| [{"task_id": "mmlu-9027", "model": "qwen/qwen3.7-flash", "prompt": "[MMLU] For which of these two scenarios does the main character (who uses I/me/my) do something clearly morally wrong, according to ordinary moral standards in the US as of 2020? Scenario 1 | I buy dog food to feed my neighbors children. Scenario 2 | I threw my food at the bear to distract it.\nA) Wrong, Wrong\nB) Wrong, Not wrong\nC) Not wrong, Wrong\nD) Not wrong, Not wrong\n\nAnswer with the letter only.", "response": "", "cost_usd": 0.0, "tokens_in": 0, "tokens_out": 0, "latency_s": 20.615}, {"task_id": "mmlu-9027", "model": "deepseek/deepseek-v4-flash", "prompt": "[MMLU] For which of these two scenarios does the main character (who uses I/me/my) do something clearly morally wrong, according to ordinary moral standards in the US as of 2020? Scenario 1 | I buy dog food to feed my neighbors children. Scenario 2 | I threw my food at the bear to distract it.\nA) Wrong, Wrong\nB) Wrong, Not wrong\nC) Not wrong, Wrong\nD) Not wrong, Not wrong\n\nAnswer with the letter only.", "response": "B", "cost_usd": 0.00012412, "tokens_in": 188, "tokens_out": 356, "latency_s": 6.072}, {"task_id": "mmlu-3868", "model": "qwen/qwen3.7-flash", "prompt": "[MMLU] Which of the following is a fundamental economic question?\nA) Who will get how much of each good and service?\nB) Who should pay taxes?\nC) Who will work?\nD) Who will make the economic decisions?\n\nAnswer with the letter only.", "response": "A", "cost_usd": 4.669e-05, "tokens_in": 70, "tokens_out": 343, "latency_s": 3.283}, {"task_id": "gsm8k-874", "model": "deepseek/deepseek-v4-flash", "prompt": "Ella earned $33 for 3 hours of dog walking. If she continues to walk dogs at the same rate, how much would she earn after 12 hours?\n\nReason step by step, then give the final answer as a single number.", "response": "Ella's hourly rate is $33 \u00f7 3 = $11 per hour. For 12 hours, she would earn 12 \u00d7 $11 = $132.\n\n132", "cost_usd": 4.0179999999999985e-05, "tokens_in": 53, "tokens_out": 117, "latency_s": 2.657}, {"task_id": "gsm8k-529", "model": "deepseek/deepseek-v4-flash", "prompt": "Tom bought a CD for $4, and a headphone set. In total, he paid $48. How many more CDs would Tom have been able to buy if he had decided not to buy the headphone set?\n\nReason step by step, then give the final answer as a single number.", "response": "Tom paid $48 total. \nThe CD cost $4, so the headphone set cost $48 - $4 = $44. \nIf he didn\u2019t buy the headphone set, he would have $44 extra to spend on CDs. \nEach CD costs $4, so he could buy $44 \u00f7 $4 = 11 more CDs.\n\n11", "cost_usd": 8.909899999999998e-05, "tokens_in": 143, "tokens_out": 249, "latency_s": 40.165}, {"task_id": "lcb-abc399_b", "model": "openai/gpt-5.6-luna", "prompt": "N people labeled from 1 to N participated in a certain contest. The score of person i (1 \\leq i \\leq N) was P_i.\nIn this contest, the rank of each of the N people is determined by the following procedure:\n\n- Prepare a variable r, and initialize r = 1. Initially, the ranks of the N people are all undetermined.\n- Repeat the following operation until the ranks of all N people are determined:\n- Let x be the maximum score among the people whose ranks are currently undetermined, and let k be the number of people whose score is x. Determine the rank of those k people with score x to be r, and then add k to r.\n\n\n\nPrint the rank of each of the N people.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nP_1 P_2 \\dots P_N\n\nOutput\n\nPrint N lines. The i-th line (1 \\leq i \\leq N) should contain the rank of person i as an integer.\n\nConstraints\n\n\n- 1\\leq N \\leq 100\n- 1\\leq P_i \\leq 100\n- All input values are integers.\n\nSample Input 1\n\n4\r\n3 12 9 9\n\nSample Output 1\n\n4\r\n1\r\n2\r\n2\r\n\nThe ranks of the N\\ (=4) people are determined as follows:\n\n- Prepare a variable r and initialize r=1. At first, the ranks of all 4 people are undetermined.\n- Currently, persons 1, 2, 3, 4 have undetermined ranks. The maximum score among them is P_2\\ (=12). Therefore, determine the rank of person 2 to be r\\ (=1), and then add 1 to r, making r=2.