Game of Connections（大数java+Catalan数）

游戏连接数的计算与Catalan数的应用

最新推荐文章于 2023-02-28 15:13:05 发布

转载最新推荐文章于 2023-02-28 15:13:05 发布 · 909 阅读

文章标签：

#ACM

ACM 同时被 3 个专栏收录

521 篇文章

订阅专栏

数论

79 篇文章

订阅专栏

poj

73 篇文章

订阅专栏

本文探讨了围绕2n个点形成圆圈，并将其编号后进行两两连线的问题，通过数学和高精度计算求解不同连接方式的数量，引入Catalan数的概念并给出递推公式及解析过程。

Link：http://poj.org/problem?id=2084

Problem：

Game of Connections

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 7497		Accepted: 3798

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

题意：有2n个点围成一个圆，现在把所有的点都和另外一个点连上线，问总共有多少种连法？

思路：数学+高精度。递推的公式：dp(n) = dp(0)*dp(n-1) + dp(1)*dp(n-2) +...+ dp(n-1)*dp(0)。

这就是Catalan数，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,...)

import java.io.*;
import java.math.*;
import java.util.*;
import java.text.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        BigInteger t1, t2, dp[] = new BigInteger[105];
        dp[1] = BigInteger.valueOf(1);
        for(int i = 2; i <= 100; i ++){
            t1 = BigInteger.valueOf(4 * i - 2);    //  要注意(4n-2)/(n+1)可能为小数。
            t2 = BigInteger.valueOf(i + 1);
            dp[i] = dp[i-1].multiply(t1).divide(t2);
        }
        while(cin.hasNext()){
            int id = cin.nextInt();
            if(id == -1) break;
            System.out.println(dp[id]);
        }
        System.exit(0);
    }
}