卡特兰数模板 hdu1023

最新推荐文章于 2021-02-23 15:33:13 发布

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本文介绍了一个火车调度问题的解决方案，利用卡特兰数计算所有可能的火车出站顺序数量。通过实现一个高效的算法来处理大规模的数据输入，并提供具体的样例输入输出以验证正确性。

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.

卡特兰数！！！

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <queue>
using namespace std;
const long long INF=0x7fffffffffffffffll;

int a[105][1000];

void katelan()
{
    memset(a,0,sizeof(a));
    a[1][1]=1;
    a[1][0]=1;
    a[2][1]=2;
    a[2][0]=1;
    int len=1,yu;
    for(int i=3;i<=100;i++)
    {
        yu=0;
        for(int j=1;j<=len;j++)
        {
         int t=a[i-1][j]*(4*i-2)+yu;
         a[i][j]=t%10;
         yu=t/10;
        }
        while(yu)
        {
            a[i][++len]=yu%10;
            yu/=10;
        }
        for(int j=len;j>0;j--)
        {
            int p=a[i][j]+yu*10;
            a[i][j]=p/(i+1);
            yu=p%(i+1);
        }
        while(!a[i][len])
        {
            len--;
        }
        a[i][0]=len;
    }
}

int main()
{
    katelan();
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=a[n][0];i>=1;i--)
         printf("%d",a[n][i]);
         puts("");
    }
    return 0;
}