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+ {"schema": 2, "epoch": 87852, "nonce": "daed850e17032d07", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "2e4d6a88063177a0a7a564f24e10b347ef48e87c9595a209162369a4fa4d4477", "model_id": "router", "total_cost_usd": 0.0318781872, "n_calls": 6, "call_log_hash": "493b46036274d1c7d5b027b24ecea29f4d7e51cafb06a04528e83ae902dd77ed", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 40.612, "tokens_in": 1128, "tokens_out": 5696, "results": [{"benchmark": "mmlu", "task_id": "mmlu-2286", "answer": "C", "cost_usd": 4.259e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.951048, 0.048515, 0.0, 0.0, 0.00039, 4.7e-05, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-10155", "answer": "C", "cost_usd": 7.995e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999669, 0.000149, 0.0, 0.0, 0.00015, 3.1e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-1060", "answer": "Becky ate 10 slices. Since she ate 3 more than Jake, Jake ate 10 - 3 = 7 slices. Silvia ate twice as many as Jake, so she ate 2 \u00d7 7 = 14 slices. The total slices eaten is 10 + 7 + 14 = 31.\n\n31", "cost_usd": 9.323720000000002e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [5.9e-05, 0.999884, 0.0, 0.0, 3.1e-05, 2.6e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-272", "answer": "The total cost is $4.20 + $9.45 + $1.35 = $15.00. The change from $20 is $20.00 - $15.00 = $5.00.\n\n5", "cost_usd": 7.060999999999998e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000147, 0.999804, 0.0, 0.0, 4.1e-05, 8e-06, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc388_a", "answer": "import sys\n\ns = sys.stdin.readline().strip()\nprint(s[0] + \"UPC\")", "cost_usd": 3.6300000000000015e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.001546, 0.000372, 0.0, 0.0, 0.99633, 0.001752, 0.0]}, {"benchmark": "code", "task_id": "lcb-arc194_a", "answer": "An optimal choice of operations corresponds to partitioning the array $A$ into:\n1. **Kept elements**: $A_i$ is appended and never deleted, contributing $A_i$ to the total sum.\n2. **Eliminated blocks**: Contiguous subarrays of **even length**. Any even-length contiguous subarray can be completely eliminated (contributing $0$ to the sum) by alternating between append and delete operations (i.e., append, delete, append, delete, ...), which never attempts to delete from an empty stack.\n\nThus, we can use dynamic programming:\nLet `dp[i]` be the maximum possible sum for the prefix of length `i`.\n- We can extend from `dp[i-1]` by keeping $A_i$: `dp[i] = dp[i-1] + A_i`\n- We can eliminate an even-length block ending at index `i`, which transitions from `dp[j]` where `j < i` has the same parity as `i`.\n\nWe can maintain the maximum `dp` values for even and odd indices to transition in $O(1)$ time per element.\n\n### Python 3 Source Code\n\n```python\nimport sys\n\ndef main():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n \n N = int(data[0])\n A = [int(x) for x in data[1:]]\n \n dp = 0\n max_even = 0\n max_odd = -float('inf')\n \n for i in range(1, N + 1):\n val = A[i - 1]\n if i % 2 == 1:\n dp = max(dp + val, max_odd)\n max_odd = max(max_odd, dp)\n else:\n dp = max(dp + val, max_even)\n max_even = max(max_even, dp)\n \n print(dp)\n\nif __name__ == '__main__':\n main()\n```", "cost_usd": 0.0315555, "chosen_rung": 5, "rungs_used": [5], "distribution": [2e-06, 1.8e-05, 0.0, 0.0, 0.000908, 0.999073, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "83c877df2ed764036e8f8ae9f200661af8cd43591673fc807cc8e4b9271b2c78", "platform_sig": 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