HDOJ 1042 N!



http://acm.hdu.edu.cn/showproblem.php?pid=1042

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 54185    Accepted Submission(s): 15370

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>
#define MAX 20000
int main()
{
	char s[100];
	int N,a[2*MAX],b[MAX],c[2*MAX],n,m,i,j,k;
	while(scanf("%d",&N)!=EOF)
	{
		if(N==0||N==1)//如果是算0,1的阶乘，则是1 
		{
			printf("1\n");
			continue;
		}
	
		memset(a,0,sizeof(a));
		a[0]=1;
		for(k=2;k<=N;k++)
		{
			memset(b,0,sizeof(b));
			sprintf(s,"%d",k);
			n=strlen(s);
			for(j=0,i=n-1;i>=0;i--,j++)
            b[j]=s[i]-'0';
			   
			for(i=2*MAX-1;i>0;i--)//a[]是k的阶乘 
			if(a[i]!=0) break;m=i;
			
	        memset(c,0,sizeof(c));
			for(i=0;i<n;i++)
			{
				for(j=0;j<=m;j++)
				c[j+i]+=a[j]*b[i];
			}
			for(i=0;i<2*MAX;i++)
			{
				if(c[i]>=10)
				{
					c[i+1]+=c[i]/10;
					c[i]=c[i]%10;
				}
			}
			
			for(i=0;i<2*MAX;i++)
			a[i]=c[i];
			     
		}
		
		for(i=2*MAX-1;i>0;i--)
		if(c[i]!=0) break;
		for(;i>=0;i--)
		printf("%d",c[i]);
		printf("\n");
	}
	return 0;
}