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.
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.
For each test case, print S?1??−S?2?? in one line.
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Thy r stdnts.
#include<bits/stdc++.h> using namespace std; typedef long long ll; #define MAXN 10005 map<char,int> mp; vector<char> vec; int main(){ string s1,s2; getline(cin,s1); getline(cin,s2); for(int i=0;i < s2.size();i++){ mp[s2[i]]++; } for(int i=0;i < s1.size();i++){ if(!mp[s1[i]]) cout << s1[i]; } return 0; }
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原文:https://www.cnblogs.com/cunyusup/p/10703361.html