OpenJudge/Poj 1517 u Calculate e

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：

 1 #include "stdio.h"

 2 //#include "stdlib.h"

 3 int main()

 4 {

 5     int tmp=1;

 6     double sum=1;

 7     int i=0;

 8     printf("n e\n");

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

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

11     for(i=1;i<10;i++)

12     {

13        tmp*=i;

14        sum+=(double)1/tmp;

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

16     }

17     //system("pause");

18     return 0;

19 }