leetcode2183 统计可以被 K 整除的下标对数目

思路：

数学题。

实现：

vector<vector<int>> d(100001, vector<int>());
bool init = false;
class Solution
{
public:
    long long countPairs(vector<int>& nums, int k)
    {
        if (!init)
        {
            for (int i = 1; i <= 100000; i++)
            {
                for (int j = i; j <= 100000; j += i)
                {
                    d[j].push_back(i);
                }
            }
            init = true;
        }
        int n = nums.size();
        unordered_map<int, int> mp;
        long long res = 0;
        for (int i = 0; i < n; i++)
        {
            int g = __gcd(nums[i], k);
            int need = k/g; 
            if (mp.count(need))
            {
                res += mp[need];
            }
            for (int j = 0; j < d[nums[i]].size(); j++)
            {
                int div = d[nums[i]][j];
                if (!mp.count(div)) mp[div] = 1;
                else mp[div]++;
            }
        }
        return res;
    }
};