PAT 1050 String Subtraction (20)

1050 String Subtraction (20)（20 分）

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 10^4. 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.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

思路：

直接用map来标记s2中出现的字符。

代码：

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>  
#include <set>
using namespace std;

map<char,int> m;

int main()
{
	string s1, s2;
	getline(cin, s1);
	getline(cin, s2);
	for (int i = 0; i < s2.length(); i++)
		m[s2[i]] = 1;
	for (int i = 0; i < s1.length(); i++)
	{
		if (m[s1[i]] != 1)
			cout << s1[i];
	}
	return 0;
}