HDOJ 1012

[align=left]Problem Description[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Output[/align]

n e
- -----------
0 1
1 2
2 2.5
3 2.666666667
4 2.708333333[code]#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;
}

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