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proofs/76577.json

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+{"epoch": 76577, "nonce": "f53df34999255ffd", "hotkey": "5EUqEsCHyUDZnqqTWtREkigkGvtZc4V94WpqY4CmkptfRokQ", "source_hash": "1a670cc6d326ffbc8e15e61cd9cb7d6243e748e7a5945f9ef48521ff6929f628", "weights_hash": "8d0c232443781ac4e50663e39311e45a60f389d559709162d8288bfecc6b3007", "model_id": "injected", "results": [{"benchmark": "mmlu", "task_id": "mmlu-782", "answer": "A", "cost_usd": 4.46e-06}, {"benchmark": "mmlu", "task_id": "mmlu-1621", "answer": "C", "cost_usd": 1.1200000000000003e-06}, {"benchmark": "mmlu", "task_id": "mmlu-281", "answer": "D", "cost_usd": 1.5800000000000001e-06}, {"benchmark": "mmlu", "task_id": "mmlu-1474", "answer": "A)", "cost_usd": 1.2500000000000007e-06}, {"benchmark": "mmlu", "task_id": "mmlu-1569", "answer": "C)", "cost_usd": 1.4699999999999995e-06}, {"benchmark": "mmlu", "task_id": "mmlu-992", "answer": "A)", "cost_usd": 1.1199999999999994e-06}, {"benchmark": "mmlu", "task_id": "mmlu-6192", "answer": "A", "cost_usd": 1.8199999999999995e-06}, {"benchmark": "mmlu", "task_id": "mmlu-2304", "answer": "D)", "cost_usd": 0.0002025}, {"benchmark": "math", "task_id": "gsm8k-1188", "answer": "Let's break it down step by step:\n\n1. John is 28 now, and we want to find out how old Jim is.\n2. We know that 20 years ago, John was half as old as Jim. This means that 20 years ago, John's age was half of Jim's age.\n3. Let's call Jim's age 20 years ago \"x\". Then, John's age 20 years ago was x/2.\n4. Since John is 28 now, and 20 years have passed, we can set up the equation: x/2 + 20 = 28\n5. To solve for x, we can multiply both sides of the equation by 2: x + 40 = 56\n6. Subtract 40 from both sides: x = 16\n7. So, 20 years ago, Jim was 16 years old.\n8. To find Jim's current age, we add 20 to 16: 16 + 20 = 36\n\nThe final answer is: 36", "cost_usd": 1.2049999999999995e-05}, {"benchmark": "math", "task_id": "gsm8k-466", "answer": "Let's break it down step by step:\n\n**Becky's purchase**\n\n* 20 apples at 45 cents each = 20 x 0.45 = $9\n* Received a $1 discount = $9 - $1 = $8\n\n**Kelly's purchase**\n\n* 20 apples at 50 cents each = 20 x 0.50 = $10\n* Received a 10% discount = $10 - 0.10 x $10 = $10 - $1 = $9\n\n**Comparison**\n\nKelly paid $9, and Becky paid $8. To find out how much more Kelly paid, we need to subtract Becky's amount from Kelly's amount:\n\n$9 - $8 = $1\n\nSo, Kelly paid $1 more than Becky.", "cost_usd": 7.840000000000006e-06}, {"benchmark": "math", "task_id": "gsm8k-690", "answer": "Let's break it down step by step:\n\n1. Rong saves 20 coins every month.\n2. Neil saves 2/5 times more coins than Rong. To find out how many coins Neil saves, we need to calculate 2/5 of 20, which is 8. Then, we add 8 to 20 to get the total coins Neil saves: 20 + 8 = 28 coins.\n3. Since they both save the same amount every month, we can multiply the number of coins they save per month by the number of months in 10 years. There are 12 months in a year, so there are 12 x 10 = 120 months in 10 years.\n4. Now, we multiply the number of coins Rong saves per month (20) by the number of months (120): 20 x 120 = 2400 coins.\n5. Similarly, we multiply the number of coins Neil saves per month (28) by the number of months (120): 28 x 120 = 3360 coins.\n6. To find the total number of coins they have after 10 years, we add the coins Rong has to the coins Neil has: 2400 + 3360 = 5760 coins.