u Calculate e

Time Limit: 2000/1000 MS
(Java/Others)    Memory Limit: 65536/32768 K
(Java/Others)
Total Submission(s): 28686    Accepted
Submission(s): 12762

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

Source

Greater
New York 2000

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　　水题。

　　切一道水题，放松下心情。

　　这道题没有输入，只有输出。

　　前3组数据由于输出格式不统一，直接输出即可。后面的数可用迭代思路求得，不用从头重新计算了。

　　代码：

 1 #include <stdio.h>

 2 int main()

 3 {

 4     int i;

 5     printf("n e\n");

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

 7     printf("0 1\n");    //没有统一格式，提前输出。

 8     printf("1 2\n");

 9     printf("2 2.5\n");

10     double ans = 1;

11     double t = 1;

12     for(i=1;i<=9;i++){

13         t = 1.0/i*t;    //迭代计算

14         ans += t;

15         if(i<3) continue;

16         printf("%d %.9lf\n",i,ans);

17     }

18     return 0;

19 }

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