【HDU】1130 - How Many Trees?（java - BigDecimal & 卡特兰大数打表）

点击打开题目

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3718    Accepted Submission(s): 2088

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

123

 

Sample Output

125

 

Source

UVA

 

题意：给你一个数字 n ，问从1~n 可以构成多少种二叉树。

这种组合数学的思想还不是很清晰，但是知道是卡特兰数后就好办了，用 java 的 BigDecimal 打表就行了。

卡特兰数的递推公式：

F ( i ) = F ( i - 1 ) * ( 4 * i - 2 ) / ( n + 1 )

代码如下：

import java.math.BigDecimal;import java.util.Scanner;public class Main{public static void main(String[] args){    Scanner sc = new Scanner(System.in);     BigDecimal one = new BigDecimal(1);    BigDecimal two = new BigDecimal(2);    BigDecimal four = new BigDecimal(4);        //三个常数    BigDecimal f[] = new BigDecimal [111];    f[0] = f[1] = new BigDecimal(1);    f[2] = new BigDecimal(2);    for (int i = 3 ; i <= 100 ; i++)    {        BigDecimal t = new BigDecimal(i);        f[i] = f[i-1].multiply(four.multiply(t).subtract(two)).divide(t.add(one));    }    while (sc.hasNext())    {        int n = sc.nextInt();        System.out.println(f[n]);    }}}