剑指offer：数值的整数幂次

 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4         bool posit = (exponent > 0) ? true : false;
 5         exponent = posit ? exponent : (-exponent);
 6         if(!exponent){
 7             return 1;
 8         }
 9         double tmp = base;
10         for(int i=1;i<exponent;++i){
11             tmp *= base;
12         }
13         if(posit){
14             return tmp;
15         }
16         else{
17             return 1/tmp;
18         }
19     }
20 };