1050 String Subtraction (20 point(s))

1050 String Subtraction (20 point(s))

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 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 S​1​​−S​2​​ in one line.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

考点：字符串处理、哈希表。

ASCⅡ共有128个（0~127），使用哈希表标记其是否属于s2的字符。如果标记，在s1中不输出，否则输出。常规方法会出现TLE。 

#include<iostream>
#include<string>
using namespace std;
const int LEN = 128;
bool mark[LEN]={false};
int main(void){
	string s1,s2;
	getline(cin,s1);
	getline(cin,s2);
	for(int i=0;i<s2.length();i++){
		mark[s2[i]]=true;
	}
	for(int i=0;i<s1.length();i++){
		if(!mark[s1[i]]) cout<<s1[i];
	}
	cout<<endl;
	return 0;
}