1 #include "000库函数.h" 2  3  4  5 //使用折半算法  牛逼算法 6 class Solution { 7 public: 8     double myPow(double x, int n) { 9         if (n == 0)return 1;10         double res = 1.0;11         for (int i = n; i != 0; i /= 2) {12             if (i % 2 != 0)13                 res *= x;14             x *= x;15         }16         return n > 0 ? res : 1 / res;17     }18 19 };20 21 22 //同样使用二分法，但是使用递归思想23 class Solution {24 public:25     double myPow(double x, int n) {26         if (n == 0)return 1;27         double res = myPow(x, n / 2);28         if (n % 2 == 0)return x * x;//是偶数，则对半乘了之后再两个大数相乘，x^n==(x^n/2)^229         if (n > 0) return res * res * x;30         return res * res / x;31     }32 };33 34 void T050() {35     Solution s;36     double x;37     int n;38     x = 0.0001;39     n = 2147483647;40     cout << s.myPow(x, n) << endl;41     x = 2.1;42     n = 3;43     cout << s.myPow(x, n) << endl;44     x = 2;45     n = -2;46     cout << s.myPow(x, n) << endl;47     x = 0.9;48     n = 2147483647;49     cout << s.myPow(x, n) << endl;50 51 52 }