u calculate e → 计算e题（Problem ID：1012）

题址：http://acm.hdu.edu.cn/showproblem.php?pid=1012

简单题

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

AC代码：#include<stdio.h>

int jieceng(int cur)
{
if(cur==0||cur==1) return 1;
else return cur*jieceng(cur-1);
}

int main()
{
double e,sum=2.5;
printf("n e\n- -----------\n");
printf("0 1\n1 2\n2 2.5\n");
for(int i=3;i<=9;i++)
{
e=1/(double)jieceng(i);
sum+=e;
printf("%d %.9lf\n",i,sum);
}

return 0;
}