POJ 2084 Catalan

Game of Connections

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 8772		Accepted: 4324

Description

This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, . . . , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. 
And, no two segments are allowed to intersect. 
It's still a simple game, isn't it? But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs? Life is harder, right?

Input

Each line of the input file will be a single positive number n, except the last line, which is a number -1. 
You may assume that 1 <= n <= 100.

Output

For each n, print in a single line the number of ways to connect the 2n numbers into pairs.

Sample Input

2
3
-1

Sample Output

2
5

Source

Shanghai 2004 Preliminary

PS：卡特兰数在ACM中比较具体的使用例子有，1括号匹配的种数。2在栈中的自然数出栈的种数。3求多边形内三角形的个数。4，n个数围城圆圈，找不相交线段的个数。5给定n个数，求组成二叉树的种数，本题为第四种。

　　h(n)=(4*n-2)/(n+1)*h(n-1).

代码：

 import java.math.*;
 import java.util.*;
 public class Main {
 public static void main(String[] args) {
         Scanner cin=new Scanner(System.in);
         int t;
         BigInteger f[]=new BigInteger[1005];
         f[1]=BigInteger.valueOf(1);
         f[2]=BigInteger.valueOf(2);
         f[3]=BigInteger.valueOf(5);
         for(int i=4;i<=100;i++){
             BigInteger a=BigInteger.valueOf(4*i-2);
             BigInteger b=BigInteger.valueOf(i+1);
             f[i]=a.multiply(f[i-1]).divide(b);
         }
         while(cin.hasNextInt())
         {
             t=cin.nextInt();
             if(t==-1) break;
             System.out.println(f[t]);
         }
 }
 }