力扣2438.二的幂数组中查询范围内的乘积

力扣2438.二的幂数组中查询范围内的乘积

• lowbit求所有2的幂

  • accumulate函数(begin,end,start,way)
  • 求和/积的方式
  • 求积并取模

•   const int N = 1e9 + 7;
    class Solution {
    public:
        int lowbit(int x)
        {
            return x & -x;
        }
        vector<int> productQueries(int n, vector<vector<int>>& queries) {
            vector<int> powers;
            while(n)
            {
                int t = lowbit(n);
                n -= t;
                powers.emplace_back(t);
            }
            vector<int> res;
            for(auto v:queries)
            {
                int l = v[0];
                int r = v[1];
                //accumulate的用法
                int product = accumulate(powers.begin() + l,powers.begin() + r + 1, 1 ,
                    [](int x,int y) {return (long long)x*y %N ;}
                );
                res.emplace_back(product);
            }
            return res;
        }
    };