Given two strings S?1?? and S?2??, S=S?1??−S?2?? is defined to be the remaining string after taking all the characters in S?2?? from S?1??. Your task is simply to calculate S?1??−S?2?? for any given strings. However, it might not be that simple to do it fast.
Each input file contains one test case. Each case consists of two lines which gives S?1?? and S?2??, respectively. The string lengths of both strings are no more than 1. 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.
For each test case, print S?1??−S?2?? in one line.
They are students.
aeiou
Thy r stdnts.
题意：
很简单的一道题目，除去第一个字符串中含有的第二个字符串的字符即可
AC代码：
#include<iostream> #include<algorithm> #include<vector> #include<queue> #include<map> #include<string> #include<cstring> using namespace std; string s1; string s2; int a[200];//标准ascii码字符集共有128个编码 int main(){ getline(cin,s1); getline(cin,s2); memset(a,0,sizeof(a)); int l1,l2; l1=s1.size(); l2=s2.size(); for(int i=0;i<l2;i++){ a[s2[i]]=1; } for(int i=0;i<l1;i++){ if(a[s1[i]]==1) continue; cout<<s1[i]; } return 0; }
使用set
#include<bits/stdc++.h> using namespace std; set<char>st; int main(){ string s1,s2; getline(cin,s1); getline(cin,s2); for(int i=0;i<s2.size();i++) st.insert(s2[i]); for(int i=0;i<s1.size();i++){ if(st.find(s1[i])==st.end()) cout<<s1[i]; } return 0; }
PAT 甲级 1050 String Subtraction (20 分) （简单送分,getline(cin,s)的使用）
原文：https://www.cnblogs.com/caiyishuai/p/11483751.html