题目链接：https://leetcode-cn.com/problems/shu-zhi-de-zheng-shu-ci-fang-lcof/
题目如下：

解一：迭代

class Solution {
public:
    double myPow(double x, int n) {
        double result=1.0;
        long long exp=abs((long long)(n));

        while(exp){
            if(exp&1) result*=x;//每次去最后一位
            x*=x;
            exp>>=1;//每次移除最后一位
        }

        if(n<0) result=1/result;

        return result;
    }
};

解二：递归

class Solution {
public:
    //注：
    //n为奇：x^n=x^(n/2)*x^(n/2)*x
    //n为偶：x^n=x^(n/2)*x^(x/2)
    double myPow(double x, int n) {
        long long N=n;
        if(N>=0) return rPow(x,N);
        else return 1.0/(rPow(x,-N));
    }
    double rPow(double x,long long n){
        if(n==0) return 1;
        double half=rPow(x,n/2);

        if(n%2==0) return half*half;
        else return half*half*x;
    }
};