数值的整次方

 2     实现函数double Power(double base,int exponent),求base的exponent次方。不得使用库函数，同时不需要考虑大数问题

  代码如下：

  5 #include<iostream>
  6 using namespace std;
  7
  8 bool g_InvaildInput = false;
  9
 10 double PowerWithUnsignedExponent(double base,unsigned int exponent)
 11 //用递归来解决大数的模运算
 12 {
 13     if(exponent == 0)
 14         return 1;
 15     if(exponent == 1)
 16         return base;
 17
 18     double result = PowerWithUnsignedExponent(base,exponent>>1);
 19     //右移一位代表的运算和处以2是等价的，但是右移运算符的效率要比除法的高
 20     result *= result;
 21     //判断是否为偶数
 22     if(exponent & 0x1 == 1)
 23         result *= base;
 24     return result;
 25 }
 26
 27 bool equal(double num1,double num2)
 28 //单精度小数的比较是和0.0000001比较才行
 29 {
 30     if( ((num1 - num2) > 0.0000001) && ((num1 - num2) < 0.0000001) )
 31         return true;
 32     else
 33         return false;
 34 }
 35
 36 double Power(double base,int exponent)
 37 {
 38     g_InvaildInput = false;
 39
 40     if(equal(base,0.0) && exponent < 0)
 41     {
 42         g_InvaildInput = true;
 43         return 0.0;
 44     }
 45
 46     unsigned int absExponent = (unsigned int) (exponent);
 47     if(exponent < 0)
 48         absExponent = (unsigned int)(-exponent);
 49
 50     double result = PowerWithUnsignedExponent(base,absExponent);
 51     if(exponent < 0)
 52         result = 1.0 / result;
 53     return result;
 54 }
 55
 56 int main()
 57 {
 58     cout<<"please input two number"<<endl;
 59     int m,n;
 60     cin>>m>>n;
 61     cout<<"Power is "<<Power(m,n)<<endl;
 62     return 0;
 63 }