原題描述：

E - Parentheses Balance

Parentheses Balance

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

題目就是想讓我們判斷一下兩種括號是否平衡。註意空字符也是平衡的。

這裏要判斷換行輸出Yes，就要用getline(cin,a).還要註意的是，前面幾組數據那裏也有個換行。要用getchar()消除掉；然後就是棧的應用了。

AC代碼：

 1 #include <iostream>
 2 #include <stack>
 3 #include <string.h>
 4 #include <stdio.h>
 5 using namespace std;
 6 int main()
 7 {
 8     int t;
 9     cin>>t;
10             getchar();
11     while (t--)
12     {
13         string a;
14         stack<char> c;
 
15         getline(cin,a);
16         if(a[0]==‘\n‘) {cout<<"Yes"<<endl;continue;}
17         c.push(‘1‘);
18         for(int i=0;i<a.size();i++)
19         {
20            if (a[i]==‘(‘||a[i]==‘[‘) c.push(a[i]);
21             else  if (a[i]==‘)‘)
22             {
23                 if(c.top()==‘(‘) c.pop();
24                 else c.push(‘0‘);
25             }
26             else
27             {
28                 if (c.top()==‘[‘) c.pop();
29                 else c.push(‘0‘);
30             }
31         }
32        if(c.top()==‘1‘)cout<<"Yes"<<endl;
33        else cout<<"No"<<endl;
34 
35     }
36     return 0;
37 }

Week 1 # E Parentheses Balance