6120： String Subtraction (20)

最新推荐文章于 2022-08-04 10:00:52 发布

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最新推荐文章于 2022-08-04 10:00:52 发布

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本文链接：https://blog.youkuaiyun.com/qq_36502291/article/details/82529638

本文介绍了一种快速计算两个字符串S1和S2的差集S的算法，即从S1中移除所有出现在S2中的字符。通过使用C++标准库中的map，该算法能有效地实现这一操作，即使对于长度达10^4的字符串也能在1秒内完成计算。

问题 D: String Subtraction (20)

时间限制: 1 Sec 内存限制: 32 MB
提交: 268 解决: 134
[提交][状态][讨论版][命题人:外部导入]

题目描述

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 - S2for any given strings. However, it might not be that simple to do it fast.

输入

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.

输出

For each test case, print S1 - S2 in one line.

样例输入

They are students.
aeiou

样例输出

Thy r stdnts.

#include<iostream>
#include<string>
#include<map>
using namespace std;
int main() {
	string s1, s2;
	while (getline(cin, s1)) {
		map<char, bool> a;
		getline(cin, s2);
		for (int i = 0; i < s2.length(); i++) {
			a[s2[i]] = true;
		}
		for (int i = 0; i < s1.length(); i++) {
			if (a[s1[i]] == false)
				cout << s1[i];
		}
		cout << endl;
	}
	return 0;
}