Corn's new language——个人c++解

迈不过友情╰ 2022-06-15 08:28 154阅读 0赞

## Description ##

Corn is going to promote programming in the campus, so he wants to add a lot of interesting ideas to make programming more attractive. One task he is working on is to develop a new programming language because he thinks all existing ones are too simple to him. The syntax rules of the language are:

1. () () is a valid program

2. (*P*) (P) is a valid program, if *P* P is a valid program

3. *P**Q* PQ is a valid program, if both *P* P and *Q* Q are valid programs

Corn wants a compiler for the language. For a program, if it is valid, the compiler should print the max depth in the program. For example "(()) (())" has a depth of 2 2, while "((())()) ((())())" has a depth of 3 3. Formally, the max depth is the max number of unmatched open brackets in any prefix.

## Input ##

Each case is a non-empty string in a single line, the string will only contain '( (' or ') )', its length will be no more than 100 100.

## Output ##

For each case, output "YES" and the integer required above if the program is valid, separated with a space. Or "NO" if the program is invalid.

## Sample Input ##

(
    )
    ()
    (())
    ((())())
    ())
    ((

## Sample Output ##

NO
    NO
    YES 1
    YES 2
    YES 3
    NO
    NO

#include<bits/stdc++.h>
    using namespace std;
    int main()
    {
        int i,j,n,l,k,m;
        char a[102];
        stack<char>s;
        while(scanf("%s",a)!=EOF)
        {
            m=0;
            k=0;
            while(!s.empty())
            {
                s.pop();
            }
            l=strlen(a);
                for(j=0;j<l;j++)
                {
                    if(a[j]=='(')
                    {
                        k=k+1;
                    }
                    if(a[j]==')')
                    {
                        if(k>m)
                        {
                            m=k;
                        }
                        k=k-1;
                    }
                    if(s.empty())
                    {
                        s.push(a[j]);
                    }
                else
                    if(s.top()+1==a[j]||s.top()+2==a[j])
                    {
                        s.pop();
                    }
                        else
                        {
                            s.push(a[j]);
                        }
                }
                if(s.empty())
                {
                    printf("YES %d\n",m);
                }
                else
                {
                    printf("NO\n");
                }
        }
        return 0;
    }