数字的整数次方

class Solution {
public:
    double Power(double base, int e) {
        typedef long long LL;
        bool flag=false;
        if(e<0)
            flag=true;
        double res=1;//存储base的2的倍数次幂
        //计算base的e次方
        for (LL i = abs((LL)e); i; i>>=1 )
        {
            if(i&1) res*=base;
            base*=base;
        }
        if(flag)    res=1/res;
        return res;
    }
};