HDOJ 1261 字串数

Posted 2020-10-01 zhchoutai

JAVA大数....

字串数

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2893    Accepted Submission(s): 679

Problem Description

一个A和两个B一共能够组成三种字符串:"ABB","BAB","BBA".
给定若干字母和它们对应的个数,计算一共能够组成多少个不同的字符串.

 

Input

每组測试数据分两行,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.測试数据以n=0为结束.

 

Output

对于每一组測试数据,输出一个m,表示一共同拥有多少种字符串.

 

Sample Input

2
1 2
3
2 2 2
0

 

Sample Output

3
90

 

Source

php?

field=problem&key=%D5%E3%BD%AD%B9%A4%D2%B5%B4%F3%D1%A7%B5%DA%CB%C4%BD%EC%B4%F3%D1%A7%C9%FA%B3%CC%D0%F2%C9%E8%BC%C6%BE%BA%C8%FC&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">浙江工业大学第四届大学生程序设计竞赛

 

import java.util.*;
import java.math.*;

public class Main
{
	public static void main(String[] args)
	{
		Scanner in=new Scanner(System.in);
		BigInteger[] jie=new BigInteger[20];
		jie[0]=BigInteger.ONE;
		BigInteger nt=BigInteger.ONE;
		for(int j=1;j<=14;j++)
		{
			jie[j]=jie[j-1].multiply(nt); 
			nt=nt.add(BigInteger.ONE);
		}
		while(in.hasNext())
		{
			int n=in.nextInt();
			if(n==0) break;
			BigInteger[] a=new BigInteger[n];
			int[] av=new int[n];
			BigInteger sum=BigInteger.ZERO;
			for(int i=0;i<n;i++)
			{
				av[i]=in.nextInt();
				a[i]=BigInteger.valueOf(av[i]);
				sum=sum.add(a[i]);
			}
			BigInteger ans=BigInteger.ONE;
			for(BigInteger i=BigInteger.ONE;i.compareTo(sum)<=0;i=i.add(BigInteger.ONE))
			{
				ans=ans.multiply(i);
			}
			for(int i=0;i<n;i++)
			{
				ans=ans.divide(jie[av[i]]);
			}
			System.out.println(ans);
		}
	}
}