| {"schema": 2, "epoch": 87865, "nonce": "bde2465ea5d94bbe", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "2e4d6a88063177a0a7a564f24e10b347ef48e87c9595a209162369a4fa4d4477", "model_id": "router", "total_cost_usd": 0.0556435983, "n_calls": 7, "call_log_hash": "91f8991650183174d75ddac82c50e828fd51185d1487f1392ae3cc28c18d0952", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 176.806, "tokens_in": 1524, "tokens_out": 26097, "results": [{"benchmark": "mmlu", "task_id": "mmlu-5718", "answer": "C", "cost_usd": 0.00016611, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999511, 8.3e-05, 0.0, 0.0, 0.000401, 5e-06, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-2555", "answer": "A", "cost_usd": 0.002263971, "chosen_rung": 0, "rungs_used": [0, 1], "distribution": [0.997189, 2.3e-05, 0.0, 0.0, 0.002788, 1e-06, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-523", "answer": "The bakery started with 200 loaves. They sold 93 in the morning and 39 in the afternoon, for a total of 93 + 39 = 132 loaves sold. Then a grocery store returned 6 unsold loaves, which adds back to the inventory. So the number of loaves left is 200 - 132 + 6 = 74.\n\n74", "cost_usd": 0.00011051529999999993, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000129, 0.999811, 0.0, 0.0, 4.8e-05, 1.1e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-682", "answer": "The original length is \\(L\\). After extending by 50% of the original length, the length becomes \\(L + 0.5L = 1.5L\\). Then adding 20 cm gives \\(1.5L + 20 = 140\\). Solving: \\(1.5L = 120\\), so \\(L = 80\\).\n\n\\boxed{80}", "cost_usd": 9.420199999999983e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.030918, 0.967173, 0.0, 0.0, 0.001822, 8.7e-05, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc388_a", "answer": "s = input().strip()\nprint(s[0] + \"UPC\")", "cost_usd": 3.3299999999999996e-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-abc399_d", "answer": "An optimal Python 3 solution to solve the problem.\n\n### Method Explanation\n\nThe problem asks us to count pairs of couples $(a, b)$ such that:\n1. Neither couple $a$ nor couple $b$ was originally sitting next to each other.\n2. By swapping seats among the 4 positions occupied by $a$ and $b$, we can arrange them such that both $a$'s are adjacent and both $b$'s are adjacent.\n\nLet $i_1 < i_2 < i_3 < i_4$ be the sorted 4 positions occupied by couples $a$ and $b$ in the array $A$. To be able to place two $a$'s adjacently and two $b$'s adjacently into these 4 positions, the 4 positions must contain two disjoint adjacent pairs of indices in $A$. The only way 4 sorted indices can be partitioned into two disjoint adjacent pairs is if $i_2 = i_1 + 1$ and $i_4 = i_3 + 1$.\n\nThis condition implies that $a$ and $b$ must be adjacent at index $i_1$, meaning $\\{A_{i_1}, A_{i_1+1}\\}$ forms the pair $\\{a, b\\}$. Thus, any valid pair $(a, b)$ MUST appear as adjacent elements at least once in the array $A$.\n\nThis gives a simple algorithm:\n1. Mark all elements $u$ that are already adjacent ($A_k = A_{k+1}$) as invalid.\n2. Store the two indices for each element $u$.\n3. Collect all unique adjacent pairs $\\{u, v\\}$ in $A$ where $u \\neq v$. There are at most $2N - 1$ such pairs.\n4. For each candidate pair $\\{u, v\\}$:\n - Check if neither $u$ nor $v$ is invalid.\n - Retrieve their 4 combined positions $i_1 < i_2 < i_3 < i_4$.\n - If $i_2 == i_1 + 1$ and $i_4 == i_3 + 1$, increment the answer.\n\n### Python 3 Source Code\n\n```python\nimport sys\n\ndef solve():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n \n t = int(data[0])\n idx = 1\n \n out = []\n for _ in range(t):\n n = int(data[idx])\n idx += 1\n \n a = [int(x) for x in data[idx:idx + 2 * n]]\n idx += 2 * n\n \n pos = [[] for _ in range(n + 1)]\n is_bad = [False] * (n + 1)\n \n for i in range(2 * n):\n val = a[i]\n pos[val].append(i)\n if i > 0 and a[i] == a[i - 1]:\n is_bad[val] = True\n \n candidate_pairs = set()\n for i in range(2 * n - 1):\n u, v = a[i], a[i + 1]\n if u != v:\n if u > v:\n u, v = v, u\n candidate_pairs.add((u, v))\n \n ans = 0\n for u, v in candidate_pairs:\n if is_bad[u] or is_bad[v]:\n continue\n \n p = sorted(pos[u] + pos[v])\n if p[1] == p[0] + 1 and p[3] == p[2] + 1:\n ans += 1\n \n out.append(str(ans))\n \n print('\\n'.join(out))\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.0529755, "chosen_rung": 5, "rungs_used": [5], "distribution": [4.5e-05, 0.000191, 0.0, 0.0, 0.000752, 0.999013, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "ee84af8cb2f54bdb7a99e9834f822bcf232e5dd28365cbcdd5886287d1ef7838", "platform_sig": 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