N!(杭电1042)（数组实现+java实现）

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 55971    Accepted Submission(s): 15886

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>
int main()
{
	int i,j,n,m;
	while(scanf("%d",&n)!=EOF)
	{
		if(n<0)
		{ 
		   continue;
	    }
		int str[10000]={0};
		str[0]=1;
		m=0;
		for(i=1;i<=n;i++)//m用来控制进位。
 
			for(j=0;j<=m;j++)
			{
				str[j]=str[j]*i;
				if(j>0&&str[j-1]>=10000)
				{
				   str[j]=str[j]+str[j-1]/10000;
				   str[j-1]=str[j-1]%10000;
			    }
			    if(str[m]>=10000)
			       m++;
			}
		printf("%d",str[m]);   
		for(i=m-1;i>=0;i--)
		    printf("%04d",str[i]);
		printf("\n");
	}
	return 0;
}
java语言实现~
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Scanner;
public class Main {


	public static void main(String[] args) {
		
		Scanner in=new Scanner(System.in);
		while(in.hasNext())
		{
			int n=in.nextInt();
			BigDecimal m=new BigDecimal(1);
		    BigDecimal a;
		    for(int i=2;i<=n;i++)
		    {
		    	a=new BigDecimal(i);
		        m=m.multiply(a);
		    }
			System.out.println(m);
			
		}
	}


}