URAL 1654. Cipher Message （STL stack）

本文详细解析了Stierlitz如何通过删除空格和标点符号，替换连续相同字母，并插入额外的字母来加密信息。通过提供一个实际的示例，解释了如何还原密文到原始消息。

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1654. Cipher Message

Time limit: 1.0 second
Memory limit: 64 MB

Müller tried to catch Stierlitz red-handed many times, but alwaysfailed because Stierlitz could ever find some excuse. Once Stierlitz waslooking through his email messages. At that moment, Müller enteredsecretly and watched a meaningless sequence of symbols appear on the screen.“A cipher message,” Müller thought.“UTF-8,” Stierlitz thought.

It is known that Stierlitz ciphers messages by the following method.

He deletes all spaces and punctuation marks.
He replaces all successive identical letters by one such letter.
He inserts two identical letters at an arbitrary place many times.

Try to restore a message as it was after the second step. For that, remove fromthe message all pairs of identical letters inserted at the third step.

Input

The only input line contains a message ciphered by Stierlitz. The messageconsists of lowercase English letters and its length is at most 200000.

Output

Output the restored message.

Sample

input	output
wwstdaadierfflitzzz	stierlitz

Problem Author: Vladimir Yakovlev (idea by Alexander Klepinin)
Problem Source: NEERC 2008, Eastern subregion quarterfinals

解析：相邻两相同字母可消去，直接进栈一遍即可，相同则出栈，不同则进栈。

AC代码：

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

stack<char> p;

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
//        freopen("out.txt", "w", stdout);
    #endif // sxk

    string s;
    while(cin>>s){
        int len = s.size();
        for(int i=0; i<len; i++){
            if(!p.empty() && p.top() == s[i]) p.pop();
            else p.push(s[i]);
        }
        s = "";
        while(!p.empty()){
            s += p.top();
            p.pop();
        }
        reverse(s.begin(), s.end());
        cout<<s<<endl;
    }
    return 0;
}