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=1048576;

int main()
{
    int T;
    cin>>T;
    while(T--)
    {
        stack<string>st;
        int flag=1;
        string s1="",s;
        cin>>s;
        for(int i=0;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(0,st.top());
                    st.pop();
                }
                st.pop();
                st.push(s2);
            }
            else if(s[i]==':')
            {
                if(!s1.empty())st.push(s1);
                int sum=s[++i]-'0';
                if((int)st.top().size()*sum>maxlen)
                {
                    flag=0;
                    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!
";
        else cout<<st.top()<<'
';
    }
    return 0;
}