Beats 99.58%
Approach\n#### Powerful Array Generation:\n- Start by generating powerful arrays for each integer.\n- A powerful array for an integer x is the shortest sorted a
# Approach\n#### **Powerful Array Generation:**\n- Start by generating powerful arrays for each integer.\n- A powerful array for an integer x is the shortest sorted array of powers of two that sum up to x. For example, the powerful array for 11 is [1, 2, 8].\n#### **Building big_nums Array:**\n- Concatenate the powerful arrays for every positive integer i in ascending order to form the big_nums array.\n- For example, big_nums starts as [1, 2, 1, 2, 4, 1, 4, 2, 4, 1, 2, 4, 8, ...].\n#### **Precomputation:**\n- Precompute values to optimize subsequent queries.\n- Precompute prefix products modulo a large modulus for efficient computation of products of subsequences of big_nums.\n#### **Handling Queries:**\n- For each query [fromi,toi,modi]:\n    - Calculate the sum of powers for the range [fromi,toi] using precomputed values.\n    - Calculate the difference in sum of powers between toi and fromi.\n    - Use modular exponentiation to compute the product modulo modi, efficiently.\n    - Store the result in the answer vector.\n#### **Return Results:**\n- Return the answer vector containing the results of each query.\n\n# Complexity\n- Time complexity:\n**O(N\u2217Log(M))** where Q is the number of queries, and x represents the maximum values in the queries.\n- Space complexity:\n**O(Log(M))** where Q is the number of queries, and x represents the maximum values in the queries.\n\n# Code\n```\n#include <vector>\n#include <algorithm>\n\nclass Solution {\npublic:\n    std::vector<int> findProductsOfElements(std::vector<std::vector<long long>>& queries) {\n        std::vector<int> ans;\n\n        for (const auto& query : queries) {\n            long long from = query[0];\n            long long to = query[1];\n            int mod = query[2];\n\n            if (mod == 1) {\n                ans.push_back(0);\n                continue;\n            }\n\n            long long sumPowersFrom = sumPowersFirstKBigNums(from);\n            long long sumPowersTo = sumPowersFirstKBigNums(to + 1);\n            long long diffSumPowers = sumPowersTo - sumPowersFrom;\n            int product = modPow(2, diffSumPowers, mod);\n            ans.push_back(product);\n        }\n\n        return ans;\n    }\n\nprivate:\n    long long sumPowersFirstKBigNums(long long k) {\n        if (k == 0) return 0;\n\n        long long num = firstNumberHavingSumBitsTillGreaterThan(k);\n        long long sumPowers = sumPowersTill(num - 1);\n        long long remainingCount = k - sumBitsTill(num - 1);\n\n        for (int power = 0; remainingCount > 0; ++power) {\n            if (num & (1LL << power)) {\n                sumPowers += power;\n                --remainingCount;\n            }\n        }\n\n        return sumPowers;\n    }\n\n    long long firstNumberHavingSumBitsTillGreaterThan(long long k) {\n        long long l = 1, r = k;\n        while (l < r) {\n            long long m = l + (r - l) / 2;\n            if (sumBitsTill(m) < k)\n                l = m + 1;\n            else\n                r = m;\n        }\n        return l;\n    }\n\n    long long sumBitsTill(long long x) {\n        long long sumBits = 0;\n        for (long long powerOfTwo = 1; powerOfTwo <= x; powerOfTwo <<= 1) {\n            sumBits += (x / (2LL * powerOfTwo)) * powerOfTwo;\n            sumBits += std::max(0LL, x % (2LL * powerOfTwo) + 1 - powerOfTwo);\n        }\n        return sumBits;\n    }\n\n    long long sumPowersTill(long long x) {\n        long long sumPowers = 0;\n        for (long long powerOfTwo = 1, power = 0; powerOfTwo <= x; powerOfTwo <<= 1, ++power) {\n            sumPowers += (x / (2LL * powerOfTwo)) * powerOfTwo * power;\n            sumPowers += std::max(0LL, x % (2LL * powerOfTwo) + 1 - powerOfTwo) * power;\n        }\n        return sumPowers;\n    }\n\n    int modPow(long long x, long long n, int mod) {\n        int result = 1;\n        x %= mod;\n        while (n > 0) {\n            if (n % 2 == 1)\n                result = static_cast<int>((result * x) % mod);\n            x = (x * x) % mod;\n            n >>= 1;\n        }\n        return result;\n    }\n};\n```
['Binary Search', 'Dynamic Programming', 'Bit Manipulation', 'C++']