N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 53785    Accepted Submission(s): 15217

Problem Description

Given an integer N(0 ≤ N ≤ 10000), your task is to calculate N!

Input

One N in one line, process to the end of file.

Output

For each N, output N! in one line.

Sample Input

1

2

3

Sample Output

1

2

6

#include<stdio.h>

#include<string.h>

const int maxn=50000;    //数组开到50000就能够满足10000的阶乘不越界

int fun[maxn];

int main()

{

	int i,j,n;

	while(~scanf("%d",&n))

	{

	   memset(fun,0,sizeof(fun));

	   fun[0]=1;

	   for(i=2;i<=n;i++)          //从2的阶乘開始，一直到指定数的阶乘

	   {

		   int c=0;

	    	for(j=0;j<maxn;j++)  //将所得阶乘数放在fun数组中，低位放在fun[0]中

	    	{

			    int s=fun[j]*i+c;

		      	fun[j] =s%10;

		      	c=s/10;

	        }

       }

	  for(j=maxn-1;j>=0;j--)     //找出该数的最高位，即数组角码最大且不为0的数

	           if(fun[j])  break;

	  for(i=j;i>=0;i--)

	          printf("%d",fun[i]);

 	  printf("\n");

	}

	return 0;

}