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 = 0; i <length; i++)
        M[S2[i]] = 1;
    length = S1.length();
    for (int i = 0; i < length; i++)
        if (!M[S1[i]])
            V.push_back(S1[i]);
    for (auto it : V)
        cout << it;
}