hdu-1012 u Calculate e

题目来源：hdu-1012

u Calculate e
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36307    Accepted Submission(s): 16399

Problem Description
A simple mathematical formula for e is

e=Σ1/i！(i=0~n);

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

Source
Greater New York 2000

题目大意：

给出一个e的表达式，即是求前n项的阶乘的导数之和。表达式为：e=1+1/1!+1/2!+1/3!+1/4!+……+1/9!；输出参考输出样例，依次输出n从0至9的结果。

解题思路：

首先按照输出及简单运算可知n等于0、1、2的时候对应结果为1、2、2.5；将此直接输出结果，此后结果含有多项小数，保留小数点后9位；对此可先求出i的阶乘，然后代入表达式，将运算结果输出即可。

AC代码：

#include<stdio.h>
double factorial(int n)     //求阶乘函数
{
double num=1;
for(int i=1;i<=n;i++)
num*=i;
return num;
}
int main()
{
double sum;             //对前面可直接计算出结果的单独输出
printf("n e\n- -----------\n0 1\n1 2\n2 2.5\n");
sum=2.5;
for(int i=3;i<=9;i++)   //将运算结果逐个计算并输出
printf("%d %.9lf\n",i,sum+=1/factorial(i));
return 0;
}