PAT A1050 String Subtraction

题目描述：

　　Given two strings S?1?? and S?2??, S=S?1??−S?2?? is defined to be the remaining string after taking all the characters in S?2?? from S?1??. Your task is simply to calculate S?1??−S?2?? for any given strings. However, it might not be that simple to do it fast.

　　Input Specification:
　　Each input file contains one test case. Each case consists of two lines which gives S?1?? and S?2??, respectively. The string lengths of both strings are no more than 10?4??. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

　　Output Specification:
　　For each test case, print S?1??−S?2?? in one line.

　　Sample Input:
　　They are students.
　　aeiou

　　Sample Output:
　　Thy r stdnts.

参考代码：

 1 /****************************************************
 2 PAT A1050 String Subtraction
 3 ****************************************************/
 4 #include <iostream>
 5 #include <string>
 6 #include <vector>
 7 
 8 using namespace std;
 9 
10 int main() {
11     string str1, str2;
12 
13     vector<bool> Hash(500, false);
14 
15     getline(cin, str1);
16     getline(cin, str2);
17 
18     for (int i = 0; i < str2.size(); ++i) {
19         Hash[str2[i]] = true;
20     }
21 
22     for (int i = 0; i < str1.size(); ++i) {
23         if (Hash[str1[i]] == false) {
24             cout << str1[i];
25         }
26     }
27 
28     return 0;
29 }

注意事项：

　　无。

PAT A1050 String Subtraction

原文：https://www.cnblogs.com/mrdragon/p/11662134.html