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;}