杭电acm1012 u Calculate e

Problem Description
A simple mathematical formula for e is

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

#include<stdio.h>
int main()
{
 double sum,t,e,i,j,n;
 printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
  t=2;
  sum=2.5;
  for(i=3;i<=9;i++)
  {
    t*=i;
    sum+=1/t;
   e=sum;
   printf("%d %.9lf\n",(int)i,e);
  }
 return 0;
}
n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
5 2.716666667
6 2.718055556
7 2.718253968
8 2.718278770
9 2.718281526

#include<stdio.h>
void main()
{
       int n,i;
       double sum,t;
       printf("n e\n");
       printf("- -----------\n");
       printf("0 1\n1 2\n2 2.5\n");
       sum=2.5;
       for(n=3;n<10;n++)
       {
          t=1.0;
          for(i=2;i<=n;i++)
          t*=1.0/i;
          sum+=t;
          printf("%d %.9f\n",n,sum);
       }
}