力扣 50.pow(x,n)
class Solution {
public:
double quickMul(double x,long long N){
if(N==0) return 1;
double value=quickMul(x,N/2);
return N%2==0?value*value:value*value*x;
}
double myPow(double x, int n) {
long long N=n;
return N>0?quickMul(x, N):1.0/quickMul(x, N);
}
};