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 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
hssaes
题目分析
输入s1(应该输入的字符串),s2(显示出来的字符串),判断键盘坏掉的键
解题思路
- 定义int asc[256],记录s2中字符出现的次数
- 定义int bk[256],记录坏键是否已打印
- 遍历s1找到在s1中未出现在s2中的字符(asc[s1[i]]==0)
- 若bk[asc[s1[i]]]==0表示未打印过,则打印,并将其值置为1,表示该坏键已打印
- 若bk[asc[s1[i]]]==1表示已打印过,不再打印
Code
#include <iostream>
#include <cstring>
using namespace std;
int main(int argc, char * argv[]) {
char s1[81],s2[81];
cin.getline(s1,81);
cin.getline(s2,81);
int asc[256]= {0},bk[256]= {0};
int len1=strlen(s1),len2=strlen(s2);
for(int i=0; i<len2; i++) {
asc[toupper(s2[i])]++;
}
for(int i=0; i<len1; i++) {
char temp = toupper(s1[i]);
if(asc[temp]==0&&bk[temp]==0){
printf("%c",temp);
bk[temp]=1;
}
}
return 0;
}
PAT Advanced 1084 Broken Keyboard (20) [Hash散列]的更多相关文章
- 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 ...
- PAT Advanced 1041 Be Unique (20) [Hash散列]
题目 Being unique is so important to people on Mars that even their lottery is designed in a unique wa ...
- PAT Advanced 1134 Vertex Cover (25) [hash散列]
题目 A vertex cover of a graph is a set of vertices such that each edge of the graph is incident to at ...
- PAT Advanced 1048 Find Coins (25) [Hash散列]
题目 Eva loves to collect coins from all over the universe, including some other planets like Mars. On ...
- PAT Basic 1047 编程团体赛(20) [Hash散列]
题目 编程团体赛的规则为:每个参赛队由若⼲队员组成:所有队员独⽴⽐赛:参赛队的成绩为所有队员的成绩和:成绩最⾼的队获胜.现给定所有队员的⽐赛成绩,请你编写程序找出冠军队. 输⼊格式: 输⼊第⼀⾏给出⼀ ...
- 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 Level) 1084. Broken Keyboard (20)
简单题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #in ...
- 【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 ...
随机推荐
- 【iOS】Swift4.0 GCD的使用笔记
https://www.jianshu.com/p/47e45367e524 前言 在Swift4.0版本中GCD的常用方法还是有比较大的改动,这里做个简单的整理汇总. GCD的队列 队列是一种遵循先 ...
- linux X64函数参数传递过程研究
基础知识 函数传参存在两种方式,一种是通过栈,一种是通过寄存器.对于x64体系结构,如果函数参数不大于6个时,使用寄存器传参,对于函数参数大于6个的函数,前六个参数使用寄存器传递,后面的使用栈传递.参 ...
- springboot - 返回JSON error 从自定义的 ErrorController
使用AbstractErrorController(是ErrorController的实现),返回json error. 1.概览 2.基于<springboot - 映射 /error 到自定 ...
- 【Android】家庭记账本手机版开发报告七
一.说在前面 昨天 实现了账单的图标显示 今天 本地化,测试APP,将工程源码放到github上 源码:https://github.com/xiaotian12-call/Android_Boo ...
- Ubuntu 14.04 搭建 ftp
一.安装ftp服务器vsftpd $sudo apt-get update $sudo apt-get install vsftpd ftp服务器使用21端口,安装成功之后查看是否打开21端口 $ s ...
- Day1-T4
原题目 Describe:注意是“两次及以上”而不是“两种及以上”!! code: #include<bits/stdc++.h> using namespace std; int k,m ...
- C++ STD Gems04
count.count_if.all_of.any_of.none_of #include <iostream> #include <vector> #include < ...
- 多线程开发之NSThrea
创建并启动 先创建线程,再启动 // 创建 NSThread *thread = [[NSThread alloc] initWithTarget:self selector:@selector( ...
- Codeforces Round #603 (Div. 2) A. Sweet Problem(水.......没做出来)+C题
Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory ...
- JZOJPJ-C 8/21题解
原题大战D1 吐槽: T1 \(O(N^2)\; N \leq 26\) N大时还要写高精, 可以增加难度 T2 不给范围 T3 居然没有完全卡掉 不对应该赞美出题人 T4 PJ考个四边形不等式?? ...