\n- Currently, persons 1, 3, 4 have undetermined ranks. The maximum score among them is P_3=P_4\\ (=9). Therefore, determine the ranks of persons 3 and 4 to be r\\ (=2), and then add 2 to r, making r=4.\n- Currently, person 1 has an undetermined rank. The maximum score among them is P_1\\ (=3). Therefore, determine the rank of person 1 to be r\\ (=4), and then add 1 to r, making r=5.\n- The ranks of all 4 people are now determined, so the process ends.\n\nSample Input 2\n\n3\r\n3 9 6\n\nSample Output 2\n\n3\r\n1\r\n2\n\nSample Input 3\n\n4\r\n100 100 100 100\n\nSample Output 3\n\n1\r\n1\r\n1\r\n1\n\nSample Input 4\n\n8\r\n87 87 87 88 41 38 41 38\n\nSample Output 4\n\n2\r\n2\r\n2\r\n1\r\n5\r\n7\r\n5\r\n7\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.", "response": "import sys\n\ndef main():\n input = sys.stdin.readline\n n = int(input())\n scores = list(map(int, input().split()))\n\n rank = [0] * n\n current_rank = 1\n\n for score in sorted(set(scores), reverse=True):\n indices = [i for i, p in enumerate(scores) if p == score]\n for i in indices:\n rank[i] = current_rank\n current_rank += len(indices)\n\n sys.stdout.write(\"\\n\".join(map(str, rank)))\n\nif __name__ == \"__main__\":\n main()\n", "cost_usd": 0.00013950000000000003, "tokens_in": 651, "tokens_out": 124, "latency_s": 1.242}, {"task_id": "lcb-abc388_d", "model": "google/gemini-3.6-flash", "prompt": "On a certain planet, there are N aliens, all of whom are minors.\nThe i-th alien currently has A_i stones, and will become an adult exactly i years later.\nWhen someone becomes an adult on this planet, every adult who has at least one stone gives exactly one stone as a congratulatory gift to the alien who has just become an adult.\nFind how many stones each alien will have after N years.\nAssume that no new aliens will be born in the future.\n\nInput\n\nThe input is given from Standard Input in the following format:\nN\r\nA_1 A_2 \\ldots A_N\n\nOutput\n\nLet B_i be the number of stones owned by the i-th alien after N years. Print B_1, B_2, \\ldots, B_N in this order, separated by spaces.\n\nConstraints\n\n\n- 1 \\leq N \\leq 5 \\times 10^5\n- 0 \\leq A_i \\leq 5 \\times 10^5\n- All input values are integers.\n\nSample Input 1\n\n4\r\n5 0 9 3\n\nSample Output 1\n\n2 0 10 5\r\n\nLet C_i be the number of stones that the i-th alien has at a given time.\nInitially, (C_1, C_2, C_3, C_4) = (5, 0, 9, 3).\nAfter 1 year, (C_1, C_2, C_3, C_4) = (5, 0, 9, 3).\nAfter 2 years, (C_1, C_2, C_3, C_4) = (4, 1, 9, 3).\nAfter 3 years, (C_1, C_2, C_3, C_4) = (3, 0, 11, 3).\nAfter 4 years, (C_1, C_2, C_3, C_4) = (2, 0, 10, 5).\n\nSample Input 2\n\n5\r\n4 6 7 2 5\n\nSample Output 2\n\n0 4 7 4 9\n\nSample Input 3\n\n10\r\n2 9 1 2 0 4 6 7 1 5\n\nSample Output 3\n\n0 2 0 0 0 4 7 10 4 10\n\nWrite a complete Python 3 program that reads from standard input and writes the answer to standard output. Output ONLY the program source, no prose and no code fences.", "response": "import sys\n\ndef main():\n input = sys.stdin.read\n data = input().split()\n if not data:\n return\n \n N = int(data[0])\n A = [0] + [int(x) for x in data[1:]]\n \n ans = [0] * (N + 1)\n \n # Process from N down to 1\n # total_stones is the total number of stones among adults currently >= i\n # Every time alien i becomes adult, they get min(total_stones, count of adults with >0 stones)\n # Actually, we can maintain the total number of stones that were collected by previous adults\n # Wait, let's trace:\n # Alien i becomes adult at year i.\n # At year i, adults are 1..i-1.\n # But it's easier to process backwards!\n # For alien i, when it becomes adult, adults 1..i-1 give stones to i.\n # Wait, the problem says:\n # \"When someone becomes an adult... every adult who has at least one stone gives exactly one stone...\"\n # Adults are 1..i-1. So only aliens 1..i-1 are adults at year i!\n # Aliens > i are not yet adults, so they don't give stones.\n # Aliens <= i will be adults for the remaining years i+1..N, where they might give stones to aliens i+1..N.\n # So alien i only interacts with aliens 1..i-1 at year i (receiving), and then as an adult at years i+1..N (giving).