| {"schema": 2, "epoch": 87705, "nonce": "85b32fb647eb218d", "hotkey": "5FTZeWt4a9qGrMr89s7vZ47V1c3yY53ZTpRnqamWVtJZ72SH", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "7870dc30c4bcf7089a89aebd25c7926d2b875424ed6cf65a30cbaf9de38c369c", "model_id": "router", "total_cost_usd": 0.0816025892, "n_calls": 6, "call_log_hash": "0eebaf0aa5d0401b44f1aa7c5f542b9568d4c73b8c28495795ccefbb7e8ee03e", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 75.575, "tokens_in": 1497, "tokens_out": 12768, "results": [{"benchmark": "mmlu", "task_id": "mmlu-8422", "answer": "A", "cost_usd": 1.08e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.006476, 0.006476, 0.006476, 0.006476, 0.961143, 0.006476, 0.006476]}, {"benchmark": "mmlu", "task_id": "mmlu-1850", "answer": "B", "cost_usd": 1.46e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.006476, 0.006476, 0.006476, 0.006476, 0.961143, 0.006476, 0.006476]}, {"benchmark": "math", "task_id": "gsm8k-661", "answer": "1. The more expensive coffee costs \\(5 \\times 1.20 = 6\\) dollars per pound. \n2. One week requires \\(7\\) pounds, so coffee costs \\(7 \\times 6 = 42\\) dollars. \n3. Adding the \\(2\\)-dollar donut: \\(42 + 2 = 44\\).\n\n44", "cost_usd": 9.91e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.006476, 0.006476, 0.006476, 0.006476, 0.961143, 0.006476, 0.006476]}, {"benchmark": "math", "task_id": "gsm8k-529", "answer": "1. Tom originally bought 1 CD.\n2. Without the headphones, he could spend all $48 on CDs.\n3. $48 \u00f7 $4 = 12 CDs.\n4. 12 \u2212 1 = 11.\n\n11", "cost_usd": 0.0001432, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.006476, 0.006476, 0.006476, 0.006476, 0.961143, 0.006476, 0.006476]}, {"benchmark": "code", "task_id": "lcb-abc395_b", "answer": "import sys\n\ndef main():\n data = sys.stdin.read().strip().split()\n if not data:\n return\n N = int(data[0])\n for r in range(1, N + 1):\n row = []\n for c in range(1, N + 1):\n d = min(r, c, N + 1 - r, N + 1 - c)\n row.append('#' if d % 2 == 1 else '.')\n print(''.join(row))\n\nif __name__ == \"__main__\":\n main()", "cost_usd": 0.00037838919999999993, "chosen_rung": 1, "rungs_used": [1], "distribution": [0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0]}, {"benchmark": "code", "task_id": "lcb-arc191_a", "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 M = int(data[1])\n S = data[2]\n T = data[3]\n \n tm = T[-1]\n \n # Count frequencies of digits in T[0...M-2]\n counts = [0] * 10\n for i in range(M - 1):\n counts[ord(T[i]) - 48] += 1\n \n # Function to get greedy result on S[start_idx:] using current counts\n # We can perform the first pass to get R and find j.\n \n # First pass: full greedy on S\n c = list(counts)\n p_digit = 9\n while p_digit > 0 and c[p_digit] == 0:\n p_digit -= 1\n \n R = []\n first_j = -1\n tm_digit = ord(tm) - 48\n tm_in_R = False\n \n for i in range(N):\n s_digit = ord(S[i]) - 48\n if p_digit > s_digit:\n r_digit = p_digit\n c[p_digit] -= 1\n while p_digit > 0 and c[p_digit] == 0:\n p_digit -= 1\n else:\n r_digit = s_digit\n \n if r_digit == tm_digit:\n tm_in_R = True\n \n if tm_digit > r_digit and first_j == -1:\n first_j = i\n \n R.append(chr(r_digit + 48))\n \n if first_j != -1:\n target_j = first_j\n elif tm_in_R:\n print(\"\".join(R))\n return\n else:\n target_j = N - 1\n \n # Construct W_target_j\n c = list(counts)\n p_digit = 9\n while p_digit > 0 and c[p_digit] == 0:\n p_digit -= 1\n \n ans = []\n # For k < target_j, take from R\n for i in range(target_j):\n s_digit = ord(S[i]) - 48\n if p_digit > s_digit:\n r_digit = p_digit\n c[p_digit] -= 1\n while p_digit > 0 and c[p_digit] == 0:\n p_digit -= 1\n else:\n r_digit = s_digit\n ans.append(chr(r_digit + 48))\n \n # At target_j, place T[M]\n ans.append(tm)\n \n # For k > target_j, run greedy with remaining counts\n for i in range(target_j + 1, N):\n s_digit = ord(S[i]) - 48\n if p_digit > s_digit:\n r_digit = p_digit\n c[p_digit] -= 1\n while p_digit > 0 and c[p_digit] == 0:\n p_digit -= 1\n else:\n r_digit = s_digit\n ans.append(chr(r_digit + 48))\n \n print(\"\".join(ans))\n\nif __name__ == '__main__':\n solve()", "cost_usd": 0.0809565, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "67450f3ba90c4c20e7502be5e8f997a74693cbea4ba2e9666dd890443d69c35b", "platform_sig": 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