hdu1042 N!(大数求阶乘)

N!

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 94583    Accepted Submission(s): 28107

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

数据范围比较大，要用到大数。

c++

#include<bits/stdc++.h>
using namespace std;
string Multiply(string s,int x)
{
    reverse(s.begin(),s.end());
    int cmp=0;
    for(int i=0; i<s.size(); i++)
    {
        cmp=(s[i]-'0')*x+cmp;
        s[i]=(cmp%10+'0');
        cmp/=10;
    }
    while(cmp)
    {
        s+=(cmp%10+'0');
        cmp/=10;
    }
    reverse(s.begin(),s.end());
    return s;
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
            string sum="1";
            for(int i=1; i<=n; i++)
            {
                sum=Multiply(sum,i);
            }
            cout<<sum<<endl;
    }
    return 0;
}

Java

import java.math.BigInteger;
import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in );
        while(cin.hasNext()) {
               int n=cin.nextInt();
                   BigInteger sum=BigInteger.valueOf(1);
                   for(int i=1;i<=n;i++)
                   {
                       BigInteger temp=BigInteger.valueOf(i);
                       sum=sum.multiply(temp);
                   }
                   System.out.println(sum);
             
        }    
    }
}

PS:这题好像还是Java比较快。。。