A1050 String Subtraction (20 point(s))

比对字符串

1. 原文

原题

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.

2. AC代码

#include<iostream>
#include<map>
#include<string>
using namespace std;

map<char,int> match;
int main()
{
	string str1,str2;
	getline(cin,str1);	
	getline(cin,str2);
	for (int i = 0; str2[i] != '\0'; ++i)
	{
		match[str2[i]]++;
	}
	for (int i = 0; str1[i] != '\0'; ++i)
	{
		if (match[str1[i]]==0)
		{
			cout<<str1[i];
		}
	}
	return 0;
}

3. AC代码2

#include<iostream>
#include<string>
using namespace std;
int main()
{
	string str1,str2;
	getline(cin,str1);	
	getline(cin,str2);
	for (int i = 0; str1[i] != '\0'; ++i)
	{
		if (str2.find(str1[i])==string::npos)
		{
			cout<<str1[i];
		}
	}
	return 0;
}