1050 String Subtraction (20分)

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 1. 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.

题目分析：利用map将S2种每个字符存入 再遍历S1进行判断

 #define _CRT_SECURE_NO_WARNINGS
 #include <climits>
 #include<iostream>
 #include<vector>
 #include<queue>
 #include<map>
 #include<set>
 #include<stack>
 #include<algorithm>
 #include<string>
 #include<cmath>
 using namespace std;
 map<char, int> M;
 vector<char> V;
 int main()
 {
     string S1,S2;
     getline(cin, S1);
     getline(cin, S2);
     int length = S2.length();
     for (int i = ; i <length; i++)
         M[S2[i]] = ;
     length = S1.length();
     for (int i = ; i < length; i++)
         if (!M[S1[i]])
             V.push_back(S1[i]);
     for (auto it : V)
         cout << it;
 }