1050 String Subtraction (20 分)（从字符串中去掉特定的字符）

原创于 2019-11-25 18:09:56 发布 · 151 阅读

0 ·

CC 4.0 BY-SA版权

PAT甲级刷题记录专栏收录该内容

38 篇文章

订阅专栏

本文介绍了一种计算两个字符串差集的高效算法，通过使用字符映射表来快速判断字符是否存在于第二个字符串中，从而实现从第一个字符串中移除所有出现在第二个字符串中的字符。此算法适用于字符串长度不超过10^4的情况，确保了处理大量字符数据时的效率。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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<bits/stdc++.h>
using namespace std;
typedef long long ll;
char book[500];
int main(){
    string s1,s2;
    getline(cin,s1);
    getline(cin,s2);
    for(int i=0;i<s2.size();i++) book[s2[i]]=true;
    for(int i=0;i<s1.size();i++) {
    	if(book[s1[i]]) continue;
    	cout<<s1[i];
    }
    return 0;
}