POJ 1517 u Calculate e

最新推荐文章于 2017-05-27 22:23:29 发布

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最新推荐文章于 2017-05-27 22:23:29 发布

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原文链接：http://www.cnblogs.com/lzj-0218/archive/2013/05/02/3055345.html

本文详细介绍了如何使用数学公式计算e的近似值，并通过代码实现，展示从n=0到n=9的计算结果。

u Calculate e

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 16932		Accepted: 10033		Special Judge

Description

A simple mathematical formula for e ise=Σ_0<=i<=n1/i!where n is allowed to go to infinity. This can actually yield very accurate approximations of e using relatively small values of n.

Input

No input

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 Input

no input

Sample Output

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333
...

这根本就是a+b的变形题么....

[C]

1 #include<stdio.h>
2 
3 int main()
4 {
5     printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n3 2.666666667\n4 2.708333333\n5 2.716666667\n6 2.718055556\n7 2.718253968\n8 2.718278770\n9 2.718281526\n");
6     return 0;
7 }