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 <iostream>
using namespace std; int main(){
string s1,s2,res="";
getline(cin,s1);
getline(cin,s2);
for(int i=;i<s1.length();i++){
if(s2.find(s1[i])==s2.npos) res+=s1[i];
}
cout<<res;
system("pause");
return ;
}
 

PAT Advanced 1050 String Subtraction (20 分)的更多相关文章

  1. 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 ...

  2. PAT Advanced 1050 String Subtraction (20) [Hash散列]

    题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer taking all th ...

  3. PAT练习--1050 String Subtraction (20 分)

    题⽬⼤意:给出两个字符串,在第⼀个字符串中删除第⼆个字符串中出现过的所有字符并输出. 这道题的思路:将哈希表里关于字符串s2的所有字符都置为true,再对s1的每个字符进行判断,若Hash[s1[i] ...

  4. 【PAT甲级】1050 String Subtraction (20 分)

    题意: 输入两个串,长度小于10000,输出第一个串去掉第二个串含有的字符的余串. trick: ascii码为0的是NULL,减去'0','a','A',均会导致可能减成负数. AAAAAccept ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

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

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

  9. PAT Advanced 1042 Shuffling Machine (20 分)(知识点:利用sstream进行转换int和string)

    Shuffling is a procedure used to randomize a deck of playing cards. Because standard shuffling techn ...

随机推荐

  1. RF快捷键

    常用快捷键 操作 键 重命名 F2 搜索关键字 F5 执行用例 F8 创建新工程 ctrl+n 创建新测试套 ctrl+shift+f 创建新用例 ctrl+shift+t 创建新关键字 ctrl+s ...

  2. koa2中间键原理

    一.koa2 const http = require('http'); const compose = require('./compose'); class Koa { constructor() ...

  3. Android Studio在Make Project时下载Grandle特别慢

    SDK下载完成了,建个工程, 又蒙了: Server returned HTTP response code: 502 for URL: https://services.gradle.org/dis ...

  4. apache用户认证

    创建一个目录abc:mkdir abc在此目录下建一个文件:12.txt正常情况下可以访问. 建立用户认证,从而使用户访问特定目录文件需要认证 在虚拟主机配置文件中即vim /usr/local/ap ...

  5. PHP加速器eAccelerator安装

    程序说明 eAccelerator是一个自由开放源码php加速器,优化和动态内容缓存,提高了php脚本的缓存性能,使得PHP脚本在编译的状态下,对 服务器的开销几乎为零. 它还有对脚本起优化作用,以加 ...

  6. java:Springmvc框架3(Validator)

    1.springmvcValidator: web.xml: <?xml version="1.0" encoding="UTF-8"?> < ...

  7. java:LeakFilling(Spring)

    1.配置文件总结: bean节点: id:用户自定义名称,用于标识当前对象,可以通过getBean(String id)从容器中获取该对象. class:要交给spring容器创建的对象的全类名(包名 ...

  8. Java多线程ThreadLocal介绍

    在Java多线程环境下ThreadLocal就像一家银行,每个线程就是银行里面的一个客户,每个客户独有一个保险箱来存放金钱,客户之间的金钱不影响. private static ThreadLocal ...

  9. DSP28335 eCAP 测频

    F28335共有6组eCAP模块,每个eCAP不但具有捕获功能,而且还可用作PWM输出功能.F28335捕获模块的主要特征如下: 1. 150MHz系统时钟的情况下,32位时基的时间分辨率为6.67n ...

  10. JWT的实现原理

    前言最近在做一个python项目的改造,将python项目重构为java项目,过程中遇到了这个知识点,觉得这个蛮实用的,所以下班后回来趁热打铁写下这篇总结,希望后面的人能够有所借鉴,少走弯路. 一.优 ...