Hdu 2510, symbol triangle, good search questions should be read frequently

Symbol triangle
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 661 Accepted Submission (s): 317

 

Problem Description
The second line of the symbolic triangle contains n symbols consisting of "+" and "-". In the future, each line has one fewer symbol than the uplink, and the two with the same number are "+", the two differences are "-". Calculate the number of different signed triangles so that they contain the same number of "+" and. The following is a signed triangle with n = 7:
++-+-++
+--+
-++-
-++-
-+-
--
+

 

Input
One positive integer n <= 24, n = 0 in each row exits.

 

Output
N and the number of signed triangles.

 

Sample Input
15
16
19
20
0

Sample Output
15 1896
16 5160
19 32757
20 59984

Source
Ecjtu2008 Autumn Contest
 

Recommend
Lcy
Plane ticket:
Http://acm.hdu.edu.cn/showproblem.php? Pid = 1, 2510
 
 
At the beginning, I wrote a few groups of data to find the rule, but I was too lazy to search for tables.
Later, Baidu used the tabulation method.
However, when Baidu came back, the brute-force search was actually stuck, and many details were not written, so it won't be processed. Later, it was done by referring to other people's code.
You need to work hard. Come on, cool.
 
Consider-'as 1,' + 'as 0. This satisfies the question.

We know that 1 ^ 1 = 0; 1 ^ 0 = 1, 0 ^ 1 = 1, 0 ^ 0 = 0;

Then we can perform an exclusive or operation.

 

Brute-force code on

[Cpp] # include <stdio. h>
# Include <string. h>
Int ans [30];
Int a [30] [30]; // The intermediate process used to save the state of the first layer of each n
Int count; // number of records 1
Void DFS (int n)
{
Int I, j;
If (n> 24) return;
For (I = 0; I <= 1; I ++)
{
A [1] [n] = I; // only the first line of brute force is required.
Count + = I;
For (j = 2; j <= n; j ++)
{
A [j] [n-j + 1] = a [J-1] [n-j + 1] ^ a [J-1] [n-j + 2];
Count + = a [j] [n-j + 1];
}
If (count * 2 = n * (n + 1)/2)
Ans [n] ++;
DFS (n + 1 );
Count-= I;
For (j = 2; j <= n; j ++)
{
A [j] [n-j + 1] = a [J-1] [n-j + 1] ^ a [J-1] [n-j + 2];
Count-= a [j] [n-j + 1];
}
 
}
}
Int main ()
{
Int I, j, k, n;
Memset (ans, 0, sizeof (ans ));
Count = 0;
DFS (1 );
While (scanf ("% d", & n )! = EOF)
{
If (! N) break;
Printf ("% d \ n", ans [n]);
 
}
Return 0;
}

# Include <stdio. h>
# Include <string. h>
Int ans [30];
Int a [30] [30]; // The intermediate process used to save the state of the first layer of each n
Int count; // number of records 1
Void DFS (int n)
{
Int I, j;
If (n> 24) return;
For (I = 0; I <= 1; I ++)
{
A [1] [n] = I; // only the first line of brute force is required.
Count + = I;
For (j = 2; j <= n; j ++)
{
A [j] [n-j + 1] = a [J-1] [n-j + 1] ^ a [J-1] [n-j + 2];
Count + = a [j] [n-j + 1];
}
If (count * 2 = n * (n + 1)/2)
Ans [n] ++;
DFS (n + 1 );
Count-= I;
For (j = 2; j <= n; j ++)
{
A [j] [n-j + 1] = a [J-1] [n-j + 1] ^ a [J-1] [n-j + 2];
Count-= a [j] [n-j + 1];
}

}
}
Int main ()
{
Int I, j, k, n;
Memset (ans, 0, sizeof (ans ));
Count = 0;
DFS (1 );
While (scanf ("% d", & n )! = EOF)
{
If (! N) break;
Printf ("% d \ n", ans [n]);

}
Return 0;
}

 

 

Code submission:

[Cpp] # include <iostream>
Using namespace std;
Int result [24] = {171,410 };
Int main ()
{
Int n;
Cin> n;
While (n! = 0)
{
Cout <n <"" <result [n-1] <endl;
Cin> n;
}
Return 0;
}

# Include <iostream>
Using namespace std;
Int result [24] = {171,410 };
Int main ()
{
Int n;
Cin> n;
While (n! = 0)
{
Cout <n <"" <result [n-1] <endl;
Cin> n;
}
Return 0;
}