快速幂

注意int范围为$[2^{31}，2^{31} - 1]$，若n为最小值，变为正数时会溢出，所以先用longlong存n比较好

class Solution {
public:
    double myPow(double x, int n) {
        int flag = 0;
        long long N = (long long)n;
        if (N < 0) flag = 1, N = -N;
        double ret = 1;
        while(N)
        {
            if(N & 1) ret *= x;
            x = x * x;
            N >>= 1;
        }
        return flag ? 1 / ret : ret;

    }
};