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：

 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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