hdu 1023 Train Problem II(java+卡特兰数)

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6442    Accepted Submission(s): 3505

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

 

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

 

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

 

Sample Input

1
2
3
10

 

Sample Output

1
2
5
16796

Hint
The result will be very large, so you may not process it by 32-bit integers.
 

 

Author

Ignatius.L

题目分析：给出了n个火车，和他们进站的顺序，问出栈的顺序有多少种，一看就是出栈入栈的问题，很容易想到解决递归问题的卡特兰数，首先我们将进栈状态看成是1，出栈状态看成是０，那么所哟状态的顺序种类是C(2*n,n)，然后去除掉非法的状态，也就是状态０比１状态多的情况，也就是在2*m+1位上累计出现了m+1个0，后面的序列中必定还有n-m-1个０和n-m+1个１，如果０和1对换不影响排列种类，所以每个不合法的顺位对应一个n+1个０，和n-1个１的序列，那么sum = c(2*n,n) -c(2*n,n+1)，正好是卡特兰数，卡特兰数在解决递归计数问题中有着不可替代的作用

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

public class Main 
{
    BigInteger [][] c = new BigInteger[202][202];
	public  void init ( )
	{
		c[1][1] = c[1][0] =c[0][0] = BigInteger.ONE;
		 for ( int i = 2 ; i <=  200 ; i++ )
			 for ( int j = 0 ; j <= i ; j++ )
				 if ( j == 0 || j == i ) c[i][j] = BigInteger.ONE;
				 else c[i][j] = c[i-1][j-1].add(c[i-1][j]);
	}
	
	public static  void main ( String args[]  )
	{
		Main a = new Main ();
		a.init();
		Scanner cin = new Scanner ( System.in );
		while ( cin.hasNext() )
		{
			int n = cin.nextInt();
			System.out.println ( a.c[2*n][n].divide(BigInteger.valueOf(n+1)) );
		}
	}
}