Parentheses matching problem (Nanyang 2) and parentheses matching Nanyang 2

Parentheses matching problem (Nanyang 2) and parentheses matching Nanyang 2
// Check the matching of parentheses. It is not too complicated to construct the stack using the stack idea. In this question, it is actually an array.
/* For example, consider the following sequence of parentheses:
[([] [])] Ask If brackets match
*/
# Include <cstdio>
# Include <cstring>
Int main ()
{
Int top, I;
Char a [1, 1010], B [2, 1010];
While (scanf ("% s", )! = EOF)
{
Top = 1; // top points to the next position under the B Array
B [top ++] = a [0];
For (I = 1; I <strlen (a); I ++)
{
If (a [I] = '(' | a [I] = '[') // in this case
B [top ++] = a [I];
Else if (a [I] = ')' & B [top-1] = '(') // overwrites the array B.
Top --;
Else if (a [I] = ']' & B [top-1] = '[')
Top --;
Else
{// Stack the remaining two cases
B [top ++] = a [I];
}
}
If (top = 1)
Printf ("matching brackets! \ N ");
Else
Printf ("Parentheses do not match !!! \ N ");
}
Return 0;
}
 
[ACM problem] Nanyang Institute of Technology Online evaluation question 15 matching brackets (2)

I roughly tested it.
This) [) (] (in this example, your answer is 6, but there are at least four, which means you have not considered completely

Another problem is that you need to save the result to an array and output the result at one time.

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;
}