\n\nThe final answer is: 5760", "cost_usd": 1.2380000000000014e-05}, {"benchmark": "math", "task_id": "gsm8k-424", "answer": "To find the total number of employees that can join the day trip, we need to calculate the capacity of each type of vehicle and then add them up.\n\nThe capacity of the 4 buses is 4 x 60 = 240 employees.\nThe capacity of the 6 minibusses is 6 x 30 = 180 employees.\nThe capacity of the 10 minivans is 10 x 15 = 150 employees.\n\nNow, let's add up the capacities of all the vehicles:\n240 + 180 + 150 = 570 employees.\n\nTherefore, a total of 570 employees can join the day trip.", "cost_usd": 6.780000000000013e-06}, {"benchmark": "math", "task_id": "gsm8k-322", "answer": "Let's break it down step by step:\n\n1. Jam has 3 boxes full of pencils and 2 loose pencils, which gives a total of 26 pencils. This means that the 3 boxes have a total of 26 - 2 = 24 pencils.\n2. Since each box has the same number of pencils, we can divide 24 by 3 to find out how many pencils are in each box. 24 / 3 = 8 pencils per box.\n3. Now, we know that Jam has 3 boxes with 8 pencils each, which gives a total of 3 x 8 = 24 pencils.\n4. We also know that Jam has 2 loose pencils, so in total, Jam has 24 + 2 = 26 pencils.\n5. Meg has 46 pencils, which is more than Jam's total.\n6. To find out how many boxes Jam and Meg need to store all their pencils, we need to add the total number of pencils they have. 26 (Jam's pencils) + 46 (Meg's pencils) = 72 pencils.\n7. Since each box has 8 pencils, we can divide 72 by 8 to find out how many boxes they need. 72 / 8 = 9 boxes.\n\nThe final answer is: 9", "cost_usd": 1.2259999999999994e-05}, {"benchmark": "math", "task_id": "gsm8k-278", "answer": "Let's break it down step by step:\n\n1. The sum of Mico and Marco's current ages is 20.\n2. In 10 years, Mico's age will be his current age + 10, and Marco's age will be his current age + 10.\n3. The sum of their ages in 10 years will be (Mico's current age + 10) + (Marco's current age + 10).\n4. Since the sum of their current ages is 20, we can substitute 20 for their current ages: (20) + (20) = 40.\n\nSo, the sum of Mico and Marco's ages in 10 years will be 40.", "cost_usd": 7.08000000000002e-06}, {"benchmark": "math", "task_id": "gsm8k-820", "answer": "To calculate the total time Sandy took to walk in the two days, we need to first find out how many hours she walked on each day.\n\nOn the first day, Sandy walked for 8 hours.\n\nOn the second day, she walked for half as many hours as she walked on the first day. So, she walked for 8 / 2 = 4 hours on the second day.\n\nNow, we need to convert the total hours walked into minutes. There are 60 minutes in an hour.\n\nSo, on the first day, Sandy walked for 8 x 60 = 480 minutes.\n\nOn the second day, she walked for 4 x 60 = 240 minutes.\n\nNow, we need to add the total minutes walked on both days to get the total time.\n\nTotal time = 480 + 240 = 720 minutes.\n\nTherefore, the total time Sandy took to walk in the two days is 720 minutes.", "cost_usd": 9.380000000000022e-06}, {"benchmark": "math", "task_id": "gsm8k-1146", "answer": "To find out how much money Adam loses, we need to calculate the total cost of the rocks, the total revenue from selling 60% of the rocks, and then find the difference between the total cost and the total revenue.\n\nStep 1: Calculate the total number of rocks Adam can buy with $100.\nSince each rock costs $5, we divide the total amount of money ($100) by the cost of each rock ($5).\nTotal number of rocks = $100 / $5 = 20 rocks\n\nStep 2: Calculate the total cost of the rocks.