PAT 1050 String Subtraction

最新推荐文章于 2021-01-27 15:06:46 发布

汪少升

最新推荐文章于 2021-01-27 15:06:46 发布

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分类专栏： pat advance practice 文章标签： pat data_stucture

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本文介绍了一种计算两个字符串差集的高效算法，并提供了完整的C++实现代码。输入为两个字符串S1和S2，输出为从S1中去除所有S2中存在的字符后的剩余字符串。文章通过示例详细解释了算法的运作过程。

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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

This is easy.

#include<iostream>
using namespace std;

int main()
{
	char ch[10001];
	bool mark[128]={0};
	int len=0;
	while(ch[len++]=cin.get(),ch[len-1]!='\n');
	char chtmp;
	while(chtmp=cin.get(),chtmp!='\n')
	{
		mark[chtmp]=1;
	}
	for(int i=0;i<len;++i)
	{
		if(mark[ch[i]]) continue;
		else cout<<ch[i];
	}
}