HDOJ 1012

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
   
   
   
   
   
   
   
   

    
    
    
    
#include <iostream>
#include <cstring>
using namespace std;
int a[]={1,1,2,6,24,120,720,5040,40320,362880};
double func(int i)
{
	double r;
	r=1.0/a[i];
	return r;
}
int main()
{
	double sum=0;
	cout<<"n e"<<endl;
	cout<<"- -----------"<<endl;
	for(int i=0;i<=9;i++)
	{
		sum+=func(i);
		//cout<<i<<" "<<sum<<endl;
		cout<<i<<" ";
		if(i>=3) printf("%.9f",sum);
		else cout<<sum;
	//	if(i!=9) 
		cout<<endl;
	}
	return 0;
}