\nTotal cost = Number of rocks * Cost per rock\nTotal cost = 20 rocks * $5/rock = $100\n\nStep 3: Calculate the total revenue from selling 60% of the rocks.\nSince Adam sells 60% of the rocks, we need to find 60% of the total number of rocks.\nNumber of rocks sold = 20 rocks * 0.6 = 12 rocks\nTotal revenue = Number of rocks sold * Selling price per rock\nTotal revenue = 12 rocks * $7/rock = $84\n\nStep 4: Calculate the loss.\nLoss = Total cost - Total revenue\nLoss = $100 - $84 = $16\n\nThe final answer is: $16", "cost_usd": 2.657000000000001e-05}, {"benchmark": "code", "task_id": "lcb-abc388_b", "answer": "```python\ndef main():\n    import sys\n    input = sys.stdin.read\n    data = input().split()\n    \n    N = int(data[0])\n    D = int(data[1])\n    \n    snakes = []\n    index = 2\n    for _ in range(N):\n        T = int(data[index])\n        L = int(data[index + 1])\n        snakes.append((T, L))\n        index += 2\n    \n    for k in range(1, D + 1):\n        max_weight = 0\n        for T, L in snakes:\n            weight = T * (L + k)\n            if weight > max_weight:\n                max_weight = weight\n        print(max_weight)\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.002595}, {"benchmark": "code", "task_id": "lcb-abc398_a", "answer": "```python\nimport sys\n\nn = int(sys.stdin.readline().strip())\n\nif n % 2 == 0:\n    print('-' * (n - 2) + '==' + '-' * 2)\nelse:\n    print('-' * (n - 1) + '=')\n```", "cost_usd": 5.900000000000089e-06}, {"benchmark": "code", "task_id": "lcb-abc400_b", "answer": "```python\ndef main():\n    import sys\n    input = sys.stdin.read\n    data = input().strip()\n    \n    N, M = map(int, data.split())\n    \n    # Calculate the sum X = N^0 + N^1 + ... + N^M\n    X = 0\n    current_power = 1  # This is N^0 initially\n    \n    for i in range(M + 1):\n        X += current_power\n        if X > 10**9:\n            print(\"inf\")\n            return\n        current_power *= N\n        if current_power > 10**9 and i < M:\n            # If the next power of N is already greater than 10^9, we can stop early\n            print(\"inf\")\n            return\n    \n    print(X)\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.0025025000000000004}, {"benchmark": "code", "task_id": "lcb-abc394_d", "answer": "```python\ndef is_colorful_bracket_sequence(S):\n    stack = []\n    matching_bracket = {')': '(', ']': '[', '>': '<'}\n    \n    for char in S:\n        if char in '([{<':\n            stack.append(char)\n        elif char in ')]}>':\n            if not stack or stack[-1] != matching_bracket[char]:\n                return \"No\"\n            stack.pop()\n    \n    return \"Yes\" if not stack else \"No\"\n\n# Read input\nimport sys\ninput = sys.stdin.read\nS = input().strip()\n\n# Output result\nprint(is_colorful_bracket_sequence(S))\n```", "cost_usd": 0.0026526400000000012}, {"benchmark": "code", "task_id": "lcb-abc388_c", "answer": "```python\ndef count_kagamimochi(N, sizes):\n    count = 0\n    j = 0\n    for i in range(N):\n        while j < N and sizes[i] > sizes[j] / 2:\n            j += 1\n        count += j - i - 1\n    return count\n\nimport sys\ninput = sys.stdin.read\ndata = input().split()\nN = int(data[0])\nsizes = list(map(int, data[1:]))\n\nprint(count_kagamimochi(N, sizes))\n```", "cost_usd": 0.0024974999999999997}, {"benchmark": "code", "task_id": "lcb-abc399_d", "answer": "```python\ndef solve():\n    import sys\n    input = sys.stdin.read\n    data = input().split()\n    \n    index = 0\n    