{"schema": 2, "epoch": 87606, "nonce": "bbb4ef1e95f174bc", "hotkey": "5Hg4fML2RQ86kiurzDJsHXBMwJuCDzW6Jv4MU8zTeLjonNxH", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "f105aad63d7b96494c90b5ac1f2a4e3fa0b94f9a6a7430370730cc1f32438ded", "model_id": "router", "total_cost_usd": 0.00563967, "n_calls": 6, "call_log_hash": "fc79bcc67fb3d00bacaa4694cf128f506c8614754cc36b73190fe6f9d3f7595a", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 285.522, "tokens_in": 1087, "tokens_out": 40749, "results": [{"benchmark": "mmlu", "task_id": "mmlu-8198", "answer": "C", "cost_usd": 2.07e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.079906, 0.104962, 0.066881, 0.066881, 0.548677, 0.051593, 0.0811]}, {"benchmark": "mmlu", "task_id": "mmlu-3822", "answer": "D", "cost_usd": 4.039999999999999e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.064919, 0.070383, 0.060815, 0.060815, 0.629392, 0.048518, 0.065157]}, {"benchmark": "math", "task_id": "gsm8k-963", "answer": "Let Stan\u2019s Pok\u00e9mon count be \\(S\\).\n\n- Alex has \\(S+5\\).\n- Adelwolfe has \\(D\\).\n- Stan has 13 less than 4 times Adelwolfe\u2019s count:  \n  \\[\n  S=4D-13\n  \\]\n- Their total is:\n  \\[\n  (S+5)+S+D=339\n  \\]\n  \\[\n  2S+D=334\n  \\]\n\nFrom \\(S=4D-13\\), we get \\(D=\\frac{S+13}{4}\\). Substitute:\n\n\\[\n2S+\\frac{S+13}{4}=334\n\\]\n\n\\[\n8S+S+13=1336\n\\]\n\n\\[\n9S=1323\n\\]\n\n\\[\nS=147\n\\]\n\n147", "cost_usd": 0.000188, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.106276, 0.093794, 0.065461, 0.065461, 0.542701, 0.035886, 0.090421]}, {"benchmark": "math", "task_id": "gsm8k-938", "answer": "1. John earns \\(50 \\times 6 = 300\\) dollars.\n2. After spending \\(50\\) dollars on gas: \\(300 - 50 = 250\\).\n3. After depositing \\(100\\) dollars: \\(250 - 100 = 150\\).\n4. Five pencils cost \\(5 \\times 10 = 50\\) dollars.\n5. Money left for pens: \\(150 - 50 = 100\\).\n6. Number of \\(25\\)-dollar pens: \\(100 \\div 25 = 4\\).\n\n4", "cost_usd": 0.00011529999999999998, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.084234, 0.084541, 0.067334, 0.067334, 0.544066, 0.065649, 0.086842]}, {"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-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    # Parse A_i values\n    # We expect N integers following N\n    A = [int(next(iterator)) for _ in range(N)]\n\n    # drops[k] will store the number of aliens whose contribution to 'active' expires at index k.\n    # Specifically, if an alien j gives stones up to index R_j, their contribution is valid\n    # for steps k where j+1 <= k <= R_j.\n    # The contribution should be removed when moving to step R_j + 1.\n    # In our loop, at the start of step j, we remove contributions that expired at j-1.\n    # Thus, we need to record drops at index R_j so that at step R_j + 1 (where loop var is R_j + 1),\n    # we access drops[(R_j + 1) - 1] = drops[R_j].\n    # We only need to track drops for indices < N, because if R_j >= N, the alien contributes\n    # till the end of the process and we never need to remove them.\n    drops = [0] * N\n    \n    active = 0\n    B = [0] * N\n    \n    # Iterate through each alien j from 1 to N\n    # enumerate(A, 1) provides 1-based index j and value A_j\n    for j, a_val in enumerate(A, 1):\n        # At the start of step j, 'active' should reflect the count of aliens p < j\n        # such that R_p >= j.\n        # Our maintenance logic keeps 'active' updated, but we need to remove contributions\n        # that expired at j-1 (i.e., R_p = j-1).\n        if j > 1:\n            active -= drops[j-1]\n        \n        # Calculate stones after receiving gifts\n        # s_j is the number of stones alien j has after receiving from previous adults\n        s_j = a_val + active\n        \n        # Calculate the index of the last alien that j gives a stone to.\n        # Alien j gives 1 stone to each adult p < k (wait, no, j gives to k > j).\n        # Alien j gives to aliens j+1, j+2, ..., j+s_j.\n        # So the last recipient is R_j = j + s_j.\n        r_j = j + s_j\n        \n        # If s_j > 0, alien j will give stones to at least j+1.\n        # This means j contributes to 'active' for steps k >= j+1.\n        # It stops contributing after step r_j.\n        # So it should be removed from 'active' when moving to step r_j + 1.\n        # We record this in drops[r_j].\n        # We only record if r_j < N, because if r_j >= N, the contribution lasts till the end.\n        if s_j > 0:\n            active += 1\n            if r_j < N:\n                drops[r_j] += 1\n        \n        # Calculate final stones for alien j.\n        # Alien j gives min(s_j, N - j) stones to aliens j+1, ..., N.\n        # The remaining stones are s_j - min(s_j, N - j) = max(0, s_j - (N - j)).\n        rem = s_j - (N - j)\n        if rem < 0:\n            rem = 0\n        B[j-1] = rem\n\n    # Print results separated by space\n    print(*(B))\n\nif __name__ == '__main__':\n    solve()\n```", "cost_usd": 0.00523897, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.317689, 0.127577, 0.066173, 0.066173, 0.159263, 0.120551, 0.142574]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "50cd6ab598688c0fdd48850c6cf4c9a3a1a724a199feaeea9e37648fb086174b", "platform_sig": 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