学步园

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 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.

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

题目不难，但是注意下当i=8时，最后一位记得补一个0给他，否则会判你Wrong Answer  很奇怪的地方~~无语~

#include<iostream>
#include<iomanip>

using namespace std;

double factorial(int i)
{
	long result = 1;
	for(int j = 1;j<=i;j++)
		result *= j;
	return result;
}
int main()
{
	cout << "n e"<<endl<<"- -----------"<<endl;
	double e = 0;
	for(int i = 0;i<=9;++i)
	{
		e += 1/factorial(i);
		cout << i <<" ";
		if(i!=8)
			cout << setprecision(10) << e <<endl;
		else cout << setprecision(10) << e <<"0" <<endl;
		
	}
	
	return 0;
}