Hdu-1023 Train Problem II（卡特兰数、大数乘除模拟）_二维数组解决卡特兰问题-优快云博客

本文链接：https://blog.youkuaiyun.com/xiaoyao_zhang/article/details/79917729

本文探讨了火车调度问题中的数学模型，利用卡特兰数解决火车按严格递增顺序进站后可能的离站顺序数量问题。提供了解决方案的C++和Java实现代码，包括大数处理技巧。

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Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10546 Accepted Submission(s): 5637

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

Sample Input

12310

Sample Output

12516796Hint
The result will be very large, so you may not process it by 32-bit integers.
 

卡特兰数递推公式： h[n]=h[n-1]*(4*n-2)/(n-1)

#include<iostream>
using namespace std;
int a[101][101];
//二维数组第i行表示第i个卡特兰数,a[i][0]表示第i个卡特兰数的位数 
void ktl()
{
	int s,c,len=1;
	a[1][0]=1,a[1][1]=1,a[2][0]=1,a[2][1]=2;
	for(int i=3;i<=100;i++)
	{
		c=0;
		for(int j=1;j<=len;j++)//大数相乘 ，从最低位个位开始依次与（4*i-2）相乘 
		{
			s=a[i-1][j]*(4*i-2)+c;
			a[i][j]=s%10;
			c=s/10;
		}
		while(c)//开拓更高位 
		{
			a[i][++len]=c%10;
			c/=10;
		}
		
		for(int j=len;j>0;j--)//大数相除 ，从最高位开始依次与（i+1）相除 
		{
			s=a[i][j]+c*10;
			c=s%(i+1);
			a[i][j]=s/(i+1);
		}
		while(!a[i][len])//确定多少位 
		len--;
		a[i][0]=len;
	}
}
int main()
{
	ktl();
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		for(int i=a[n][0];i>0;i--)
		cout<<a[n][i];
		cout<<endl;
	}
}

Java语言，很6

import java.io.*;
import java.util.*;
import java.math.BigInteger;
public class Main
{
    public static void main(String args[])
    {
        BigInteger[] a=new BigInteger[101];
        a[0]=BigInteger.ZERO;
        a[1]=BigInteger.valueOf(1);
        for(int i=2;i<=100;++i)
            a[i]=a[i-1].multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));
        Scanner in=new Scanner(System.in);
        int n;
        while(in.hasNext())
        {
            n=in.nextInt();
            System.out.println(a[n]);
        }
    }
}