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=;
     for(int i=; i<s.size(); i++)
     {
         cmp=(s[i]-'')*x+cmp;
         s[i]=(cmp%+'');
         cmp/=;
     }
     while(cmp)
     {
         s+=(cmp%+'');
         cmp/=;
     }
     reverse(s.begin(),s.end());
     return s;
 }
 int main()
 {
     int n;
     while(~scanf("%d",&n))
     {
             string sum="";
             for(int i=; i<=n; i++)
             {
                 sum=Multiply(sum,i);
             }
             cout<<sum<<endl;
     }
     return ;
 }

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();
                    for(int i=;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。。。