关于#c语言#的问题：语言括号匹配

Question

一个括号匹配问题，不知道哪里错了，请指出
You are given a string consisting of parentheses () and []. A string of this type is said to be correct:
(a) if it is the empty string
(b) if A and B are correct, AB is correct,
(c) if A is correct, (A) and [A] is correct.
Write a program that takes a sequence of strings of this type and check their correctness. Your
program can assume that the maximum string length is 128.
Input
The file contains a positive integer n and a sequence of n strings of parentheses ‘()’ and ‘[]’, one string
a line.
Output
A sequence of ‘Yes’ or ‘No’ on the output file.
Sample Input
3
([])
(([()])))
([()])()
Sample Output
Yes
No
Yes

#include<bits/stdc++.h>
using namespace std;
char bi(char a,char b){
    if((a=='('&&b==')')||(a=='['&&b==']')||(a=='#'&&b=='#')) return '>';
    else if((a=='('&&b==']')||(a=='['&&b==')')) return '=';
    else if(a==')'||a==']') return '=';
    else if(a=='#'&&(b==')'||b==']')) return '=';
    else return '<';
}
int main(){
    stack<char>s;
    int n,i,m,j;
    scanf("%d",&n);
    getchar();
    for(i=0;i<n;i++){
        char a[10005]={0};
        gets(a);
        if(strcmp(a,"\n")==0){
            printf("YES\n");
        }else{
        char b='#';
        s.push(b);
        m=strlen(a);
        a[m]='#';
        for(j=0;j<=m;j++){
            if(bi(s.top(),a[j])=='<'){
                s.push(a[j]);
            }else if(bi(s.top(),a[j])=='>'){
                s.pop();
            }else if(bi(s.top(),a[j])=='='){
                printf("NO\n");
                break;
            }
        }
        if(s.empty()) printf("YES\n");
        else {
            for(;!s.empty();){
                s.pop();
            }
        }
}}
}

Answer 1

括号匹配哪里需要那么麻烦
很简单，定义一整数，开始为0
遇到( +1
遇到) -1，如果出现负数，或者遍历结尾不是0，就是不匹配，否则就是匹配。