每天一道算法题【蓝桥杯】【Pow（x，n）】

思路

分情况讨论

n大于零

return 1.0/pow(x,-n);

n小于零

return pow（x/n）；

n为奇数

pow(x,n/2)*pow(x,n/2)*x

n 为偶数

pow(x,n/2)*pow(x,n/2)

#define _CRT_SECURE_NO_WARNINGS 1
class Solution {
public:double myPow(double x, int n) {if (n < 0)return 1.0 / pow(x, -(long long)n);//判断大于零还是小于零else return pow(x, n);};double pow(double x, int n){if (n == 0)return 1.0;double tmp = pow(x, n / 2);       //快速幂:把n次方不断拆分return n % 2 == 0 ? tmp * tmp : tmp * tmp * x;};};