杭电1012-u Calculate e

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#杭电1012 #u Calculate e #打表 #规律

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本文详细介绍了如何通过一种简单的数学公式计算e值，并使用打表法进行快速逼近。通过实现一个高效的算法，可以准确地生成从n=0到n=9的e值近似，展示其在数学计算领域的应用与优势。

u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 36343 Accepted Submission(s): 16422

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

这个题主要是找到前后规律，然后打表即可！

AC代码：

#include<cstdio>
int jc(int x)
{
	int sum=1;
	for(int i=x;i>=1;i--)
	{
		sum*=i;
	}
	return sum;
}
int main()
{ 
	double a[11];
	int i;
	a[0]=1;
	for(i=1;i<=9;i++)
		a[i]=a[i-1]+1.0/jc(i);
	printf("n e\n- -----------\n");
	printf("0 1\n1 2\n2 2.5\n");
	for(i=3;i<=9;i++)
	printf("%d %.9lf\n",i,a[i]);
 }