奋战杭电ACM（DAY16）1023

原创于 2013-09-08 09:18:51 发布 · 526 阅读

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本文探讨了一个火车调度问题，即所有火车严格按升序进入车站后可能的离站顺序数量。通过使用卡塔兰数解决该问题，并分享了实现过程中遇到的大数运算挑战及解决方法。

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被这道题虐死了……

先是完全混乱，再一搜，卡塔兰数，数学没学过怎么办……

有了递推式，开始写代码，大数乘除又虐了一次……

全部写完了，提交——RE，泪……结果是数组开小了……尼玛开100还小！！开200过了……

Train Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4661 Accepted Submission(s): 2544

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

Sample Output

1
2
5
16796

Hint
The result will be very large, so you may not process it by 32-bit integers.
 
代码：
#include <iostream>
using namespace std;

#define MAXN 200

int main()
{
	int catalan[MAXN][MAXN],N;
	memset(catalan,0,sizeof(catalan));
	while(cin >> N)
	{
		catalan[1][MAXN-1]=1;
		bool flag= false;
		int i,j;
		if(N>=2)
		{
			for(i=2; i<=N; i++)
			{
				//乘法
				int temp=0,r=0;
				for(j=MAXN-1; j>0; j--)
				{
					temp=catalan[i-1][j]*(4*i-2)+r;
					catalan[i][j]=temp%10;
					r=temp/10;
				}
				//除法
				temp=0,r=0;
				for(j=0; j<MAXN; j++)
				{
					if(catalan[i][j]!=0) flag=true;
					if(flag==true){
					temp=catalan[i][j]+r*10;
					catalan[i][j]=temp/(i+1);
					r=temp%(i+1);}
				}
			}
		}
		flag=false;
		for(j=0; j<MAXN; j++)
		{
			if(catalan[N][j]!=0) flag=true;
			if(flag==true)
				cout << catalan[N][j];
		}
		cout << endl;
	}
	return 0;
}