Pow(x, n)

  double res = 1.0;
        bool flag = false;
        unsigned int newn =n;  //这一步比较神奇，为了照顾最小的负数
        if(n<0) {newn=-n;flag=true;}
        while(newn){
        if(newn & 0x01){
           res*=x;
           }
           newn>>=1;
           x*=x;
        }
        
        if(flag) return 1/res;
        return res;

 double myPow(double x, int n) {
    if (n == 0) return 1.0;
    double half = pow(x, n/2);
    if (n%2 == 0)
    {
        return half*half;
    }
    else if (n>0)
    {
        return half*half*x;
    }
    else
    {
        return half/x*half;
    }