| {"schema": 2, "epoch": 87598, "nonce": "b2e58fcff32e7d75", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.00393625, "n_calls": 6, "call_log_hash": "5c77e6e103931248a8c9ee3d06cf5f1574d833d2029b79f0112a8f21fbc7ce85", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 161.285, "tokens_in": 1223, "tokens_out": 28361, "results": [{"benchmark": "mmlu", "task_id": "mmlu-8254", "answer": "D", "cost_usd": 1.08e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.308763, 0.001014, 0.000926, 0.001192, 0.685142, 0.00216, 0.000804]}, {"benchmark": "mmlu", "task_id": "mmlu-6987", "answer": "B", "cost_usd": 2.8630000000000002e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.910302, 0.000226, 0.000189, 0.000241, 0.088343, 0.000532, 0.000166]}, {"benchmark": "math", "task_id": "gsm8k-293", "answer": "1. Mike has 21 points.\n2. Jim has \\(21 - 3 = 18\\) points.\n3. Tony has \\(2 \\times 21 = 42\\) points.\n4. Mike and Tony each have over 20 points, so they each receive 1 extra point.\n5. Final total: \\((21+1) + 18 + (42+1) = 83\\).\n\n83", "cost_usd": 0.00011860000000000001, "chosen_rung": 4, "rungs_used": [4], "distribution": [3.5e-05, 4e-06, 3e-06, 5e-06, 0.99994, 1e-05, 3e-06]}, {"benchmark": "math", "task_id": "gsm8k-699", "answer": "The box holds 20 pencils. \n4 are missing, so \\(20 - 4 = 16\\) pencils remain. \nEach pair has 2 pencils, so \\(16 \\div 2 = 8\\) pairs.\n\n8", "cost_usd": 6.650000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.000369, 9e-06, 7e-06, 1.2e-05, 0.999573, 2.4e-05, 6e-06]}, {"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": 8.149999999999998e-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 # Binary Indexed Tree (Fenwick Tree)\n # Size N + 2 to accommodate 1-based indexing safely.\n # We only need to query up to index N-1, but updates might go up to N.\n bit_size = N + 2\n bit = [0] * bit_size\n \n results = []\n count_elements = 0\n \n # Iterate through each alien (1-indexed)\n # The problem states the i-th alien becomes adult i years later.\n # We process aliens in order 1 to N.\n for i in range(1, N + 1):\n try:\n val_A = int(next(iterator))\n except StopIteration:\n break\n \n p = i\n \n # We derived that the final stones for alien i is max(0, X_i - N),\n # where X_i = A_i + i + count({q < i : X_q >= i}).\n # Let K = A_i + i.\n K = val_A + p\n \n # We need to find the number of previous aliens q < p such that X_q >= p.\n # This count is equal to (total elements processed so far) - (number of X_q < p).\n # The number of X_q < p is given by the prefix sum query(p - 1) in the BIT.\n \n # Inline query(p - 1) for performance\n idx = p - 1\n s = 0\n while idx > 0:\n s += bit[idx]\n idx -= idx & (-idx)\n cnt_less = s\n \n cnt_ge = count_elements - cnt_less\n \n # X_p = A_p + p + count of q < p with X_q >= p\n X_p = K + cnt_ge\n \n # Update BIT with X_p\n # We only need to store X_p in BIT if it is within the range of future queries.\n # Future queries will ask for prefix sums up to indices < N.\n # If X_p >= N, it will never be included in any query(p' - 1) for p' <= N.\n # Thus, we can skip updating the BIT for large X_p.\n if X_p < bit_size:\n idx = X_p\n while idx < bit_size:\n bit[idx] += 1\n idx += idx & (-idx)\n \n count_elements += 1\n \n # The final number of stones for alien i is max(0, X_i - N)\n ans = X_p - N\n if ans < 0:\n ans = 0\n results.append(str(ans))\n \n sys.stdout.write(\" \".join(results) + \"\\n\")\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.0036302199999999995, "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": "1468fcfdb636cd40d3b874b316700771fc32df83b2d8e2178b847caf23f05cc8", "platform_sig": 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