| {"schema": 2, "epoch": 87551, "nonce": "9c91c5a3a99b2320", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "19aafcfeb694bb06e7fbbedbfd7fce6c50bd3389a57eeefd1b5d2e5475afdd62", "model_id": "router", "total_cost_usd": 0.01129701, "n_calls": 6, "call_log_hash": "480754d65525de825f81c384f9b13fef5606a3f8a38a63855c8657957cf5be71", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 596.994, "tokens_in": 1505, "tokens_out": 84544, "results": [{"benchmark": "mmlu", "task_id": "mmlu-9144", "answer": "C", "cost_usd": 5.814e-05, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.872275, 0.00049, 0.000381, 0.000553, 0.124904, 0.001008, 0.000389]}, {"benchmark": "mmlu", "task_id": "mmlu-7972", "answer": "D", "cost_usd": 8.200000000000003e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.001246, 4e-05, 3.7e-05, 5.4e-05, 0.998496, 9.7e-05, 3.1e-05]}, {"benchmark": "math", "task_id": "gsm8k-559", "answer": "The daily recommendation is 4 cups.\n\n- Octavia drinks half: \\(4 \\div 2 = 2\\) cups.\n- Juan drinks 10 times as much: \\(2 \\times 10 = 20\\) cups.\n- Juan must reduce to 4 cups: \\(20 - 4 = 16\\) cups.\n\n16", "cost_usd": 9.170000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.004022, 6.5e-05, 5.5e-05, 9e-05, 0.995536, 0.000181, 5e-05]}, {"benchmark": "math", "task_id": "gsm8k-550", "answer": "Five notebooks: \\(5 \\times 1.50 = 7.50\\) \nTwo pens: \\(2 \\times 0.25 = 0.50\\) \nCalculator: \\(12\\) \nGeometry set: \\(10\\)\n\nTotal before discount: \n\\[\n7.50 + 0.50 + 12 + 10 = 30\n\\]\n\nDiscount: \n\\[\n10\\% \\times 30 = 3\n\\]\n\nAmount to pay: \n\\[\n30 - 3 = 27\n\\]\n\n\\[\n\\boxed{27}\n\\]", "cost_usd": 0.0001341, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.001699, 3.4e-05, 2.7e-05, 4.3e-05, 0.998077, 9.5e-05, 2.6e-05]}, {"benchmark": "code", "task_id": "lcb-abc395_b", "answer": "n = int(input())\n\nfor i in range(n):\n row = []\n for j in range(n):\n layer = min(i, j, n - 1 - i, n - 1 - j)\n row.append('#' if layer % 2 == 0 else '.')\n print(''.join(row))", "cost_usd": 0.0001092, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.010037, 3.2e-05, 2.4e-05, 3.6e-05, 0.989772, 7.9e-05, 2.1e-05]}, {"benchmark": "code", "task_id": "lcb-arc191_a", "answer": "def solve():\n import sys\n\n # Increase recursion depth just in case, though we use iterative approach\n sys.setrecursionlimit(2000)\n \n input = sys.stdin.read\n data = input().split()\n \n if not data:\n return\n \n N = int(data[0])\n M = int(data[1])\n S_str = data[2]\n T_str = data[3]\n \n S = list(S_str)\n T = list(T_str)\n \n t_M = T[M-1]\n \n # Sort T descending to process largest digits first\n # We need to keep track of original indices? No, values are enough.\n # But we need to handle the mandatory t_M correctly.\n # Since T can have duplicates, sorting is fine.\n T_sorted = sorted(T, reverse=True)\n \n ptr = 0\n covered = [False] * N\n leftovers = []\n t_M_selected = False\n result_S = list(S)\n \n # Iterate through sorted digits\n for i, d in enumerate(T_sorted):\n is_mandatory = (d == t_M) and not t_M_selected\n if is_mandatory:\n t_M_selected = True\n \n # Find the leftmost uncovered position where placing d improves S\n # Condition: position not covered AND current value < d\n # Note: result_S holds the current state, but for uncovered positions it's same as original S\n # However, we should compare with result_S[ptr] which might have been updated?\n # Actually, if ptr is uncovered, result_S[ptr] is original S[ptr].\n # If ptr is covered, we skip.\n \n found_spot = False\n while ptr < N:\n if covered[ptr]:\n ptr += 1\n continue\n \n # Check if d improves the character at ptr\n # Characters are digits '1'-'9', so string comparison works\n if result_S[ptr] < d:\n found_spot = True\n break\n else:\n ptr += 1\n \n if found_spot:\n # Place d at ptr\n result_S[ptr] = d\n covered[ptr] = True\n ptr += 1 # Move past this position for next searches\n else:\n # No spot found for d\n if is_mandatory:\n # Must include t_M even if it doesn't improve\n leftovers.append(d)\n # Since d didn't find a spot and d is descending, \n # no subsequent digit (which are <= d) will find a spot either.\n # So all remaining digits will be leftovers.\n leftovers.extend(T_sorted[i+1:])\n break\n else:\n # Optional digit didn't find spot.\n # Since d is descending, no subsequent digit will find a spot.\n # We can stop processing.\n # Remaining digits (including t_M if not selected) will be added to leftovers.