TZOJ 4621 Grammar(STL模拟)

描述

Our strings only contain letters(maybe the string contains nothing).

Now we define the production as follows:

1. (C) --> C

2. C --> C

3. (C:num)-->repeat C num times.

Illustration: (C) or C stands for a string only contains letters. (C:num) means that we should repeat string C num times (1<=num<=9).

For example: (((ab)(cd:2)):2) --> abcdcdabcdcd.

If the length of result exceed 2^20 characters print "Too Long!" in one line.

输入

There are multiple test cases. The first is a positive integer T, indicating the number of test cases.

For each test case, there is only one line contains a justifiable string.

输出

Print the result after converting. If the length of result exceed 2^20 characters print "Too Long!" in one line.

样例输入

7
(abc)
(abc:5)
((ab)(cd:2))
(((ab)(cd:2)):2)
()
(aa(bb:3)(cc:2))
(((((((((((((((((((((((uNVZgs:2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2):2)

样例输出

abc
abcabcabcabcabc
abcdcd
abcdcdabcdcd

aabbbbbbcccc
Too Long!

题意

1. (C) --> C

2. C --> C

3. (C:num)-->repeat C num times.

题解

类似于括号匹配，如果出现右括号就把中间的字符串接起来重新插入

代码

 #include<bits/stdc++.h>
 using namespace std;

 const int maxlen=;

 int main()
 {
     int T;
     cin>>T;
     while(T--)
     {
         stack<string>st;
         int flag=;
         string s1="",s;
         cin>>s;
         for(int i=;i<(int)s.size();i++)
         {
             if(s[i]=='(')
             {
                 if(!s1.empty())st.push(s1);
                 s1="(";
                 st.push(s1);
                 s1.clear();
             }
             else if(s[i]==')')
             {
                 string s2="";
                 if(!s1.empty())st.push(s1);
                 s1.clear();
                 while(st.top()!="(")
                 {
                     s2.insert(,st.top());
                     st.pop();
                 }
                 st.pop();
                 st.push(s2);
             }
             else if(s[i]==':')
             {
                 if(!s1.empty())st.push(s1);
                 int sum=s[++i]-'';
                 if((int)st.top().size()*sum>maxlen)
                 {
                     flag=;
                     break;
                 }
                 s1=st.top();st.pop();
                 string s3="";
                 while(sum--)s3+=s1;
                 st.push(s3);
                 s1.clear();
             }
             else
                 s1+=s[i];
         }
         if(!flag||(int)st.top().size()>maxlen)cout<<"Too Long!\n";
         else cout<<st.top()<<'\n';
     }
     return ;
 }