#include "000库函数.h"

 //使用折半算法  牛逼算法

 class Solution {

 public:

     double myPow(double x, int n) {

         if (n == 0)return 1;

         double res = 1.0;

         for (int i = n; i != 0; i /= 2) {

             if (i % 2 != 0)

                 res *= x;

             x *= x;

         }

         return n > 0 ? res : 1 / res;

     }

 };

 //同样使用二分法，但是使用递归思想

 class Solution {

 public:

     double myPow(double x, int n) {

         if (n == 0)return 1;

         double res = myPow(x, n / 2);

         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)^2

         if (n > 0) return res * res * x;

         return res * res / x;

     }

 };

 void T050() {

     Solution s;

     double x;

     int n;

     x = 0.0001;

     n = 2147483647;

     cout << s.myPow(x, n) << endl;

     x = 2.1;

     n = 3;

     cout << s.myPow(x, n) << endl;

     x = 2;

     n = -2;

     cout << s.myPow(x, n) << endl;

     x = 0.9;

     n = 2147483647;

     cout << s.myPow(x, n) << endl;

 }