hdu 1012:u Calculate e（数学题，水题）

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：

 #include <stdio.h>
 int main()
 {
     int i;
     printf("n e\n");
     printf("- -----------\n");
     printf("0 1\n");    //没有统一格式，提前输出。
     printf("1 2\n");
     printf("2 2.5\n");
     double ans = ;
     double t = ;
     for(i=;i<=;i++){
         t = 1.0/i*t;    //迭代计算
         ans += t;
         if(i<) continue;
         printf("%d %.9lf\n",i,ans);
     }
     return ;
 }

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