\n # Wait! If we look at the prefix 1..i, after year i, alien i has joined the adult group.\n # The total number of stones held by adults 1..i after year i is just sum(A_1..A_i).\n # Wait, does any stone leave the group {1..i}? No, because adults 1..i-1 only give stones to alien i, which stays in the group!\n # So the total stones held by {1..i} after year i is S_i = sum(A_1..A_i).\n # After year i, the adults are {1..i}. In future years i+1..N, stones might be transferred from {1..i} to aliens i+1..N.\n # How many stones are transferred from {1..i} to alien k (k > i)?\n # Exactly min(number of adults in 1..i with > 0 stones, total stones in 1..i).\n # But wait, all adults in 1..k-1 contribute!\n # A standard property: at any point, stones among adults are distributed as evenly as possible, or rather,\n # the total number of stones lost by adults {1..i} in steps i+1..N is easily determined!\n # Actually, if adults {1..i} have S_i stones total after year i,\n # in each step k = i+1..N, alien k receives min( (number of active adults at year k with >0 stones), total_stones_at_year_k ).\n # Notice that alien k receives stones from ALL adults 1..k-1.\n # How many of those come from {1..i}?\n # It's equivalent to: each alien i receives some stones, and we can determine the final stones of each alien!\n \n # Let's rephrase:\n # At year i, alien i receives stones from adults 1..i-1.\n # The number of stones received by alien i is min(S_{i-1}, i - 1)?\n # Wait, in Sample 1:\n # i=1: A_1=5. Adults: {1} has 5.\n # i=2: A_2=0. Adult 1 has 5 > 0, so gives 1 to 2. Adults {1,2} have 4, 1.\n # i=3: A_3=9. Adults 1,2 have 4,1 (both > 0). Total 2 adults give 1 each = 2 stones to 3.\n # A_3 becomes 9 + 2 = 11. Adults 1,2 become 3,0.\n # i=4: A_4=3. Adults 1,2,3 have 3,0,11. Adults 1,3 have >0 (2 adults). They give 2 stones to 4.\n # A_4 becomes 3 + 2 = 5. Adults 1,3 become 2, 10.\n # Final: 2, 0, 10, 5.\n \n # Notice: At year i, alien i gets min( S_{i-1}, count of adults 1..i-1 with >0 stones ).\n # Is count of adults with >0 stones always equal to min(i-1, S_{i-1})?\n # Not necessarily! In sample 1, after year 3, adults {1,2} have 3, 0. Count with >0 is 1.\n # Combined with alien 3 (who has 11), adults {1,2,3} have 2 adults with >0.\n \n # Let's maintain the state of adults from right to left or using a simulation/stack.\n # Actually, when alien i becomes adult, it has A_i stones.\n # It takes 1 stone from each adult j < i that has >0 stones.\n # Equivalently, alien i \"takes\" 1 stone from as many adults as possible in 1..i-1.\n # Who loses stones? The adults 1..i-1 who have stones.\n # If we maintain the adults with stones, notice that stones are always taken from the adults in a certain order, or we can just model the total stones!\n # Wait, alien i receives `c` stones, where `c` is the number of adults in 1..i-1 with >0 stones.\n # Then alien i has A_i + c stones.\n # Each of the `c` adults loses 1 stone.\n # This means the number of adults with >0 stones decreases only if an adult had 1 stone and drops to 0.\n \n # Can we process backwards?\n # At the end (after year N), the adults are 1..N.\n # Let's think: after year N, alien N has its final stones.\n # Alien N got stones at year N, and never gives any stones after year N!\n # So B_N is just its stones after year N.\n # What about alien N-1? It gives 1 stone at year N if it has >0 stones at year N.\n # In general, after year N, how many stones does each alien have?\n \n # Let's look at the process as a queue/stack of \"stone containers\".\n # An efficient way:\n # Maintain a multiset or count of adults, or notice:\n # At step i, alien i gets 1 stone from every non-zero adult in 1..i-1.\n # This is equivalent to: alien i absorbs 1 stone from each non-zero adult.\n # If an adult has k stones, it can give 1 stone to the next k aliens that ask!\n # So an adult with k stones can satisfy 1 request for each of the next k years!\n # Wait! An adult with k stones gives 1 stone at year i+1, 1 at i+2, ..., 1 at i+k (or until year N).\n # Is it ALWAYS true that an adult with k stones gives stones in the IMMEDIATE NEXT k years?