HDU 1023 Train Problem II ——————Catalan 数

本文探讨了火车问题II的解决方案,通过计算Catalan数来确定所有火车能够以严格递增顺序离开轨道的方式数量。采用C++编程实现,并提供了完整的代码示例。

Train Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11325    Accepted Submission(s): 6002


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
1
2
3
10
 

Sample Output
1
2
5
16796

Hint

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

 


Catalan 数打表

#include<bits/stdc++.h>
using namespace std;
const int MAXN=110;
int a[MAXN][MAXN];//a[i][0] save the length

void GetCatalans()
{
    int carry=0;
    int len=1;
    a[1][1]=1; a[1][0]=1;
    a[2][1]=2; a[2][0]=1;
    for(int i=3;i<MAXN;i++)
    {

        for(int j=1;j<=len;j++)
        {
            int sum = a[i-1][j]*(i*4-2) + carry;
              carry = sum / 10;
            a[i][j] = sum%10;
        }
        while(carry)
        {
            a[i][++len] = carry%10;
            carry /= 10;
        }


        for(int j=len;j>0;j--)
        {
            int sum = a[i][j] + carry*10;
            a[i][j] = sum / (i+1);
              carry = sum % (i+1);
        }
        while(a[i][len]==0) len--;
        a[i][0]=len;
    }
}
int main()
{
    int n;
    GetCatalans();
    while(~scanf("%d",&n))
    {
        for(int i=a[n][0];i>0;i--)
            printf("%d",a[n][i]);
        printf("\n");
    }
    return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值