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

1050 String Subtraction (20 分)

 

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.

Input Specification:

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.

Output Specification:

For each test case, print S​1​​−S​2​​ 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[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;
}

 

转载于:https://www.cnblogs.com/caiyishuai/p/11483751.html