作业9——Implement pow(x, n), which calculates x raised to the power n (xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000
Example 2:

Input: 2.10000, 3
Output: 9.26100
Example 3:

Input: 2.00000, -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25
Note:

-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−2^31, 2^31-1]

计算机数学

在此输入正文

/*
*File:pow.c
*Author:Hongpei lin
*Date:20181014
*Purpose:to solve the problem in Leetcode
*https://leetcode.com/problems/powx-n/description/ 
*/
/**
*Declaration of functions.
*Input a double x and integer n
*Output x^n
*Method:In normal circumstances,make a circulate,let x*x execute n times
*In special circumstances,make special method
**/
double myPow(double x, int n) {
  double k=1; //k memory the result
  //if x=1,find that x^n=1
  if(x==1) return 1.0;
  //if x=-1,and n is a even,find that x^n=1
  //if x=-1,and n is an odd,find that x^n=-1
  if(x==-1)
  {
    if(n%2==0) return 1.0;
    else return -1.0;
  }
  //if n=0,find that x^n=1
  if(n==0) return 1.0;
  //if n=-(2^31),find the result will overflow
  if(n==-2147483648) return 0.0;
  else if(n>0)
  {
    //the process of pow
    while(n!=0)
    {
      //if result is infinitely great,it will overflow
      if(k==NAN) return 0.0;
      //if n is a odd,result multiply a longly x
      if(n%2==1) {k*=x;}
      //n divide 2
      n=n/2;
      //x=x*x
      x*=x;
    }
  }
  else
  {
    //if n<0,let x=1/x,the result is (1/x)^|n|
    x=1/x;
    n=-n;
    while(n!=0)
    {
      //if result is infinitely small,it will overflow
      if(k==INFINITY) return 0.0;
      //the same process
      if(n%2==1) {k*=x;}
      n=n/2;
      x*=x;
    }
  }
  return k;
}