u Calculate e解题报告

最新推荐文章于 2019-01-11 16:02:37 发布

原创最新推荐文章于 2019-01-11 16:02:37 发布 · 577 阅读

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本文介绍了一个简单数学公式用于近似计算e值，并通过代码实现，展示从n=0到n=9的计算结果。

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u Calculate e

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16844 Accepted Submission(s): 7307

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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#include<iostream>
using namespace std;
double pow(double i)
{
	double sum=1,j;
	for(j=1;j<=i;j++)
	{
		sum*=j;
	}
	return 1/sum;
}
double he(double i)
{
	double sun=0,sum=0,j;
	for(j=1;j<=i;j++)
	{
		sum+=pow(j);
	}
	return sum;
}
int main()
{
	double a,i,j,sum=1,e;
	cout<<'n'<<' '<<'e'<<endl;
	cout<<'-'<<' '<<"-----------"<<endl;
	cout<<'0'<<' '<<'1'<<endl;
	cout<<'1'<<' '<<'2'<<endl;
	cout<<'2'<<' '<<"2.5"<<endl;

	for(i=3;i<=9;i++)
	{
		cout<<i<<' ';
		e=he(i);
		printf("%.9f\n",e+1);
	}
	return 0;
}