PAT甲级 1050. String Subtraction (20)

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本文介绍了一个简单的字符串处理问题——计算两个字符串的差集，并提供了一段C++代码实现。输入为两个字符串S1和S2，输出为S1中去除所有S2中存在的字符后的结果。该任务挑战在于如何高效地移除所有匹配的字符。

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.

Sample Input:

They are students.
aeiou

Sample Output:

Thy r stdnts.

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

char s[10001];
char s1[10001];
int flag[200];
int main()
{
	gets(s);
	gets(s1);
	for(int i=0;s1[i];i++)
	{
		flag[s1[i]]=1;
	}
	for(int i=0;s[i];i++)
	{
		if(!flag[s[i]])
			cout<<s[i];
	}
	return 0;
}