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：

#include "stdio.h"
//#include "stdlib.h"
int main()
{
    int tmp=1;
    double sum=1;
    int i=0;
    printf("n e\n");
    printf("- -----------\n");
    printf("%d %.10g\n",i,sum);
    for(i=1;i<10;i++)
    {
       tmp*=i;
       sum+=(double)1/tmp;
       printf("%d %.10g\n",i,sum);
    }
    //system("pause");
    return 0;
}