Problem Description：

A simple mathematical formula for e is

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where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Output：

Output the approximations of e generated by the above formula for the values of n from 0 to 9. The beginning of your output should appear similar to that shown below.

Sample Output：

n e

- -----------

0 1

1 2

2 2.5

3 2.666666667

4 2.708333333

程序代码：

#include<stdio.h>
int f(int j)
{
  int i,sum=1;
  for(i=1;i<=j;i++)
    sum*=i;
  return sum;
}
int main()
{
  int i;
  double e,num=2.5;
  printf("n e\n");
  printf("- -----------\n");
  printf("0 1\n");
  printf("1 2\n");
  printf("2 2.5\n");
  for(i=3;i<=9;i++)
  {
    e=1.0*1/f(i);
    num+=e;
    printf("%d %.9f\n",i,num);
  }
  return 0;
}