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

朱雀 2022-05-16 11:22 179阅读 0赞

# 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;
    }