UVA 673.Parentheses Balance（栈）

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

题意：括号配对，只有两种括号（）和【】，给你一串字符串，判断括号是否配对

题本身并不难，一直卡到我的是这句话“if it is the empty string”，说明在输入里可能会有空行，这就牵扯到用哪个输入问题了，刚开始一直wa，是因为用了scanf，经过多次尝试发现，cin和scanf都会wa，getline和gets就ok了，是因为scanf和cin遇到空格都不处理，这次算是长记性了

#include<cstdio>
#include<iostream>
#include<algorithm>
#include <cstring>
#include<string>
#include<stack>
using namespace std;
stack<char>s;

int main() {
    int n;
    scanf("%d", &n);
    getchar();
    while(n--){
        char str[];
        gets(str);      //注意输入问题
        int len = strlen(str);
        if(len%!=){
            printf("No\n");
            continue;
        }
        int flag = ;
        for(int i = ; i < len; i++){
            if(str[i] == '(' || str[i] == '[')
                s.push(str[i]);
            else if(!s.empty() && s.top()=='(' && str[i]==')')
                    s.pop();
            else if(!s.empty() && s.top()=='[' && str[i]==']')
                    s.pop();
            else{
                flag = ;
                break;
            }

        } 
        if(flag&&s.empty()) printf("Yes\n");
        else    printf("No\n");

        while(!s.empty())   s.pop();    //一定要把栈给清空
    }
    return ;
}