Given two strings S₁ and S₂, S = S₁ - S₂ is defined to be the remaining string after taking all the characters in S₂ from S₁. Your task is simply to calculate S₁ - S₂ 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 S₁ and S₂, respectively. The string lengths of both strings are no more than 10⁴. 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 S₁ - S₂ in one line.

They are students.
aeiou

Thy r stdnts.


思路
水题，map模拟一个字典记录要删除的字母，遍历打印s1时不打印字典上的字母就行。

另外:吐槽一下pat好多题一个文件对应一个测试用例，不像acm一个文件有多个用例需要重复读取==。

代码

#include<iostream>
#include<string>
#include<unordered_map>
using namespace std;
int main()
{
    string s1,s2;
    getline(cin,s1);
    getline(cin,s2);
    unordered_map<char,int> dic;
    for(int i = 0;i <s2.size();i++)
    {
        dic.insert(pair<char,int>(s2[i],1));
    }

    for(int i = 0;i < s1.size();i++)
    {
        if(dic.count(s1[i]) > 0)
            continue;
        else
            cout << s1[i];
    }
}