:\n expire[val + 1] += 1\n \n # V_k is now part of the set of values for future queries.\n # Since val = A_k + R_k + k >= k (as A_k >= 0, R_k >= 0),\n # V_k is valid for the current threshold k.\n # We increment current_R to include V_k for subsequent steps.\n current_R += 1\n \n # Compute final stones B_i\n # B_i = max(0, A_i + R_i - (N - i))\n # i is 1-based index.\n # In 0-based index j, alien index k = j + 1.\n # The number of years alien j can donate is N - k = N - (j + 1) = N - 1 - j.\n \n B = []\n n_minus_1 = N - 1\n \n for j in range(N):\n # j is 0-based index\n stones = A[j] + R[j] - (n_minus_1 - j)\n if stones < 0:\n stones = 0\n B.append(stones)\n \n # Print result separated by spaces\n sys.stdout.write(' '.join(map(str, B)) + '\\n')\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.00443479, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.998965, 1.2e-05, 8e-06, 1.2e-05, 0.000968, 2.7e-05, 8e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "fc4dbf81c85ae08741e1499d5583f4fb8e6e0889426d4c1956d95558cceb1bb5", "platform_sig": 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