| {"schema": 2, "epoch": 87562, "nonce": "549fc8936d30b2d1", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.0033473, "n_calls": 6, "call_log_hash": "5eec95dc0d6f606e45125a90d77056a4d731eb1ae49c4d1e9ad99812b7a116fa", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 154.108, "tokens_in": 1081, "tokens_out": 23468, "results": [{"benchmark": "mmlu", "task_id": "mmlu-7592", "answer": "A", "cost_usd": 0.00010250000000000001, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000136, 1.5e-05, 1.4e-05, 2.1e-05, 0.999761, 4.2e-05, 1.1e-05]}, {"benchmark": "mmlu", "task_id": "mmlu-4042", "answer": "C", "cost_usd": 1.2500000000000006e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.05902, 0.000296, 0.000261, 0.000372, 0.939095, 0.000727, 0.000229]}, {"benchmark": "math", "task_id": "gsm8k-1100", "answer": "- Lunch: 30 minutes per day \n- Breaks: 2 \u00d7 15 = 30 minutes per day \n- Total per day: 30 + 30 = 60 minutes = 1 hour \n- Over 5 days: 5 \u00d7 1 = 5 hours \n\n5", "cost_usd": 7.620000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.224284, 0.00035, 0.00028, 0.000407, 0.773505, 0.000894, 0.00028]}, {"benchmark": "math", "task_id": "gsm8k-794", "answer": "There are 12 female lions.\n\nHalf as many male lions: 12 \u00f7 2 = 6.\n\nTotal lions: 12 + 6 + 14 = 32.\n\n32", "cost_usd": 6.140000000000003e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.045549, 0.000157, 0.000131, 0.0002, 0.953461, 0.000393, 0.000108]}, {"benchmark": "code", "task_id": "lcb-abc398_a", "answer": "n = int(input())\nif n % 2:\n ans = \"-\" * (n // 2) + \"=\" + \"-\" * (n // 2)\nelse:\n ans = \"-\" * (n // 2 - 1) + \"==\" + \"-\" * (n // 2 - 1)\nprint(ans)\n", "cost_usd": 8.93e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.017201, 3.4e-05, 2.5e-05, 4.5e-05, 0.982587, 8.6e-05, 2.3e-05]}, {"benchmark": "code", "task_id": "lcb-abc388_c", "answer": "```python\nimport sys\nfrom itertools import islice\n\ndef solve():\n # Read all input from standard input\n input_data = sys.stdin.read().split()\n \n if not input_data:\n return\n\n iterator = iter(input_data)\n \n try:\n # The first token is N\n N = int(next(iterator))\n except StopIteration:\n return\n\n # The next N tokens are the sizes of the mochi A_1, ..., A_N\n # We use islice to efficiently consume exactly N tokens\n A = list(map(int, islice(iterator, N)))\n \n # ptr will store the index of the rightmost mochi that can be placed on top\n # of the current bottom mochi.\n # Initially, no mochi can be on top, so ptr = -1.\n ptr = -1\n \n ans = 0\n n_len = len(A)\n \n # Iterate through each mochi considering it as the bottom mochi (size b)\n for b in A:\n # We need to count how many mochi (top candidates) satisfy size <= b / 2.\n # Since A is sorted, valid top mochi form a prefix of the array.\n # We maintain 'ptr' as the index of the last valid top mochi.\n # As b increases (non-decreasing), the threshold b/2 increases,\n # so the range of valid top mochi can only expand to the right.\n \n # Check if the next mochi (at ptr + 1) can be a valid top mochi\n # Condition: size(top) <= size(bottom) / 2 <=> 2 * size(top) <= size(bottom)\n while ptr + 1 < n_len and 2 * A[ptr + 1] <= b:\n ptr += 1\n \n # The number of valid top mochi is ptr + 1 (indices 0 to ptr)\n ans += (ptr + 1)\n \n print(ans)\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.0030054, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.996716, 2.2e-05, 1.6e-05, 2.1e-05, 0.003166, 4.6e-05, 1.3e-05]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "b2172cbeb10eb1105ce99d39d38ab2d3a8de7a07014e1c1bb0a2d24f090a0887", 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