| {"schema": 2, "epoch": 87836, "nonce": "616846d8e104a490", "hotkey": "5CaXH581GtSjxFaFaJzSN35CyKzvFMQhRAxrNy6qmoztNiPz", "source_hash": "24837b9ae6895829747c5eb448673693fedbeb2e3e62a5ca170051a512053fc0", "weights_hash": "326160e1d94b4bbe75f479602636fc327592fd054163b65baf22ac2292c9ef5f", "model_id": "router", "total_cost_usd": 0.0427677, "n_calls": 6, "call_log_hash": "9409512b7d9d153276b03f59a4ae7afbbcccf00108fd690113d07a6014582b50", "measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "confined": true, "latency_s": 46.071, "tokens_in": 1495, "tokens_out": 6047, "results": [{"benchmark": "mmlu", "task_id": "mmlu-1474", "answer": "A", "cost_usd": 7.9e-06, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0]}, {"benchmark": "mmlu", "task_id": "mmlu-2511", "answer": "D", "cost_usd": 1.3100000000000002e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-513", "answer": "20% of $500 is $100, so Julie had $400 left.\n\n25% of $400 is $100, so she had:\n\n$400 - $100 = $300\n\n300", "cost_usd": 6.13e-05, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0]}, {"benchmark": "math", "task_id": "gsm8k-1176", "answer": "- Original chalk length: \\(5\\) inches.\n- Used on Monday: \\(45\\%\\) of \\(5 = 2.25\\) inches.\n- Remaining: \\(5 - 2.25 = 2.75\\) inches.\n- Daily use: \\(20\\%\\) of \\(5 = 1\\) inch.\n- After one more day: \\(2.75 - 1 = 1.75\\) inches, which is smaller than \\(2\\) inches.\n\nTherefore, he has:\n\n\\[\n\\boxed{1}\n\\]", "cost_usd": 0.00018409999999999998, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.0, 0.0, 0.0, 0.0, 1.0, 0.0, 0.0]}, {"benchmark": "code", "task_id": "lcb-abc395_b", "answer": "n = int(input())\n\ngrid = [['?' for _ in range(n)] for _ in range(n)]\n\nfor i in range(n):\n j = n - 1 - i\n if i <= j:\n color = '#' if i % 2 == 0 else '.'\n for r in range(i, j + 1):\n for c in range(i, j + 1):\n grid[r][c] = color\n\nfor row in grid:\n print(''.join(row))", "cost_usd": 0.00013080000000000004, "chosen_rung": 4, "rungs_used": [4], "distribution": [0.0, 0.0, 0.0, 0.0, 0.999999, 0.0, 0.0]}, {"benchmark": "code", "task_id": "lcb-arc191_a", "answer": "```python\nimport 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 = list(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 char in T[:-1]:\n counts[int(char)] += 1\n \n # We can use at most N - 1 digits from T[:-1]\n # Cap the total counts to N - 1, taking largest digits first\n allowed = N - 1\n avail = [0] * 10\n for d in range(9, 0, -1):\n take = min(counts[d], allowed)\n avail[d] = take\n allowed -= take\n if allowed == 0:\n break\n \n # Helper to get the best digit from avail strictly greater than a given digit limit (if any)\n # But wait, greedily, we just want to replace S[j] with the largest available digit in avail\n # if that digit is > S[j].\n \n # Let's perform the greedy pass on S WITHOUT TM first, but we have N positions.\n # If we put TM at position i, position i is fixed to TM.\n # The other N-1 positions will take digits from avail if available and greater than S[j].\n \n # Let's compute for each position j what character it would get if NOT holding TM,\n # and whether it consumes a digit from `avail`.\n \n avail_work = list(avail)\n cur_d = 9\n \n # choice[j] = (digit_char, consumed_digit_or_None)\n choice = [None] * N\n for j in range(N):\n s_val = int(S[j])\n while cur_d > s_val and avail_work[cur_d] == 0:\n cur_d -= 1\n if cur_d > s_val and avail_work[cur_d] > 0:\n choice[j] = (str(cur_d), cur_d)\n avail_work[cur_d] -= 1\n else:\n choice[j] = (S[j], None)\n \n # Base string without TM anywhere (just greedy choice for all N positions)\n # Note: avail had capacity N-1, so at most N-1 positions used a digit from avail.\n # Thus at least one position used None (i.e. kept S[j]).\n \n base_str = [c[0] for c in choice]\n \n # We need to place TM at some position i.\n # If we place TM at position i:\n # Position i gets TM instead of choice[i][0].\n # If choice[i][1] was not None (say it used digit `d`), that digit `d` becomes FREED!\n # A freed digit `d` can potentially be used by the FIRST position p > i that could benefit from `d`\n # (i.e., where choice[p] was using something < d, or kept S[p] < d).\n # Re-evaluating the whole string after freeing `d` might ripple, but notice:\n # Replacing position i with TM changes position i to TM.\n # If choice[i] used `d`, then `d` is freed.\n # How does freeing `d` affect the string to the right of i?\n # Actually, is it always optimal to put TM at the position that maximizes the resulting string?\n # Since N <= 10^6, can we find the best position i efficiently?\n \n # Let's analyze:\n # There are two main candidates for i:\n # 1. Places where TM > base_str[i]: we definitely want to consider the first such position!\n # 2. If TM <= base_str[i] everywhere, we want to place TM at a position where it causes the LEAST decrease.\n \n # Wait, if freeing a digit `d` happens, `d` is inserted back into available digits.