hdu 1133 Buy the Ticket（Catalan）

Buy the Ticket

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5651    Accepted Submission(s): 2357

Problem Description

The "Harry Potter and the Goblet of Fire" will be on show in the next few days. As a crazy fan of Harry Potter, you will go to the cinema and have the first sight, won’t you?

Suppose the cinema only has one ticket-office and the price for per-ticket is 50 dollars. The queue for buying the tickets is consisted of m + n persons (m persons each only has the 50-dollar bill and n persons each only has the 100-dollar bill).

Now the problem for you is to calculate the number of different ways of the queue that the buying process won't be stopped from the first person till the last person. 
Note: initially the ticket-office has no money. 

The buying process will be stopped on the occasion that the ticket-office has no 50-dollar bill but the first person of the queue only has the 100-dollar bill.

 

Input

The input file contains several test cases. Each test case is made up of two integer numbers: m and n. It is terminated by m = n = 0. Otherwise, m, n <=100.

 

Output

For each test case, first print the test number (counting from 1) in one line, then output the number of different ways in another line.

 

Sample Input

   
   
   
   

    
    
    
    3 0
3 1
3 3
0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    Test #1:
6
Test #2:
18
Test #3:
180
   
   
   
   

 

Author

HUANG, Ninghai

题意：

m個人拿50，n個人拿100，而且售票处没有零钱。求这排人最终能够能全买到票的方案数

（如果没有50零钱找给拿一百的人则停）。

思路：

排列的所有可能情况：C(n+m,m)

假设一个数列 0110010当买到第三个人的时候便停止了，将后面的数1->0,0->1便能得到 0111101

反之亦然，于是我们可以得出他们是一一对应的。

于是我们用所有可能减去不成立的

(C(n+m,m) - C(n+m,m+1))*m!*n! = (n+m)! * (m-n+1) / (m+1); 

然后再利用JAVA的大整数即可 

import java.math.BigInteger;
import java.util.Scanner;

public class Main {

	static BigInteger []F =  new BigInteger[105];
	static BigInteger []A =  new BigInteger[205];
	public static void ini()
	{
		F[1] = BigInteger.valueOf(1);
		A[1] = BigInteger.valueOf(1);
		A[0] = BigInteger.valueOf(0);
		F[0] = F[1];
		for(int i = 2;i < 105;i++)
		{
			F[i] = F[i-1].multiply(BigInteger.valueOf(i*4-2)).divide(BigInteger.valueOf(i+1));
		}
		for(int i = 2;i <= 203;i++)
		{
			A[i] = A[i-1].multiply(BigInteger.valueOf(i));
		}
	}
	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ini();
		Scanner Reader = new Scanner(System.in);
		int x,y;
		int cas = 1;
		BigInteger ans;
		while(true)
		{
			x = Reader.nextInt();
			y = Reader.nextInt();
			
			if(x == 0 && y == 0)
				break;
			System.out.println("Test #"+cas+":");
			cas++;
			if(x < y){
				System.out.println("0");
			}
			else
			{
			  int t = x+y;
			  ans = A[t].multiply(BigInteger.valueOf(x-y+1)).divide(BigInteger.valueOf(x+1));
						  System.out.println(ans);
			}
		}     
	}

}