hdu1261 字串数(排列组合、大整数)



Problem Description

一个A和两个B一共可以组成三种字符串:"ABB","BAB","BBA".
给定若干字母和它们相应的个数,计算一共可以组成多少个不同的字符串.

 

Input

每组测试数据分两行,第一行为n(1<=n<=26),表示不同字母的个数,第二行为n个数A1,A2,...,An(1<=Ai<=12),表示每种字母的个数.测试数据以n=0为结束.

 

Output

对于每一组测试数据,输出一个m,表示一共有多少种字符串.

 

Sample Input

2 1 2 3 2 2 2 0

 

Sample Output

3 90

题意：给出n个字母及每个字母的个数，求能构成多少组不同的字符串

思路：假设有k个个字母，每个字母的个数分别为n1,n2,...,nk,则相应的解为

(n1+n2+...+nk)! / n1! / n2! .../nk!，在计算时，涉及到大整数乘法和除法

代码如下：

import java.io.FileInputStream;
import java.io.OutputStreamWriter;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
import java.math.BigInteger;

public class Main implements Runnable
{  
    private static final boolean DEBUG = false;
    private Scanner cin;
    private PrintWriter cout;
    private static final int N = 26;
    private static final int M = 13;
    private int n;
    private final int[] c = new int[N];
    private int sum;
    private BigInteger[] big = new BigInteger[M];
    
    private void init()
    {
        try {
            if (DEBUG) {
                cin = new Scanner(new InputStreamReader(new FileInputStream("f:\\OJ\\uva_in.txt")));
            } else {
                cin = new Scanner(new InputStreamReader(System.in));
            }
            
            cout = new PrintWriter(new OutputStreamWriter(System.out));
        } catch (Exception e) {
            e.printStackTrace();
        }
        
        big[0] = BigInteger.ZERO;
        big[1] = BigInteger.ONE;
        for (int i = 2; i < M; i++) {
            big[i] = big[i - 1].multiply(BigInteger.valueOf(i));
        }
    }
    
    private boolean input()
    {
        n = cin.nextInt();
        if (n == 0) return false;
        
        sum = 0;
        for (int i = 0; i < n; i++) {
            c[i] = cin.nextInt();
            sum += c[i];
        }
        
        return true;
    }
    
    
    private void solve()
    {
        BigInteger ans = BigInteger.ONE;
        for (int i = 2; i <= sum; i++) {
            ans = ans.multiply(BigInteger.valueOf(i));
        }
        
        for (int i = 0; i < n; i++) {
            ans = ans.divide(big[c[i]]);
        }
        
        cout.println(ans.toString());
        cout.flush();
    }
    
    @Override
    public void run()
    {
        init();
        while (input()) 
        {
            solve();
        }
    }
    
    public static void main(String[] args) 
    {
        // TODO code application logic here
       new Thread(new Main()).start();
    }
}