class Solution {
public:
    double myPow(double x, int n) {
        if(n<0)
        {
            long long a = -(long long)n;
            double temp = dfs(x,a);
            return 1.0/temp;
        }
        else
        {
        double temp = dfs(x,n);
        return temp;
        }
    }
    double dfs(double x,int n)//给一个数，给一个n，求出x的n次幂
    {
        //递归结束条件
        if(n == 0) return 1;
        //子问题
        double temp = dfs(x,n/2);//求出一半的幂
        //返回相乘结果 
        if(n%2 == 0) return temp*temp;
        else return temp*temp*x;
    }
};