HDU u Calculate e

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

超级水题，无聊上一下代码：

AC代码：

# include <cstdio>
using namespace std;
int main(){
	int i, j, k;
	printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
	double ans=2.5;
	double jie=0.5;
	for(int i=3; i<=9; i++){
		printf("%d ", i);
		ans=ans+jie/i;
		jie=jie/i;
		printf("%.9lf\n", ans);
	}
	return 0;
}