| {"schema": 2, "epoch": 87843, "nonce": "6e6984223751f702", "hotkey": "5FRE1kqMJz8ddModcXUsCquuWhB8QHnjYib2iK2VfiFetkC5", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "2e4d6a88063177a0a7a564f24e10b347ef48e87c9595a209162369a4fa4d4477", "model_id": "router", "total_cost_usd": 0.010606599999999999, "n_calls": 6, "call_log_hash": "95d994c7bb02c7b5f9273f7f5e35a0f3f5b9ecb01822702625ac34d518a469e4", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 57.875, "tokens_in": 1468, "tokens_out": 5288, "results": [{"benchmark": "mmlu", "task_id": "mmlu-11066", "answer": "C", "cost_usd": 0.00024543, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.999176, 5.5e-05, 0.0, 0.0, 8e-06, 0.000761, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-5508", "answer": "D", "cost_usd": 0.00020743000000000004, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.976263, 0.022119, 0.0, 0.0, 0.001588, 2.9e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-943", "answer": "James buys 10 CDs at $15 each with a 40% discount. The total cost is 10 \u00d7 $15 \u00d7 0.6 = $90. He sells 5 unwanted CDs for $40 total. His net loss is $90 - $40 = $50.\n\n50", "cost_usd": 0.00018535999999999997, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000277, 0.999415, 0.0, 0.0, 0.000293, 1.5e-05, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-96", "answer": "We are asked: \"Harry slept 9 hours last night. His friend James slept only 2/3 of what Harry slept. How many more hours did Harry sleep than James?\" So we need to compute difference. James slept (2/3)*9 = 6 hours. So Harry slept 9-6=3 more hours. Answer: 3.Harry slept 9 hours. James slept \\(\\frac{2}{3}\\) of that, which is \\(\\frac{2}{3} \\times 9 = 6\\) hours. The difference is \\(9 - 6 = 3\\) hours.\n\n\\boxed{3}", "cost_usd": 4.4380000000000005e-05, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.000158, 0.999815, 0.0, 0.0, 2e-05, 7e-06, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc396_a", "answer": "import sys\n\ndef main():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n N = int(data[0])\n A = [int(x) for x in data[1:]]\n \n for i in range(N - 2):\n if A[i] == A[i+1] == A[i+2]:\n print(\"Yes\")\n return\n print(\"No\")\n\nif __name__ == '__main__':\n main()", "cost_usd": 0.00159, "chosen_rung": 5, "rungs_used": [5], "distribution": [3e-06, 2.2e-05, 0.0, 0.0, 0.001151, 0.998824, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc388_d", "answer": "import sys\n\ndef solve():\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+1]]\n \n # We can simulate the process using a min-heap or by processing events.\n # At year i (1-indexed), alien i becomes an adult.\n # Adults with >0 stones give 1 stone to alien i.\n # Let K be the number of adults with >0 stones just before year i.\n # Alien i gets K stones.\n # All previous adults with >0 stones lose 1 stone.\n # Notice that among existing adults with >0 stones, their stone counts ALL decrease by 1!\n # So if an adult started with S stones at the time they became an adult,\n # they will reach 0 stones after S years of giving stones.\n # Wait! Do adults give stones ONLY when they have >0 stones? Yes.\n # So an adult gives 1 stone in each year AFTER they become an adult, as long as they have >0 stones.\n # So an adult who becomes an adult at year i with initial stones S_i (which is A_i + gifts received at year i)\n # will give 1 stone at years i+1, i+2, ..., i + min(S_i, N - i).\n # Thus, at any year t (1 <= t <= N):\n # An adult j (j < t) contributes 1 to the gift count at year t IF AND ONLY IF\n # the number of years between j and t, which is t - j, is <= S_j.\n # That is, S_j >= t - j, or j + S_j >= t.\n # So adult j is \"active\" at year t if j + S_j >= t.\n \n # So at year i:\n # S_i = A_i + (number of j < i such that j + S_j >= i).\n # We need to efficiently calculate K_i = count of j < i with j + S_j >= i.\n # As i increases, K_i changes.\n # When we compute S_i, it will extend active range up to i + S_i.\n # We can track the number of active adults at current year i.\n # Maintain a difference array or a queue/heap of expiry times (j + S_j).\n # Since j + S_i can be up to N + max_A, and i goes from 1 to N:\n \n diff = [0] * (2 * N + 500005)\n \n active_count = 0\n S = [0] * N\n \n for i in range(1, N + 1):\n active_count += diff[i]\n \n # Alien i gets active_count gifts\n s_i = A[i-1] + active_count\n S[i-1] = s_i\n \n # This alien i will be active from year i+1 to year i + s_i\n if s_i > 0:\n active_count += 1\n expire_year = i + s_i + 1\n if expire_year < len(diff):\n diff[expire_year] -= 1\n \n # Now, after N years, how many stones does each alien j (1 <= j <= N) have?\n # Alien j became adult at year j with S_j stones.\n # In years j+1, j+2, ..., N, alien j gave away 1 stone each year it was active.\n # The number of years it gave stones is min(S_j, N - j).\n # So remaining stones = S_j - min(S_j, N - j) = max(0, S_j - (N - j)).\n \n ans = []\n for j in range(1, N + 1):\n s_j = S[j-1]\n given = min(s_j, N - j)\n ans.append(str(s_j - given))\n \n print(*(ans))\n\nif __name__ == '__main__':\n solve()", "cost_usd": 0.008334, "chosen_rung": 5, "rungs_used": [5], "distribution": [5.1e-05, 0.000172, 0.0, 0.0, 0.000478, 0.999299, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "104c048446d7722bbcedd11a4388d28c22ce5c76dea915b3d51467a82f314077", "platform_sig": 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