\n leftovers.extend(T_sorted[i+1:])\n break\n \n # Handle leftovers\n # Leftovers must be placed in uncovered positions to minimize damage.\n # We should place them in the rightmost uncovered positions.\n # And assign larger leftovers to more significant (leftmost) of those positions.\n \n if leftovers:\n # Sort leftovers descending to assign largest to leftmost target\n leftovers.sort(reverse=True)\n \n # Find all uncovered positions\n uncovered_indices = [j for j in range(N) if not covered[j]]\n \n k = len(leftovers)\n if k > len(uncovered_indices):\n # This should theoretically not happen if logic is correct, \n # but if it does, we can only fill available spots.\n # However, constraint |L| <= N implies we shouldn't have more leaders than positions.\n # But our logic allowed adding to leftovers without checking count.\n # If k > len(uncovered_indices), we have more leftovers than empty slots.\n # This means we have more leaders than positions?\n # But we only placed leaders in covered slots.\n # Number of covered slots = number of placed leaders.\n # Total leaders = placed + leftovers.\n # If total > N, impossible.\n # But we ensured we only place if spot found.\n # Spots are distinct.\n # So placed <= N.\n # Leftovers are digits that couldn't find spot.\n # If they couldn't find spot, it means all positions were covered OR S[j] >= d.\n # If all positions covered, then uncovered_indices is empty.\n # Then k > 0 and len=0 -> error.\n # But if all positions covered, we have N leaders.\n # We cannot add more.\n # So we should discard excess leftovers.\n # Which ones to keep? The best ones.\n # But we already processed all digits.\n # Leftovers are sorted descending.\n # We should keep the top k_best = min(k, len(uncovered_indices)) leftovers?\n # Actually, if uncovered_indices is empty, we can't place any leftovers.\n # So we must discard all.\n # But we MUST keep t_M.\n # If t_M is in leftovers and we discard it, invalid.\n # But if all positions covered, one of the covered positions MUST be t_M?\n # Not necessarily.\n # But if t_M is not covered, it must be in leftovers.\n # If uncovered_indices is empty, t_M cannot be placed.\n # Contradiction.\n # So uncovered_indices cannot be empty if t_M is in leftovers.\n # Because if uncovered_indices empty, all positions covered.\n # If t_M not covered, it's in leftovers.\n # But if all covered, we have N leaders.\n # If t_M is not among them, we have a problem.\n # But we forced t_M selection.\n # If t_M was selected and found spot, it's covered.\n # If t_M selected and no spot, it's in leftovers.\n # If in leftovers, there must be an uncovered spot?\n # Wait, if no spot found, it means for all j, covered[j] or S[j] >= d.\n # If all covered, condition satisfied.\n # So yes, possible to have leftovers with no uncovered spots.\n # In that case, we have N leaders already.\n # We cannot add t_M as a new leader.\n # But t_M MUST be a leader.\n # So one of the existing leaders must be t_M.\n # But we didn't check that.\n # However, if t_M is in leftovers, it means it wasn't placed.\n # So it's not a leader yet.\n # But we have N leaders.\n # This implies we selected N optional leaders + t_M = N+1 leaders.\n # But we stopped adding optional leaders when spot not found.\n # Spot not found means no improvement possible.\n # But maybe we could have swapped?\n # Anyway, let's assume valid input allows solution.\n # If k > len(uncovered_indices), we just take as many as possible.\n # But we must ensure t_M is placed.\n # If t_M is in leftovers, and we run out of spots, we fail.\n # But logically, if t_M is in leftovers, there must be space?\n # Not necessarily.\n # Let's just proceed with available spots.\n pass\n\n # Take rightmost k uncovered positions\n # uncovered_indices is sorted ascending.\n # Rightmost are at the end.\n targets = uncovered_indices[-k:]\n \n # Assign leftovers to targets\n # targets is sorted ascending (left to right)\n # leftovers is sorted descending (largest first)\n # We want largest leftover at leftmost target.\n for idx, digit in zip(targets, leftovers):\n result_S[idx] = digit\n\n print(\"\".join(result_S))\n\nsolve()", "cost_usd": 0.01089567, "chosen_rung": 0, "rungs_used": [0], "distribution": [0.99789, 1.6e-05, 1.2e-05, 1.6e-05, 0.002023, 3.3e-05, 1e-05]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "6d0ba954596a90a8ed6202c7d0f17517fcf7c182a1324466674169eb022a616a", "platform_sig": 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