PAT.A1050. String Subtraction

最新推荐文章于 2024-05-17 06:07:48 发布

克拉拉123

最新推荐文章于 2024-05-17 06:07:48 发布

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本文链接：https://blog.youkuaiyun.com/mokena1/article/details/79682850

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。输入包含两个字符串S1和S2，输出为S1中去除所有S2中存在的字符后的剩余部分。文章通过使用哈希表来提高查找效率。

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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.

#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
using namespace std;
int main(){
	char s1[10010];
	char s2[1010];
	gets_s(s1);
	gets_s(s2);
	int l1 = strlen(s1);
	int l2 = strlen(s2);
	int hash[128] = { 0 };
	for (int i = 0; i < l2; i++)
		hash[s2[i]]++;
	for (int i = 0; i < l1; i++)
		if (hash[s1[i]] == 0)
			printf("%c", s1[i]);
	return 0;
}