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就可以满足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的阶乘开始，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;}

HDoj-1042 大数阶乘