hdu 1042 N!

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 83969    Accepted Submission(s): 24758

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

 

 

Author

JGShining（极光炫影）

 

 

题解：也是一道大整数和整数的乘法     可以使用java来做

java对于大整数的题目  相比C，C++来说    有很大的优势    毕竟它自带了大整数0.0

以下是代码：

特别注意以下  java  一定要定义成Main

import java.math.BigInteger;
import java.util.Scanner;
 
 
public class Main {

    public static void main(String[] args) {
        Scanner cin = new Scanner ( System.in );
        BigInteger one = BigInteger.ONE;
        while ( cin.hasNext() ) {
            BigInteger N = cin.nextBigInteger();
            BigInteger sum = new BigInteger ( "1" );
            for ( BigInteger i = BigInteger.ONE; i.compareTo(N) <= 0; i = i.add(one) ) {
                sum = sum.multiply( i );
            }
            System.out.println( sum );
        }
    }
 
}

也可以使用C，C++来敲的

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int a[100002];
int main()
{
    int n;
    int i,j,k,count,temp;
    while(cin>>n)
    {
        a[0]=1;
        count=1;
        for(i=1;i<=n;i++)
        {
            k=0;
            for(j=0;j<count;j++)
            {
                temp=a[j]*i+k;
                a[j]=temp%10;
                k=temp/10;
            }
            while(k)//记录进位
             {
                a[count++]=k%10;
                k/=10;
            }
        }
        for(j=100001;j>=0;j--)
            if(a[j])
                break;//忽略前导0
            for(i=count-1;i>=0;i--)
                cout<<a[i];
            cout<<endl;
    }
    return 0;
}