题目来源
String Subtraction - PTA
String Subtraction - 牛客
Description
Given two strings S 1 S_1 S1and S 2 , S = S 1 − S 2 S_2,S=S_1−S_2 S2,S=S1−S2is defined to be the remaining string after taking all the characters in S 2 S_2 S2from S 1 S_1 S1. Your task is simply to calculate S 1 − S 2 S_1−S_2 S1−S2for 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 S_1 S1and S 2 S_2 S2, respectively. The string lengths of both strings are no more than 10 4 10^4 104. 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 S 1 − S 2 S_1−S_2 S1−S2in one line.
Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.
题目大意
给定两个字符串 S 1 , S 2 S_1,S_2 S1,S2 ,定义 S 1 − S 2 S_1-S_2 S1−S2 是字符串 S 1 S_1 S1 中去掉所有出现在 S 2 S_2 S2 的字符得到的字符串,要求输出这个字符串
思路简介
因为字符用 ASCII 码表示,共 128 个,可以用一个桶记录某个字符是否在 S 2 S_2 S2 出现,如果出现就不输出这个字符
遇到的问题
代码
cpp
/**
* String Subtraction
* https://pintia.cn/problem-sets/994805342720868352/exam/problems/type/7?problemSetProblemId=994805429018673152
* https://www.nowcoder.com/pat/5/problem/4089
* 桶排
*/
#include<bits/stdc++.h>
using namespace std;
int s[300];
void solve(){
string s1,s2;
getline(cin,s1);
getline(cin,s2);
int len1=s1.size(),len2=s2.size();
for(int i=0;i<len2;++i){
s[s2[i]-'\0']=1;
}
for(int i=0;i<len1;++i){
if(!s[s1[i]-'\0'])cout<<s1[i];
}
cout<<'\n';
}
int main(){
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
//fstream in("in.txt",ios::in);cin.rdbuf(in.rdbuf());
int T=1;
//cin>>T;
while(T--){
solve();
}
return 0;
}