杭电 ACM 1012

http://acm.hdu.edu.cn/showproblem.php?pid=1013

简单的小题；

代码：

#include<stdio.h>
#include<math.h>
double multi(int n)
{
int i;
double mul = 1;
for(i = 0;i <= n;i++)
{
if(i == 0)
{
mul = 1;
continue;
}
mul = mul*i;
}
return mul;
}
int main(void)
{
int n = 10;
int i;
double sum = 0;;
printf("n e\n- -----------\n");
for(i = 0;i < n;i++)
{
sum += 1/multi(i);
if(i < 2)
printf("%d %.0lf\n", i, sum);
if(i == 2)
printf("%d %.1lf\n", i, sum);
if(i > 2)
printf("%d %.9lf\n", i, sum);
}
getchar();
return 0;
}