杭电1012--u Calculate e

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原文链接：http://www.cnblogs.com/soTired/p/4596615.html

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#java

本文介绍了一种通过数学公式计算自然常数e的近似值的方法，并提供了一个C语言实现的例子，该程序能够输出当n从0到9时e的不同近似值。

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 35178 Accepted Submission(s): 15843

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

//水题，格式；

 1 #include<stdio.h>
 2 double sieve[100];
 3 int main()
 4 {
 5     int i, total=1; sieve[0]=1;
 6     for(i=1;i<10;i++)
 7     {
 8         total*=i;
 9         sieve[i]=sieve[i-1]+1.0/total;    
10     }
11     printf("n e\n");
12     printf("- -----------\n");
13     int n;
14     for(i=0;i<10;i++)
15     {
16         printf("%d ",i);
17         if(i<=1)
18         printf("%d\n",(int)sieve[i]);
19         else if(i==2)
20         printf("%.1lf\n",sieve[i]);
21         else
22         printf("%.9lf\n",sieve[i]);
23     }
24     return 0;
25 }