Question about the number of matching brackets in Huawei OJ training, Huawei oj

Question about the number of matching brackets in Huawei OJ training, Huawei oj

The questions are as follows:

Parentheses matching

Input n parentheses and output n parentheses that can be combined. For example --

Only () is input;

When the input is 2, there are two types: () and;

When 3 is input, there are (), () and ((())), five types.

There are 14 input values .. And so on.

I think of stupid methods, and so on, because I believe there must be a formula. After a while. I did not summarize it, so I read the book because I have read similar questions in my impressions.

Then the formula is found. Haha.

import java.util.Scanner;  public class Main {        public static void main(String[] args) {            Scanner scanner = new Scanner(System.in);          int n = scanner.nextInt();          scanner.close();          if (n == 0) {        System.out.println(0);}else{  System.out.println((int) (jiecheng(2 * n) / (jiecheng(n) * jiecheng(n) * (n + 1))));  }          }        private static long jiecheng(int n) {          long result = 1;          for (int i = 1; i <= n; i++) {              result *= i;          }          return result;      }  }  

Brackets matching

# Include <stdio. h>
# Deprecision MAX 100
Int match (char * str)
{
Char stack [MAX], * p = stack;
While (* str)
{
Switch (* str)
{
Case '(':
{
* P ++ = * str;
Break;
}
Case ')':
{
If (* -- p! = '(')
Return 0;
Break;
}
Case '[':
{
* P ++ = * str;
Break;
}
Case ']':
{
If (* -- p! = '[')
Return 0;
Break;
}
Case '{':
{
* P ++ = * str;
Break;
}
Case '}':
{
If (* -- p! = '{')
Return 0;
Break;
}
}
Str ++;
}
If (stack = p)
Return 1;
Else
Return 0;
}

Int main ()
{
Char str [MAX];
Gets (str );
If (match (str ))
{
Printf ("match \ n ");
}
Else
{
Printf ("not match \ n ");
}
Return 0;
}

Brackets matching

# Include <stdio. h>
# Include <string. h>
# Define Max size 1000
Typedef char datatype;

Typedef struct
{
Datatype s [MAXSIZE];
Int top;
} Stack;

Void init (stack * s)
{
S-> top = 0;
}

Void push (stack * s, datatype c)
{
S-> s [s-> top ++] = c;
}

Int empty (stack * s)
{
Return s-> top = 0;
}

Datatype pop (stack * s)
{
If (! Empty (s ))
Return s-> s [-- s-> top];
Return 0;
}

Int main ()
{
Stack s;
Char input [1000];
Int len, OK, I;
Char temp;
While (gets (input ))
{
OK = 1;
Init (& s );
Len = strlen (input );
For (I = 0; OK & I <len; I ++)
{
Switch (input [I])
{
Case '(':
Case '[':
Case '{':
Push (& s, input [I]); break;
Case ')':
Temp = pop (& s );
If (temp! = '(')
OK = 0;
Break;
Case ']':
Temp = pop (& s );
If (temp! = '[')
... The remaining full text>