hdu 1130，hdu 1131(卡特兰数,大数)

How Many Trees?

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3382    Accepted Submission(s): 1960

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

 

题意：由 n个结点组成二叉树的种数

卡特兰数+BigInteger

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

 
public class Main {
    public static void main(String[] args) {
        BigInteger [] h = new BigInteger[101];
        h[1] = new BigInteger("1");
        for(int i=2;i<=100;i++){
            h[i] = h[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
        }
        Scanner sc =new Scanner (System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            System.out.println(h[n]);
        }
    }
}

 hdu:1131  由n个带编号的结点组成二叉树的个数

思路：卡特兰数乘上编号的全排列

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

 
public class Main {
    public static void main(String[] args) {
        BigInteger [] h = new BigInteger[101];
        h[1] = new BigInteger("1");
        for(int i=2;i<=100;i++){
            h[i] = h[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
        }
        Scanner sc =new Scanner (System.in);
        while(sc.hasNext()){
            int n =sc.nextInt();
            if(n==0)break;
            System.out.println(h[n].multiply(fac(n)));
        }
    }

    private static BigInteger fac(int n) {
        BigInteger sum = BigInteger.valueOf(1);
        for(int i=1;i<=n;i++) sum=sum.multiply(BigInteger.valueOf(i));
        return sum;
    }
    
}