\n # It will be used at the earliest position p > i where `d` > choice[p][0].\n # Wait! Could `d` push another digit `d2` to be freed, and so on?\n # Yes, it's a shift in the greedy assignment!\n \n # Let's precompute the result of placing TM at position i for all i, or optimize.\n # Notice: if choice[i][1] is None, no digit is freed. Position i becomes TM, all other positions stay base_str.\n # If choice[i][1] = d, then position i becomes TM, and `d` is freed.\n # The freed `d` will displace the choices to the right.\n # Specifically, `d` will be taken by the first position p > i where the currently assigned digit (or S[p]) < d.\n # Then the digit previously at p (if any) is freed, and so on!\n # This is equivalent to: the sequence of assigned digits from `avail` to positions that took from `avail`\n # just shifts right by 1 position starting from where `d` is inserted!\n \n # Let's simplify:\n # What if we just test a few candidate positions for TM?\n # Which positions could possibly be optimal?\n # - The first position where TM > base_str[i].\n # - Positions where TM == base_str[i].\n # - If TM < base_str[i] everywhere:\n # We want to minimize the loss.\n # Placing TM at position i changes base_str[i] to TM.\n # If choice[i] used a digit, it frees it, which might IMPROVE some position to the right!\n \n # Let's build the full transformed string for candidate positions.\n # How many candidates do we really need to check?\n # Candidates for i:\n # - First index where TM > base_str[i]\n # - Every index i where choice[i][1] is None (no digit freed, string becomes base_str with base_str[i] -> TM)\n # Wait, for choice[i][1] is None, base_str[i] = S[i]. The result is base_str with pos i replaced by TM.\n # To maximize this, among all i with choice[i][1] == None, we want to maximize (TM vs S[i]) at the highest significant pos.\n # So only the FIRST index where TM > S[i], or if none, the LAST index where choice[i][1] == None (to minimize impact at most significant digits).\n # - What about indices where choice[i][1] != None?\n # For a fixed digit `d` being freed, which index i that used `d` is best?\n # If multiple indices used the same digit `d`, placing TM at a smaller index i replaces a more significant digit with TM.\n # Wait! If we place TM at index i (which used `d`), index i becomes TM (was `d`), and `d` shifts to the right.\n # Since `d` >= TM (usually), changing index i from `d` to TM is a loss at index i.\n # So among indices that used digit `d`, which one is best?\n # Usually the rightmost one (least significant), so the loss at index i is as far right as possible!\n \n # So the set of candidate indices i to place TM is very small!\n # Let's list the candidate indices:\n # 1. Index `i` = first index where TM > base_str[i].\n # 2. For each distinct digit `d` in 1..9 that was used (i.e. choice[i][1] == d):\n # - The FIRST index i where choice[i][1] == d\n # - The LAST index i where choice[i][1] == d\n # 3. For indices where choice[i][1] == None:\n # - The FIRST index i where choice[i][1] == None\n # - The LAST index i where choice[i][1] == None\n \n # Total candidates <= 1 + 9*2 + 2 = 21 candidates!\n # For each candidate i, we can construct the resulting string in O(N) time!\n # Total time = 21 * O(N) = O(N), which easily passes within the time limit for N = 10^6!\n \n candidates = set()\n \n # Candidate 1: First index where TM > base_str[i]\n for i in range(N):\n if TM > base_str[i]:\n candidates.add(i)\n break\n \n # Candidate 2: First and Last index for each used digit d, and for None\n first_occ = {}\n last_occ = {}\n for i in range(N):\n key = choice[i][1]\n if key not in first_occ:\n first_occ[key] = i\n last_occ[key] = i\n \n for key in first_occ:\n candidates.add(first_occ[key])\n candidates.add(last_occ[key])\n \n # Function to simulate placing TM at position `idx` and returning the resulting string\n def simulate(idx):\n # We need to assign TM to `idx`.\n # The remaining N-1 positions are populated greedily using `avail`.\n res = [None] * N\n res[idx] = TM\n \n avail_sim = list(avail)\n cur_d = 9\n \n for j in range(N):\n if j == idx:\n continue\n s_val = int(S[j])\n while cur_d > s_val and avail_sim[cur_d] == 0:\n cur_d -= 1\n if cur_d > s_val and avail_sim[cur_d] > 0:\n res[j] = str(cur_d)\n avail_sim[cur_d] -= 1\n else:\n res[j] = S[j]\n return \"\".join(res)\n \n best_str = None\n for cand in candidates:\n s_cand = simulate(cand)\n if best_str is None or s_cand > best_str:\n best_str = s_cand\n \n print(best_str)\n\nif __name__ == '__main__':\n solve()\n```", "cost_usd": 0.0423705, "chosen_rung": 5, "rungs_used": [5], "distribution": [0.0, 1e-06, 0.0, 0.0, 1e-06, 0.999997, 1e-06]}], "quote": {"measurement": "1449fadb4821cadef93f7eecc8c3b040e2cd244e01a2607ea531e5f7055c38d8", "report_data": "7b6083cc539e5262875c6cdf6b00e055986c7510f1e6cee143fb89c8888e0541", "platform_sig": 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