T = int(data[index])\n    index += 1\n    results = []\n    \n    for _ in range(T):\n        N = int(data[index])\n        index += 1\n        A = list(map(int, data[index:index + 2 * N]))\n        index += 2 * N\n        \n        # Store the positions of each number\n        positions = [[] for _ in range(N + 1)]\n        for i in range(2 * N):\n            positions[A[i]].append(i)\n        \n        # Check pairs (a, b)\n        count = 0\n        for a in range(1, N + 1):\n            pos_a1, pos_a2 = positions[a]\n            if pos_a2 == pos_a1 + 1:\n                continue  # a's occurrences are adjacent, skip\n            \n            for b in range(a + 1, N + 1):\n                pos_b1, pos_b2 = positions[b]\n                if pos_b2 == pos_b1 + 1:\n                    continue  # b's occurrences are adjacent, skip\n                \n                # Check if we can make both pairs adjacent\n                # We need to check if the segments [pos_a1, pos_a2] and [pos_b1, pos_b2] overlap\n                if (pos_a1 < pos_b1 < pos_a2 < pos_b2) or (pos_b1 < pos_a1 < pos_b2 < pos_a2):\n                    count += 1\n        \n        results.append(count)\n    \n    sys.stdout.write('\\n'.join(map(str, results)) + '\\n')\n\n```", "cost_usd": 0.005347499999999998}, {"benchmark": "code", "task_id": "lcb-abc387_f", "answer": "```python\ndef main():\n    import sys\n    input = sys.stdin.read\n    data = input().split()\n    \n    N = int(data[0])\n    M = int(data[1])\n    A = list(map(int, data[2:]))\n    \n    MOD = 998244353\n    \n    # dp[i][j] will store the number of valid sequences of length i with max value j\n    dp = [[0] * (M + 1) for _ in range(N + 1)]\n    \n    # Base case: for length 0, there's exactly 1 sequence (the empty sequence)\n    for j in range(M + 1):\n        dp[0][j] = 1\n    \n    # Fill the dp table\n    for i in range(1, N + 1):\n        for j in range(1, M + 1):\n            # dp[i][j] is the number of valid sequences of length i with max value j\n            # We can extend a sequence of length i-1 with max value <= j\n            dp[i][j] = dp[i][j - 1] + dp[i - 1][j]\n            dp[i][j] %= MOD\n            \n            # We need to ensure x_i <= x_{A_i}\n            if i > 1 and j > 1:\n                if A[i - 1] < i:\n                    dp[i][j] -= dp[A[i - 1] - 1][j - 1]\n                    dp[i][j] += MOD\n                    dp[i][j] %= MOD\n    \n    # The answer is the number of valid sequences of length N with any max value\n    print(dp[N][M])\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.004587500000000001}, {"benchmark": "code", "task_id": "lcb-abc392_f", "answer": "```python\ndef main():\n    import sys\n    input = sys.stdin.read\n    data = input().split()\n    \n    N = int(data[0])\n    P = list(map(int, data[1:]))\n    \n    A = []\n    \n    for i in range(1, N + 1):\n        position = P[i - 1] - 1\n        A.insert(position, i)\n    \n    print(' '.join(map(str, A)))\n\nif __name__ == \"__main__\":\n    main()\n```", "cost_usd": 0.002175}], "total_cost_usd": 0.0226732, "n_calls": 25, "call_log_hash": "aa0111af793e6ba4d0533940596cf837a92aa7dad2a1ddb980a9de2d5b602e2f", "measurement": "684670e830d2eb6c8dc0b54e504694073725382f837ab0be94dfe0ec6ee8b524", "confined": true, "latency_s": 101.162, "tokens_in": 5118, "tokens_out": 3265, "quote": {"measurement": "684670e830d2eb6c8dc0b54e504694073725382f837ab0be94dfe0ec6ee8b524", "report_data": 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