1050 String Subtraction (20)

最新推荐文章于 2024-05-29 11:50:32 发布

running987

最新推荐文章于 2024-05-29 11:50:32 发布

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本文介绍了一种计算两字符串差集的高效算法，并提供了一个C++实现示例。该算法通过遍历第二个字符串来标记字符，然后从第一个字符串中移除这些被标记的字符。

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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.

思路：

给出两个字符串，在第一个字符串中删除第二个字符串中出现过的所有字符并输出。

C++：

#include "cstring"
#include <cstdio>
#include "iostream"
using namespace std;
char s1[10001];
char s2[10001];
int main(){
	int len1,len2;
	cin.getline(s1,10001);
	cin.getline(s2,10001);
	len1=strlen(s1);
	len2=strlen(s2);
	bool flag[256]={false};
	for (int i=0;i<len2;i++)
	{
		flag[s2[i]]=true;
	}
	for (int i=0;i<len1;i++)
	{
		if (!flag[s1[i]])
		{
			printf("%c",s1[i]);
		}
	}
	return 0;
}