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++

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 string Multiply(string s,int x)
 4 {
 5     reverse(s.begin(),s.end());
 6     int cmp=0;
 7     for(int i=0; i<s.size(); i++)
 8     {
 9         cmp=(s[i]-'0')*x+cmp;
10         s[i]=(cmp%10+'0');
11         cmp/=10;
12     }
13     while(cmp)
14     {
15         s+=(cmp%10+'0');
16         cmp/=10;
17     }
18     reverse(s.begin(),s.end());
19     return s;
20 }
21 int main()
22 {
23     int n;
24     while(~scanf("%d",&n))
25     {
26             string sum="1";
27             for(int i=1; i<=n; i++)
28             {
29                 sum=Multiply(sum,i);
30             }
31             cout<<sum<<endl;
32     }
33     return 0;
34 }

View Code

Java

 1 import java.math.BigInteger;
 2 import java.util.Scanner;
 3 public class Main {
 4     public static void main(String[] args) {
 5         Scanner cin = new Scanner(System.in );
 6         while(cin.hasNext()) {
 7                int n=cin.nextInt();
 8                    BigInteger sum=BigInteger.valueOf(1);
 9                    for(int i=1;i<=n;i++)
10                    {
11                        BigInteger temp=BigInteger.valueOf(i);
12                        sum=sum.multiply(temp);
13                    }
14                    System.out.println(sum);
15              
16         }    
17     }
18 }

View Code

PS:这题好像还是Java比较快。。。