C++ | Intuition | Detailed Solution
Intuition\nThe problem involves finding the number of pairs of elements in an array that have a given GCD. By understanding how GCD works, we can determine the 
# Intuition\nThe problem involves finding the number of pairs of elements in an array that have a given GCD. By understanding how GCD works, we can determine the pairs based on multiples. The approach will involve counting how many numbers can contribute to a specific GCD.\n\n\n# Code\n```cpp []\n#include <bits/stdc++.h>\nusing namespace std;\n\nclass Solution {\npublic:\n    vector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {\n        // Determine the maximum value in nums to size the gdcmulip array\n        int maxVal = *max_element(nums.begin(), nums.end()) + 1;\n        vector<long long> gdcmulip(maxVal, 0); // To hold count of pairs with GCD i\n        vector<long long> cummsum(maxVal, 0);      // Cumulative sums of pairs\n\n        // Count occurrences of each number in nums\n        for (int num : nums) {\n            gdcmulip[num]++;\n        }\n\n        // Calculate pairs for each possible GCD\n        for (int i = 1; i < maxVal; ++i) {\n            long long count = 0;\n            for (int j = i; j < maxVal; j += i) {\n                count += gdcmulip[j]; // Sum counts of multiples of i\n            }\n            gdcmulip[i] = count * (count - 1) / 2; // Calculate pairs\n        }\n\n        // Adjust counts to ensure gdcmulip[i] holds pairs with GCD exactly i\n        for (int i = maxVal - 1; i >= 1; --i) {\n            for (int j = i + i; j < maxVal; j += i) {\n                gdcmulip[i] -= gdcmulip[j]; // Remove multiples from count\n            }\n        }\n\n        // Calculate cumulative sums of pairs\n        for (int i = 1; i < maxVal; ++i) {\n            cummsum[i] = cummsum[i - 1] + gdcmulip[i];\n        }\n\n        vector<int> results; // To store results for each query\n        for (long long query : queries) {\n            // Use binary search to find the smallest k satisfying the query\n            int k = lower_bound(cummsum.begin() + 1, cummsum.end(), query + 1) - cummsum.begin();\n            results.push_back(k);\n        }\n        return results; // Return the results for all queries\n    }\n};\n\n\n```
['C++']