HDU 1130 How Many Trees?

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1130

题面：

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3426    Accepted Submission(s): 1976

Problem Description

A binary search tree is a binary tree with root k such that any node v reachable from its left has label (v) <label (k) and any node w reachable from its right has label (w) > label (k). It is a search structure which can find a node with label x in O(n log n) average time, where n is the size of the tree (number of vertices). 

Given a number n, can you tell how many different binary search trees may be constructed with a set of numbers of size n such that each element of the set will be associated to the label of exactly one node in a binary search tree? 

 

Input

The input will contain a number 1 <= i <= 100 per line representing the number of elements of the set.

 

Output

You have to print a line in the output for each entry with the answer to the previous question.

 

Sample Input

   
   
   
   

    
    
    
    1
2
3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
5
   
   
   
   

 

Source

UVA

 

题目大意：

    问给定不同的数，能组成多少组有序二叉树。

解题：

    实质就是求卡特兰数， h(n)=(4*n-2)/(n+1)*h(n-1);数字较大，得用Java或者C++大数

代码：

import java.math.BigInteger;
import java.util.Scanner;
public class Main
{
	public static void main(String []args)
	{
		Scanner cin=new Scanner(System.in);
		BigInteger h[]=new BigInteger[105];
		h[0]=new BigInteger("1");
		h[1]=new BigInteger("1");
		BigInteger one=new BigInteger("1");
		BigInteger two=new BigInteger("2");
		BigInteger four=new BigInteger("4");		
		for(int i=2;i<105;i++)
		{
			BigInteger tmp1=BigInteger.valueOf(4*i-2);
			BigInteger tmp2=BigInteger.valueOf(i+1);
			h[i]=h[i-1].multiply(tmp1); 
			h[i]=h[i].divide(tmp2);
		}
		while(cin.hasNext())
		{
			int n=cin.nextInt();
			System.out.println(h[n]);
		}
	}
}