class Solution {
public:
double Power(double base, int exponent) {
//快速幂不太熟,下次回来看看
typedef long long LL;
bool is_minus = exponent < 0;
double res = 1;
for(LL k = abs(LL(exponent));k;k >>= 1){
if(k & 1) res *= base;
base *= base;
}
if(is_minus) res = 1 / res;
return res;
}
};