50 Pow(x, n)(求x的n次方Medium)

class Solution {
public:
double myPow(double x, int n) {
if(x==1.0)return x;
else if(x==-1.0){
if(n%2==0)return 1.0;
else return -1.0;
}
double ans=1.0;
int flag=abs(n);
while(flag--&&abs(ans)>0.0000001)ans*=x;
if(n>0)return ans;
else return 1.0/ans;
}
};