Hdu 2067 bunny Board (catlan number)

 

Rabbit Board

 

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

Total Submission (s): 2323 Accepted Submission (s): 1360

Problem Description

The uncle of the rabbit came back from outside and brought her a gift. The rabbit happily ran back to his room and opened the board. The rabbit was disappointed. But a few days later I found the fun of the Board. The number of shortest paths from the start point (0, 0) to the end point (n, n) is C (2n, n ), now, the rabbit wants to determine the number of such paths without traversing the diagonal line (but can touch the grid points on the diagonal line? I have never thought about it for a long time. Now I want to ask you to help me solve this problem. It should be difficult for you!

 

Input

Input n (1 <= n <= 35) at a time, and end the input when n is equal to-1.

 

Output

For the number of output paths for each input data, see Sample for the specific format.

 

Sample Input

1

3

12

-1

 

Sample Output

1 1 2

2 3 10

3 12 416024

The most basic formula for finding other methods on the Internet is the catlan number (not a large number). I will summarize the catlan number first.

 

Code:

Java code

# Include <iostream>

# Include <stdio. h>

# Include <memory. h>

# Include <cmath>

Using namespace std;

 

Long h [45];

Long f [45] [45];

 

Void catalan2 () // catlan number: method 2

{

Int I, j;

For (I = 1; I <= 36; I ++)

{

F [0] [I] = 1;

}

For (I = 1; I <36; I ++)

{

For (j = I; j <= 36; j ++)

{

If (I = j)

{

F [I] [j] = f [I-1] [j];

}

Else

{

F [I] [j] = f [I-1] [j] + f [I] [J-1];

}

}

}

}

 

Void catalan () // catlan number: method 1

{

Int I, j;

H [0] = 1;

For (I = 1; I <36; I ++)

{

H [I] = 0;

For (j = 0; j <= I; j ++)

{

H [I] + = h [j] * h [i-j-1];

}

}

}

 

Int main ()

{

Int n, zz = 1;

Catalan ();

Catalan2 ();

While (scanf ("% d", & n )! = EOF)

{

If (n =-1) break;

// H [n] and f [n] [n] are all catlan numbers

// Printf ("% d % I64d \ n", zz ++, n, h [n] * 2 );

Printf ("% d % I64d \ n", zz ++, n, f [n] [n] * 2 );

}

 

Return 0;

}