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.
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MAXN 10005 map<char,int> mp;
vector<char> vec; int main(){
string s1,s2;
getline(cin,s1);
getline(cin,s2); for(int i=;i < s2.size();i++){
mp[s2[i]]++;
} for(int i=;i < s1.size();i++){
if(!mp[s1[i]]) cout << s1[i];
}
return ;
}

——

 

PAT 1050 String Subtraction的更多相关文章

  1. pat 1050 String Subtraction(20 分)

    1050 String Subtraction(20 分) Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the ...

  2. PAT 解题报告 1050. String Subtraction (20)

    1050. String Subtraction (20) Given two strings S1 and S2, S = S1 - S2 is defined to be the remainin ...

  3. 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 t ...

  4. PAT甲级——1050 String Subtraction

    1050 String Subtraction Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the remain ...

  5. 1050 String Subtraction (20 分)

    1050 String Subtraction (20 分) Given two strings S​1​​ and S​2​​, S=S​1​​−S​2​​ is defined to be the ...

  6. PAT (Advanced Level) 1050. String Subtraction (20)

    简单题. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> ...

  7. PAT甲题题解-1050. String Subtraction (20)-水题

    #include <iostream> #include <cstdio> #include <string.h> #include <algorithm&g ...

  8. PAT 甲级 1050 String Subtraction

    https://pintia.cn/problem-sets/994805342720868352/problems/994805429018673152 Given two strings S~1~ ...

  9. PAT Advanced 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 ...

随机推荐

  1. (转载)C#语言开发规范

    1.  命名规范a) 类[规则1-1]使用Pascal规则命名类名,即首字母要大写.eg:Class Test{...}[规则1-2]使用能够反映类功能的名词或名词短语命名类.[规则1-3]不要使用“ ...

  2. 解决: docker pull registry.access.redhat.com/rhel7/pod-infrastructure:latest

    直接获取 rpm文件 wget http://mirror.centos.org/centos/7/os/x86_64/Packages/python-rhsm-certificates-1.19.1 ...

  3. B树,B+树比较

    首先注意:B树就是B-树,"-"是个连字符号,不是减号.也就是B-树其实就是B树 B-树是一种平衡的多路查找(又称排序)树,在文件系统中有所应用.主要用作文件的索引.其中的B就表示 ...

  4. 【C#】委托的发展

    "用事件去处理程序, 进而解决问题" ---- 委托的目的 委托早在C#2的时候就已经初具模型, 但是并不是特别灵活, 制止C#3才在代码中被广泛使用. C#4中泛型委托, C#5 ...

  5. --HTML标签2

    表单元素: <input>标签 搜集用户信息 属性:type=" " text 默认值 size 长度 value 规定值 readonly 规定值 placehold ...

  6. 17秋 SDN课程 第四次上机作业

    1.建立以下拓扑,并连接上ODL控制器. 2.利用ODL下发流表,使得h3在10s内ping不通h1,10s后恢复. 3.借助Postman通过ODL的北向接口下发流表,再利用ODL北向接口查看已下发 ...

  7. POJ 3414 Pots(罐子)

    POJ 3414 Pots(罐子) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 You are given two po ...

  8. Lintcode452-Remove Linked List Elements-Easy

    Remove Linked List Elements Remove all elements from a linked list of integers that have value val. ...

  9. 2017-2018-2 20165306 实验四《Android开发基础》实验报告

    实验四<Android开发基础>实验报告 实验报告封面 实验内容 Android程序设计-1 实验要求: 参考<Java和Android开发学习指南(第二版)(EPUBIT,Java ...

  10. windows下远程连接Mysql

    使用“Ctrl + R”组合键快速打开cmd窗口,并输入“cmd”命令,打开cmd窗口. 使用“mysql -uroot -proot”命令可以连接到本地的mysql服务. 使用“use mysql” ...