1050 String Subtraction

最新推荐文章于 2024-05-17 06:07:48 发布

三块不一样的石头

最新推荐文章于 2024-05-17 06:07:48 发布

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分类专栏： PAT (Advanced Level) Practice 文章标签：算法 PTA

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这篇博客介绍如何在给定两个字符串S1和S2的情况下，通过高效的算法计算S1中移除S2中所有字符后的剩余字符串。通过实例和C++代码展示了实现思路，适合字符串操作的编程挑战。

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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.

题目大意

给你俩个字符串，分别为s1,s2

在s2中出现的字符，要在s1中删除

输出删除指定字符的s1

思路

见代码

C/C++

#include<bits/stdc++.h>
using namespace std;
int main()
{
    map<char,bool> op;
    string s1,s2;
    getline(cin,s1);
    getline(cin,s2);
    for(char x:s2) op[x] = true;
    for(char x:s1) if(!op[x]) putchar(x);
    return 0;
}