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

 1 #define _CRT_SECURE_NO_WARNINGS
 2 #include <climits>
 3 #include<iostream>
 4 #include<vector>
 5 #include<queue>
 6 #include<map>
 7 #include<set>
 8 #include<stack>
 9 #include<algorithm>
10 #include<string>
11 #include<cmath>
12 using namespace std;
13 map<char, int> M;
14 vector<char> V;
15 int main()
16 {
17     string S1,S2;
18     getline(cin, S1);
19     getline(cin, S2);
20     int length = S2.length();
21     for (int i = 0; i <length; i++)
22         M[S2[i]] = 1;
23     length = S1.length();
24     for (int i = 0; i < length; i++)
25         if (!M[S1[i]])
26             V.push_back(S1[i]);
27     for (auto it : V)
28         cout << it;
29 }

View Code

1050 String Subtraction (20分)

原文：https://www.cnblogs.com/57one/p/12044533.html