PAT 甲级 1050 String Subtraction (20 分) （简单送分,getline(cin,s)的使用）

提示：扫一扫查出行
【扫一扫了解最新限行尾号】
复制提示

1050 String Subtraction (20 分)

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 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.

Output Specification:

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

Sample Input:

They are students. aeiou

Sample Output:

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[];//标准ascii码字符集共有128个编码
int main(){
getline(cin,s1);
getline(cin,s2);
memset(a,,sizeof(a));
int l1,l2;
l1=s1.size();
l2=s2.size();
for(int i=;i<l2;i++){
a[s2[i]]=;
}
for(int i=;i<l1;i++){
if(a[s1[i]]==) continue;
cout<<s1[i];
}
return ;
}

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=;i<s2.size();i++)
st.insert(s2[i]);
for(int i=;i<s1.size();i++){
if(st.find(s1[i])==st.end())
cout<<s1[i];
}
return ;
}