1.链接地址：

http://bailian.openjudge.cn/practice/1517

http://poj.org/problem?id=1517

2.题目：

总时间限制:

1000ms

内存限制:

65536kB

描述

A simple mathematical formula for e is
e=Σ_0<=i<=n1/i!
where
n is allowed to go to infinity. This can actually yield very accurate
approximations of e using relatively small values of n.

输入

No input

输出

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.

样例输入

no input

样例输出

n e

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

0 1

1 2

2 2.5

3 2.666666667

4 2.708333333

...

来源

Greater New York 2000

3.思路：

4.代码：

 #include "stdio.h"

 //#include "stdlib.h"

 int main()

 {

     int tmp=;

     double sum=;

     int i=;

     printf("n e\n");

     printf("- -----------\n");

     printf("%d %.10g\n",i,sum);

     for(i=;i<;i++)

     {

        tmp*=i;

        sum+=(double)/tmp;

        printf("%d %.10g\n",i,sum);

     }

     //system("pause");

     return ;

 }