C语言测试n的阶乘和x的n次方

/*************************************************************************
    > File Name: 2.c
    > Author: 
    > Mail: 
    > Created Time: Wed 12 Dec 2018 09:03:22 AM CST
 ************************************************************************/
#include<stdio.h>
double fact(int n)
{
    double s = 1.0;
  for(int i=1; i<= n; i++)
    {
        s=s*i;
    } 
    return s;
}
double mypow(double x,int n)
{         
  double s = 1.0;
  //printf("%lf %dn",x,n);
  if(n == 0)  
  {       
   return 1.0; 
  }            
  if(n == 1)  
  {       
   return x;                                        
  }
  s = x;
  for(int i =0;i<n-1;i++)
  {       
   s = x*s;  
  }
  //printf("%lf n",s);
  return s;   
}
void main(void)
{
  double x = 0.0;
  int n = 0;
  double s;
  scanf("%lf %d",&x,&n);
  //printf("%lfn",mypow(-1.0,2));
  if(n == 1)
  {
    s = x;
  }
  else
  {
    s = x;
    for(int i=2;i<=n;i++)
    {
      s = s+ mypow(-1.0,i-1)*mypow(x,i)/fact(i);
    }
  }
  printf("%.4lfn",s);
}

#include <stdio.h>
int main ()
{
double x, ret, tmp1, tmp2;
int n, i, j;
while (~scanf("%lf %d", &x, &n))
{
ret = 0;
for (i = 1; i <= n; i++)
{
tmp1 = 1;
for (j = 1; j <= i; j++)
{
tmp1 *= x;
}
tmp2 = 1;
for (j = 1; j <= i; j++)
{
tmp2 *= j;
}
if (i % 2 == 1)
{
ret += tmp1 / tmp2;
}
else
{
ret -= tmp1 / tmp2;
}
}
printf("%.04fn", ret);
}
return 0;
}