hdu 1012

最新推荐文章于 2025-09-05 16:34:31 发布

转载最新推荐文章于 2025-09-05 16:34:31 发布 · 51 阅读

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原文链接：http://www.cnblogs.com/hpuzyf/p/3365439.html

文章标签：

#java #php

本文介绍了一种通过数学公式计算自然对数的底数e的近似值的方法，并给出了一段C语言代码实现，该代码能够计算并输出当n从0到9时e的不同近似值。

http://acm.hdu.edu.cn/showproblem.php?pid=1012

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25495 Accepted Submission(s): 11313

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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 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 
 4 int main()
 5 {
 6     printf("n e\n");
 7     printf("- -----------\n");
 8     printf("0 1\n");
 9     printf("1 2\n");
10     printf("2 2.5\n");
11     
12     
13     double m=2.0,i,j,s=2.5;
14     for(i=3,j=3;i<10;i++,j++)
15     {
16       m*=j;  
17       //printf("%.1f\n",m);         
18       s=s+1/m;
19       printf("%.0f %.9f\n",i,s);
20     }
21     system("pause");
22     return 0;
23     
24 }