7-2 Symbol pairing

Please write a program to check C Whether the following symbols are paired in the language source program ：/* And */、( And )、[ And ]、{ And }.

Input format :

Input as a C Language source . When you read a line with only one period . And a carriage return , It marks the end of the input . In the program, you need to check that the paired symbols do not exceed 100 individual .

Output format :

First , If all the symbols match correctly , Then output... In the first line YES, Otherwise output NO. Then point out in the second line The first one is not paired The symbol of ： If the left symbol is missing , The output ?- Right sign ; If the right symbol is missing , The output Left sign -?.

sample input 1：

void test()
{
    int i, A[10];
    for (i=0; i<10; i++) { /*/
        A[i] = i;
}
.

sample output 1：

NO
/*-?

Answer key

There is an important point in the output format of this question , If there is a mismatch , Pointed out that The first one is not paired The symbol of , therefore , We need to use the stack for this problem

First , We can use a string to filter out the questions from several input strings Detected characters , Considering that only

/* and */ It's two characters , We can , When this pair is detected , use < and > Storage , Convenient judgment , Then use stack , The left side of the filtered string push Into the stack , And decide , Um. , There are many judgments on this question ...

#include<bits/stdc++.h>

using namespace std;

int main()
{
    stack<char> C;
    string ss;
    string s = ""; //  Store the characters to be detected in the input 
    while(true)
    {
        getline(cin, ss);
        if(ss == ".")
            break;
        for(int i = 0; i < ss.length(); i++)
        {
            if(ss[i] == '['|| ss[i] == '{'||ss[i] == '('||ss[i] == ']'||ss[i] == '}'||ss[i] == ')')
                s += ss[i];
            else if(ss[i] == '/' && ss[i+1] == '*')
            {
                s += '<';
                i++;
            }
            else if(ss[i] == '*' && ss[i+1] == '/')
            {
                s += '>';
                i++;
            }
        }
    }
    for(int i = 0; i < s.length() ; i++)
    {
        if(s[i] == '<'||s[i] == '{'||s[i] == '('||s[i] == '[')
            C.push(s[i]);
        else if(C.empty())
        {
            if(s[i] == '}'||s[i] == ')'||s[i] == ']')
            {
                cout << "NO" << endl;
                cout << "?-" << s[i] << endl;
                return 0;
            }
            else if(s[i] == '>')
            {
                cout << "NO" << endl;
                cout << "?-*/" << endl;
                return 0;
            }
        }
        else if(C.top() == '<')
        {
            if(s[i] == '>')
                C.pop();
            else
            {
                cout << "NO" << endl;
                cout << "/*-?" << endl;
                return 0;
            }
        }
        else if(C.top() == '{')
        {
            if(s[i] == '}')
                C.pop();
            else
            {
                cout << "NO" << endl;
                cout << C.top() << "-?" << endl;
                return 0;
            }
        }
        else if(C.top() == '(')
        {
            if(s[i] == ')')
                C.pop();
            else
            {
                cout << "NO" << endl;
                cout << C.top() << "-?" << endl;
                return 0;
            }
        }
        else if(C.top() == '[')
        {
            if(s[i] == ']')
                C.pop();
            else
            {
                cout << "NO" << endl;
                cout << C.top() << "-?" << endl;
                return 0;
            }
        }
    }
    if(!C.empty())
    {
        if(C.top() == '<')
        {
            cout << "NO" << endl;
            cout << "/*-?" << endl;
        }
        else
        {
            cout << "NO" << endl;
            cout << C.top() << "-?" << endl;
        }
    }
    else
        cout << "YES" << endl;
    return 0;
}