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

123

 

Sample Output

126
#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);

}
}


}