poj2084

package middle;


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

/**
 * 令h(0)＝1,h(1)＝1，catalan数满足递归式：
 * h(n)= h(0)*h(n-1) + h(1)*h(n-2) + ... + h(n-1)h(0) (其中n>=2),这是n阶递推关系;
 * 还可以化简为1阶递推关系: 如h(n)=(4n-2)/(n+1)*h(n-1)(n>1) h(0)=1
 * 该递推关系的解为：h(n)=c(2n,n)/(n+1) (n=1,2,3,...)//2n!/n!/n!
 * 卡 塔兰数例的前几项为(sequence A000108 in OEIS) [注： n = 0, 1, 2, 3, … n]
 * 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440,
 * 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020,
 * 91482563640, 343059613650, 1289904147324, 4861946401452, …
 * poj2084
 * @author NC
 */
public class Poj2084 {

    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedInputStream(System.in));
        while (scan.hasNext()) {
            BigInteger n = scan.nextBigInteger();
            if (n.compareTo(BigInteger.ZERO) == -1) {
                break;
            }
            BigInteger ss = BigInteger.ONE;
            BigInteger c = BigInteger.ONE;
            BigInteger s = null ;
            for (int i = 1; i <= 2 * n.intValue(); i++) {
                ss = ss.multiply(c);
                c=c.add(BigInteger.ONE);
                if(i==n.intValue()){
                s = ss;
                }
            }
            System.out.println(ss.divide(s).divide(s).divide(n.add(BigInteger.ONE)));

        }
    }
}