Pat1084:Broken Keyboard

Posted 2020-10-09 0kk470

1084. Broken Keyboard (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.

Now given a string that you are supposed to type, and the string that you actually type out, please list those keys which are for sure worn out.

Input Specification:

Each input file contains one test case. For each case, the 1st line contains the original string, and the 2nd line contains the typed-out string. Each string contains no more than 80 characters which are either English letters [A-Z] (case insensitive), digital numbers [0-9], or "_" (representing the space). It is guaranteed that both strings are non-empty.

Output Specification:

For each test case, print in one line the keys that are worn out, in the order of being detected. The English letters must be capitalized. Each worn out key must be printed once only. It is guaranteed that there is at least one worn out key.

Sample Input:

7_This_is_a_test
_hs_s_a_es

Sample Output:

7TI

思路
1.map模拟一个字典dic存放键位，队列q按照键位的第一次输入依次存放键，然后根据实际的输入效果确定损坏的键位。
2.当队列不为空时，依次取出队列的首元素，检查是否在字典中已遍历，没遍历表示缺失，直接打印。

注：map会自动根据键值排序，所以用队列先记录下键位的输入顺序。。。另外这道题20分ac了19分，一个特殊用例死活过不去，暂且先放着。

代码

#include<iostream>
#include<map>
#include<queue>
using namespace std;
int main()
{
    string s;
    string r;
    queue<char> q;
    map<char,int> dic;
    while(cin >> s >> r)
    {
       for(int i = 0;i < s.size();i++)
       {
           if(s[i] == ‘_‘ || dic.count(toupper(s[i])) > 0)
             continue;
           else
           {
             dic.insert(pair<char,int>(toupper(s[i]),1));
             q.push(toupper(s[i]));
           }
       }

       for(int i = 0;i < r.size();i++)
       {
           if(r[i] == ‘_‘)
            continue;
           else
           {
             dic[toupper(r[i])] = -1;
           }
       }

       while(!q.empty())
       {
           if(dic[q.front()] > 0)
             cout << q.front();
           q.pop();
       }
       cout << endl;
    }
}