1084 Broken Keyboard (20 分)
On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters corresponding to those keys will not appear on screen.
Now given a string that you are supposed to type, and the string that you actually type out, please list those keys which are for sure worn out.
Input Specification:
Each input file contains one test case. For each case, the 1st line contains the original string, and the 2nd line contains the typed-out string. Each string contains no more than 80 characters which are either English letters [A-Z] (case insensitive), digital numbers [0-9], or _ (representing the space). It is guaranteed that both strings are non-empty.
Output Specification:
For each test case, print in one line the keys that are worn out, in the order of being detected. The English letters must be capitalized. Each worn out key must be printed once only. It is guaranteed that there is at least one worn out key.
Sample Input:
7_This_is_a_test
_hs_s_a_es
Sample Output:
7TI
分析:注意字符串结尾是string::npos, 7行搞定!
/**
* Copyright(c)
* All rights reserved.
* Author : Mered1th
* Date : 2019-02-25-19.55.33
* Description : A1084
*/
#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<string>
#include<unordered_set>
#include<map>
#include<vector>
#include<set>
#include<unordered_map>
using namespace std;
int main(){
#ifdef ONLINE_JUDGE
#else
freopen("1.txt", "r", stdin);
#endif
string s1,s2,ans="";
cin>>s1>>s2;
;i<s1.length();i++){
if(s2.find(s1[i])==string::npos &&ans.find(toupper(s1[i]))==string::npos){
ans+=toupper(s1[i]);
}
}
cout<<ans;
;
}
1084 Broken Keyboard (20 分)的更多相关文章
- 【PAT甲级】1084 Broken Keyboard (20 分)
题意: 输入两行字符串,输出第一行有而第二行没有的字符(对大小写不敏感且全部以大写输出). AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #inclu ...
- 1084. Broken Keyboard (20)【字符串操作】——PAT (Advanced Level) Practise
题目信息 1084. Broken Keyboard (20) 时间限制200 ms 内存限制65536 kB 代码长度限制16000 B On a broken keyboard, some of ...
- 1084. Broken Keyboard (20)
On a broken keyboard, some of the keys are worn out. So when you type some sentences, the characters ...
- PAT Advanced 1084 Broken Keyboard (20) [Hash散列]
题目 On a broken keyboard, some of the keys are worn out. So when you type some sentences, the charact ...
- PAT (Advanced Level) 1084. Broken Keyboard (20)
简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...
- 1084. Broken Keyboard (20)-水题
#include <iostream> #include <cstdio> #include <string.h> #include <algorithm&g ...
- PAT 1084 Broken Keyboard
1084 Broken Keyboard (20 分) On a broken keyboard, some of the keys are worn out. So when you type ...
- pat1084. Broken Keyboard (20)
1084. Broken Keyboard (20) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue On a ...
- pat 1084 Broken Keyboard(20 分)
1084 Broken Keyboard(20 分) On a broken keyboard, some of the keys are worn out. So when you type som ...
随机推荐
- 2019.1.22 zigbee test
1传输测试 频谱仪设置: sigfox 模块串口设置: 自动选择对应型号 Test step: PS:发送TX指令 AT$cw=波特率,通道,uint 这里有个问题--不应该只发送一次 ------- ...
- Andriod Studio 解决问题 Failed to resolve: com.android.support:appcompat-v7:28.+
Andriod Studio报错提示: Error:(26, 13) Failed to resolve: com.android.support:appcompat-v7:28.+ 原因:Andri ...
- iOS 开发 Framework
制作Framework 的好处和缺点 好处: 1.如果模块间接口定义的比较完善,模块化的程序具有很好的可扩展性与内聚性: 2.物理上的模块化便于开发过程的管理与测试,尤其是在程 ...
- Qt QML referenceexamples attached Demo hacking
/********************************************************************************************* * Qt ...
- POJ 2367:Genealogical tree(拓扑排序模板)
Genealogical tree Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7285 Accepted: 4704 ...
- SUST OJ 1674: 入侵与反击(最长不下降子序列)
1674: 入侵与反击 时间限制: 1 Sec 内存限制: 128 MB提交: 229 解决: 28[提交][状态][讨论版] 题目描述 A国部署的反导系统遇到了一个致命BUG,那就是每一次发射的 ...
- .NET 中什么样的类是可使用 await 异步等待的?
我们已经知道 Task 是可等待的,但是去看看 Task 类的实现,几乎找不到哪个基类.接口或者方法属性能够告诉我们与 await 相关. 而本文将探索什么样的类是可使用 await 异步等待的? D ...
- hot load那点事
热加载,最初接触的时候是使用create-react-app的时候,创建一个项目出来,修改一点代码,页面自动刷新了,贫道当时就感叹,这是造福开发者的事情. 再后来编写静态页面的时候使用 VS Code ...
- 1050. 螺旋矩阵(25) pat乙级题
本题要求将给定的N个正整数按非递增的顺序,填入“螺旋矩阵”.所谓“螺旋矩阵”,是指从左上角第1个格子开始,按顺时针螺旋方向填充.要求矩阵的规模为m行n列,满足条件:m*n等于N:m>=n:且m- ...
- Jquery中.attr与.prop的区别
☆ http://www.jb51.net/article/114876.htm http://www.365mini.com/page/jquery-attr-vs-prop.htm https:/ ...