浙江大学PAT_甲级_1050. String Subtraction (20)

字符串差集计算

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本文介绍了一个简单的字符串处理问题，即如何快速计算两个字符串S1和S2的差集S1-S2，通过示例和C++代码展示了具体的实现方法。

题目链接：点击打开链接

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.

我的C++程序：

#include<iostream>
#include<string>
#include<vector>
using namespace std;
int main()
{
	int i = 0,j=0;
	string s1, s2;
	vector<bool>not_exist(1000,true);
	getline(cin, s1);//s1,s2可能含有空格所以用getline()
	getline(cin, s2);
	for (i = 0;i < s2.size();i++)
	{
		not_exist[s2[i]] = false;//记录s2存在的字符
	}
	for (i = 0;i < s1.size();i++)
	{
		if (not_exist[s1[i]] == true)//输出s1里不含s2的字符
			printf("%c", s1[i]);
	}
	//system("pause");
	return 0;
}