simple solution using prefixsum and lowerbound approach
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n     // Upper bound for nums[i]\n\nvector<int> gcdValues(vector<int>& nums, vector<long long>& queries) {\n    // Step 1: Frequency array for nums\n    const int MAX_NUM = *max_element(nums.begin() , nums.end());\n    vector<long long> freq(MAX_NUM + 1, 0) , ans(MAX_NUM +1 , 0);\n    for (int num : nums) {\n        freq[num]++;\n    }\n\n    // Step 2: GCD pair count\n    // vector<pair<int, long long>> gcdPairs;\n    for (int g = 1; g <= MAX_NUM; ++g) {\n        // Count how many numbers are divisible by g\n        long long count = 0;\n        for (int mult = g; mult <= MAX_NUM; mult += g) {\n            count += freq[mult];\n        }\n\n        // If there are at least 2 numbers divisible by g, calculate the number of pairs\n        if (count >= 2) {\n            long long pairCount = (count * 1ll*(count - 1)) / 2;\n            ans[g]=pairCount;\n        }\n    }\n    \n    // Step 3: Sort the GCDs by ascending order and their pair counts\n\n    for(int i = MAX_NUM ; i>1 ;i--){\n        long long x=ans[i];\n        ans[1]-=x;\n        for(int j=2 ;j*j <= i ; j++){\n            if(i%j==0){\n                ans[j]-=x;\n                if(i/j!=j)ans[i/j]-=x;\n            }\n        }\n    }\n    // sort(gcdPairs.begin(), gcdPairs.end());\n    \n    // Step 4: Create the prefix sum array\n    vector<long long> prefixSum;\n    long long totalPairs = 0;\n    for (int i=1 ; i<=MAX_NUM ;i++) {\n        totalPairs += ans[i];\n        prefixSum.push_back(totalPairs);\n    }\n    for(auto a: prefixSum) cout<<a<<" ";\n    cout<<endl;\n    // Step 5: Answer each query\n    vector<int> result;\n    for (long long query : queries) {\n        // Binary search to find the k-th smallest GCD\n        auto it = lower_bound(prefixSum.begin(), prefixSum.end(), query+1);\n        int idx = distance(prefixSum.begin(), it);\n        result.push_back(idx+1);\n        // cout<<idx<<" "<<idx+1<<endl;\n    }\n\n    return result;\n}\n};\n```
['C++']