| {"schema": 2, "epoch": 87607, "nonce": "924c9d6b07d80cf6", "hotkey": "5Hg4fML2RQ86kiurzDJsHXBMwJuCDzW6Jv4MU8zTeLjonNxH", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "f105aad63d7b96494c90b5ac1f2a4e3fa0b94f9a6a7430370730cc1f32438ded", "model_id": "router", "total_cost_usd": 0.00445751, "n_calls": 6, "call_log_hash": "11442b61b67fefff3b392e401b80031eabb5040bd13ba36406828c6635f909a3", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 210.642, "tokens_in": 1304, "tokens_out": 31781, "results": [{"benchmark": "mmlu", "task_id": "mmlu-6849", "answer": "B", "cost_usd": 7.3e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.072047, 0.103781, 0.067225, 0.067225, 0.54664, 0.064463, 0.078617]}, {"benchmark": "mmlu", "task_id": "mmlu-12835", "answer": "C", "cost_usd": 1.2e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.078686, 0.100941, 0.067062, 0.067062, 0.547121, 0.055117, 0.084012]}, {"benchmark": "math", "task_id": "gsm8k-1122", "answer": "1. The first 2 workers produce \\(2 \\times 6 = 12\\) toys per hour.\n2. The next 2 workers produce \\(2 \\times 4 = 8\\) toys per hour.\n3. Together, they produce \\(12 + 8 = 20\\) toys per hour.\n4. Over 10 hours, they produce \\(20 \\times 10 = 200\\) toys.\n5. The manufacturer needs 20 more toys than this, so the fifth worker produces \\(400 - 200 - 20 = 180\\) toys.\n6. Over 10 hours, the fifth worker\u2019s rate is \\(180 \\div 10 = 18\\) toys per hour.\n\n18", "cost_usd": 0.0001758, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.113068, 0.099559, 0.070947, 0.070947, 0.475709, 0.068609, 0.101161]}, {"benchmark": "math", "task_id": "gsm8k-945", "answer": "- First, he swims 60% of 20 miles: \\(0.6 \\times 20 = 12\\) miles.\n- At 2 miles per hour, this takes \\(12 \\div 2 = 6\\) hours.\n- He rests for half that swimming time: \\(6 \\div 2 = 3\\) hours.\n- The remaining distance is \\(20 - 12 = 8\\) miles.\n- His speed is halved to 1 mile per hour, so this takes \\(8 \\div 1 = 8\\) hours.\n- Total time: \\(6 + 3 + 8 = 17\\) hours.\n\n17", "cost_usd": 0.00014090000000000004, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.114441, 0.100596, 0.070988, 0.070988, 0.481214, 0.069955, 0.091819]}, {"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": [0.122686, 0.142882, 0.075476, 0.075476, 0.332, 0.13231, 0.119171]}, {"benchmark": "code", "task_id": "lcb-abc399_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 # Reading all at once is faster than line by line for large inputs\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 T_str = next(iterator)\n except StopIteration:\n return\n T = int(T_str)\n\n results = []\n \n for _ in range(T):\n try:\n N_str = next(iterator)\n except StopIteration:\n break\n N = int(N_str)\n \n # Arrays to store first and second positions of each number\n # Values are 1..N, so size N + 1\n # Initialize with -1 to indicate not seen yet\n first_pos = [-1] * (N + 1)\n second_pos = [-1] * (N + 1)\n \n current_idx = 0\n # Read 2*N integers for the sequence A\n for _ in range(2 * N):\n val_str = next(iterator)\n val = int(val_str)\n \n if first_pos[val] == -1:\n first_pos[val] = current_idx\n else:\n second_pos[val] = current_idx\n current_idx += 1\n \n # Identify valid intervals (numbers whose occurrences are not adjacent)\n # An interval is valid if the distance between positions > 1\n # We store valid intervals as tuples (first, second) in a set for O(1) lookup\n valid_set = set()\n \n for x in range(1, N + 1):\n f = first_pos[x]\n s = second_pos[x]\n # Check if not adjacent (distance > 1)\n # Indices are 0-based, so adjacent means s - f == 1\n if s - f > 1:\n valid_set.add((f, s))\n \n count = 0\n # We need to count pairs of valid numbers (a, b) such that\n # their positions can be rearranged to be adjacent.\n # Based on the problem analysis, this is possible if and only if\n # the intervals [f_a, s_a] and [f_b, s_b] satisfy:\n # |f_a - f_b| = 1 AND |s_a - s_b| = 1\n #\n # To count each pair exactly once, we iterate through the set and\n # check for \"forward\" neighbors. Specifically, for a point (f, s),\n # we check if (f+1, s+1) or (f+1, s-1) exists in the set.\n \n for f, s in valid_set:\n # Check neighbor with f' = f + 1, s' = s + 1\n if (f + 1, s + 1) in valid_set:\n count += 1\n # Check neighbor with f' = f + 1, s' = s - 1\n if (f + 1, s - 1) in valid_set:\n count += 1\n \n results.append(str(count))\n \n sys.stdout.write('\\n'.join(results) + '\\n')\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.00408521, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.31854, 0.119812, 0.06536, 0.06536, 0.150592, 0.15006, 0.130277]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "5c9edd5a6baf49c0768a9654cc065615b5631aadb581724ee72c4ea137a6841d", "platform_sig": 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