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.

7_This_is_a_test_hs_s_a_es

7TI
地址： http://pat.zju.edu.cn/contests/pat-a-practise/1084
算法：此题属于简单的模拟类型题。我的方法是，先把所有字母都转为大写字母，然后从源字符串和输入字符串开始，若当前两个字符相等，说明这个键没有坏，二者同时向后移动一位；若两个字符串不相等，说明这个键是坏的，若该键未记录则记录，
然后源串向后移动一位，输入串不移动。代码：

 1 #include<string> 2 #include<iostream> 3 #include<vector> 4 using namespace std; 5  6 int main() 7 { 8   string originalStr, typeStr; 9   while(cin >> originalStr >> typeStr){10     int originalN = originalStr.size();11     int typeN = typeStr.size();12     int i = 0, j = 0;13     for(i = 0; i < originalN; ++i){14       if(originalStr[i] >= ‘a‘ && originalStr[i] <= ‘z‘)15         originalStr[i] = originalStr[i] - ‘a‘ + ‘A‘;16     }17     for(i = 0; i < typeN; ++i){18       if(typeStr[i] >= ‘a‘ && typeStr[i] <= ‘z‘)19         typeStr[i] = typeStr[i] - ‘a‘ + ‘A‘;20     }21     vector<bool> isBroken(256,false);22     string result;23     for(i = 0; i < originalN; ++i){24       if(originalStr[i] == typeStr[j]){25         ++j;26       }else{27         if(!isBroken[originalStr[i]]){28           isBroken[originalStr[i]] = true;29           result.push_back(originalStr[i]);30         }31       }32     }33     cout << result << endl;34   }35   return 0;36 }