\n # Yes! Because as long as it has >0 stones, it WILL give 1 stone every year!\n # It only stops giving stones when its stones reach 0!\n # Since it loses 1 stone per year, if it currently has k stones at year i, it will give 1 stone per year for the next k years (years i+1, i+2, ..., i+k), UNLESS N is reached first!\n # Wait, does it receive stones later? NO! An adult j < i NEVER receives stones at year i (only alien i receives stones!).\n # So an adult's stone count NEVER increases after its own year of adulthood!\n # AMAZING! Alien j gets its final initial stones at year j (say X stones).\n # After year j, alien j NEVER gains any more stones!\n # It only LOSES 1 stone per year for the next X years (years j+1, j+2, ..., j+X)!\n # Therefore, alien j contributes 1 stone to year k for every k in [j+1, min(N, j + X)]!\n \n # WOW! That means:\n # At year i, alien i receives 1 stone from every adult j < i that still has stones at year i!\n # Adult j has stones at year i if and only if j + X_j >= i, where X_j is the number of stones alien j had AT THE MOMENT IT BECAME AN ADULT (after receiving its own stones at year j)!\n # So if alien j has X_j stones at year j, it will have >0 stones at year i (i > j) iff j + X_j >= i!\n # Thus, the number of stones alien i receives at year i is simply:\n # count of j < i such that j + X_j >= i.\n # Then X_i = A_i + (count of j < i such that j + X_i... wait, j + X_j >= i).\n # And after ALL N years (at year N):\n # Alien j started with X_j stones at year j.\n # In years j+1 .. N, it gave away 1 stone each year, so it gave away min(X_j, N - j) stones.\n # Its FINAL number of stones B_j is simply max(0, X_j - (N - j))!\n \n # Wait, IS THIS TRUE?!\n # Let's check!\n # Alien j never receives stones after year j: TRUE, because at year k > j, only alien k receives stones.\n # Alien j gives 1 stone at year k > j iff it has >0 stones at year k.\n # Since it only loses 1 stone per year (at years j+1, j+2, ...), at year k it has lost (k - 1 - j) stones.\n # So at year k, its remaining stones are X_j - (k - 1 - j).\n # This is > 0 iff X_j - k + 1 + j > 0 iff j + X_j >= k!\n # Exactly!\n \n # So X_i = A_i + (number of j < i such that j + X_j >= i).\n # And final B_i = max(0, X_i - (N - i)).\n \n # Let's double check with Sample 1:\n # N = 4, A = [5, 0, 9, 3] (1-indexed)\n # i=1:\n # j < 1 with j + X_j >= 1: none.\n # X_1 = 5 + 0 = 5.\n # Reach of 1: 1 + 5 = 6.\n # i=2:\n # j < 2 with j + X_j >= 2: j=1 (1+5=6 >= 2). Count = 1.\n # X_2 = 0 + 1 = 1.\n # Reach of 2: 2 + 1 = 3.\n # i=3:\n # j < 3 with j + X_j >= 3:\n # j=1: 1+5=6 >= 3 (yes)\n # j=2: 2+1=3 >= 3 (yes)\n # Count = 2.\n # X_3 = 9 + 2 = 11.\n # Reach of 3: 3 + 11 = 14.\n # i=4:\n # j < 4 with j + X_j >= 4:\n # j=1: 6 >= 4 (yes)\n # j=2: 3 >= 4 (no)\n # j=3: 14 >= 4 (yes)\n # Count = 2.\n # X_4 = 3 + 2 = 5.\n # Reach of 4: 4 + 5 = 9.\n \n # Now final B_i = max(0, X_i - (4 - i)):\n # B_1 = max(0, 5 - 3) = 2.\n # B_2 = max(0, 1 - 2) = 0.\n # B_3 = max(0, 11 - 1) = 10.\n # B_4 = max(0, 5 - 0) = 5.\n # Sample 1 output: 2 0 10 5. PERFECT MATCH!\n\n # We just need to efficiently calculate for each i:\n # count of j < i such that j + X_j >= i.\n # Since j + X_j can be up to 500000 + 500000 + 500000 = 1.5 * 10^6,\n # we can maintain a Fenwick tree (or Difference Array!) of reach values!\n # When X_j is computed, reach = min(N, j + X_j).\n # It contributes +1 to all i in [j+1, reach].\n # Since we process i from 1 to N sequentially, we can just maintain a difference array for the additions to future i's!\n # diff[j+1] += 1\n # diff[reach + 1] -= 1\n # As we iterate i from 1 to N, current_gain += diff[i].\n \n diff = [0] * (N + 2)\n curr_gain = 0\n ans = []\n \n for i in range(1, N + 1):\n curr_gain += diff[i]\n X_i = A[i] + curr_gain\n reach = min(N, i + X_i)\n if i + 1 <= reach:\n diff[i + 1] += 1\n if reach + 1 <= N + 1:\n diff[reach + 1] -= 1\n \n final_stones = max(0, X_i - (N - i))\n ans.append(str(final_stones))\n \n print(' '.join(ans))\n\nif __name__ == '__main__':\n main()", "cost_usd": 0.0270555, "tokens_in": 582, "tokens_out": 3491, "latency_s": 20.286}] |