A. Keyboard
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:

qwertyuiop
asdfghjkl;
zxcvbnm,./

Unfortunately Mole is blind, so sometimes it is problem for him to put his hands accurately. He accidentally moved both his hands with one position to the left or to the right. That means that now he presses not a button he wants, but one neighboring button
(left or right, as specified in input).

We have a sequence of characters he has typed and we want to find the original message.

Input

First line of the input contains one letter describing direction of shifting ('L' or 'R' respectively
for left or right).

Second line contains a sequence of characters written by Mole. The size of this sequence will be no more than 100. Sequence contains only symbols that appear on Mole's
keyboard. It doesn't contain spaces as there is no space on Mole's keyboard.

It is guaranteed that even though Mole hands are moved, he is still pressing buttons on keyboard and not hitting outside it.

Output

Print a line that contains the original message.

Sample test(s)
input
R
s;;upimrrfod;pbr
output
allyouneedislove


代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
char s[100]={'q','w','e','r','t','y','u','i','o','p','a','s','d','f','g','h','j','k','l',';','z','x','c','v','b','n','m',',','.','/'};
char s2[200];
int main()
{
char d[10];
scanf("%s",d);
scanf("%s",s2);
if(d[0]=='L')
{
int n=strlen(s2);
for(int i=0;i<n;i++)
{
for(int j=0;j<30;j++)
{
if(s2[i]==s[j])
{
printf("%c",s[j+1]);
break;
}
}
}
printf("\n");
}
else
{
int n=strlen(s2);
for(int i=0;i<n;i++)
{
for(int j=0;j<30;j++)
{
if(s2[i]==s[j])
{
printf("%c",s[j-1]);
break;
}
}
}
printf("\n");
}
return 0;
}


A. Keyboard Codeforces Round #271(div2)的更多相关文章

  1. D. Flowers Codeforces Round #271(div2)

    D. Flowers time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input ...

  2. B. Worms Codeforces Round #271 (div2)

    B. Worms time limit per test 1 second memory limit per test 256 megabytes input standard input outpu ...

  3. Codeforces Round #271 (Div. 2)题解【ABCDEF】

    Codeforces Round #271 (Div. 2) A - Keyboard 题意 给你一个字符串,问你这个字符串在键盘的位置往左边挪一位,或者往右边挪一位字符,这个字符串是什么样子 题解 ...

  4. Codeforces Round #539 div2

    Codeforces Round #539 div2 abstract I 离散化三连 sort(pos.begin(), pos.end()); pos.erase(unique(pos.begin ...

  5. 【前行】◇第3站◇ Codeforces Round #512 Div2

    [第3站]Codeforces Round #512 Div2 第三题莫名卡半天……一堆细节没处理,改一个发现还有一个……然后就炸了,罚了一啪啦时间 Rating又掉了……但是没什么,比上一次好多了: ...

  6. Codeforces Round#320 Div2 解题报告

    Codeforces Round#320 Div2 先做个标题党,骗骗访问量,结束后再来写咯. codeforces 579A Raising Bacteria codeforces 579B Fin ...

  7. Codeforces Round #564(div2)

    Codeforces Round #564(div2) 本来以为是送分场,结果成了送命场. 菜是原罪 A SB题,上来读不懂题就交WA了一发,代码就不粘了 B 简单构造 很明显,\(n*n\)的矩阵可 ...

  8. Codeforces Round #361 div2

    ProblemA(Codeforces Round 689A): 题意: 给一个手势, 问这个手势是否是唯一. 思路: 暴力, 模拟将这个手势上下左右移动一次看是否还在键盘上即可. 代码: #incl ...

  9. Codeforces Round #626 Div2 D,E

    比赛链接: Codeforces Round #626 (Div. 2, based on Moscow Open Olympiad in Informatics) D.Present 题意: 给定大 ...

随机推荐

  1. idea+spring4+springmvc+mybatis+maven实现简单增删改查CRUD

    在学习spring4+springmvc+mybatis的ssm框架,idea整合简单实现增删改查功能,在这里记录一下. 原文在这里:https://my.oschina.net/finchxu/bl ...

  2. [NOI2015]品酒大会(SA数组)

    [NOI2015]品酒大会 题目描述 一年一度的"幻影阁夏日品酒大会"隆重开幕了.大会包含品尝和趣味挑战 两个环节,分别向优胜者颁发"首席品酒家"和" ...

  3. Spring Tool Suit安装virgo server插件、virgo的下载

    virgo-tomcat原先是Spring DM Server,后来转eclipse社区维护 安装教程:http://osgi.com.cn/article/7289514 virgo-tomcat各 ...

  4. HDU2452 Navy maneuvers 记忆化搜索

    这题目意思能忍?读了半年,乱七八糟的 记忆化搜索 拖拖的,dp[i][0]代表以获得最小值为目标的船以i为起点.dp[i][1]代表以获得最大值为目标的船以i为起点.接下来暴力枚举入度为0的点为起点, ...

  5. hdu-1342 Lotto

    http://acm.hdu.edu.cn/showproblem.php? pid=1342 题意:以升序的形式给定k个数.输出从中挑选6个数满足升序的全部情况. 思路:两个參数.第一个保存当前搜索 ...

  6. rac重新启动遭遇ORA-01078、ORA-01565、ORA-17503、ORA-12547

    今天測试环境server重新启动导致一个节点集群无法重新启动,遭遇ORA-12547错误.详细例如以下: server重新启动后,rac1集群无法启动,rac2正常启动: [root@rac1 ~]# ...

  7. hdu 1695 GCD (欧拉函数、容斥原理)

    GCD Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submis ...

  8. Hadoop2.2集群安装配置-Spark集群安装部署

    配置安装Hadoop2.2.0 部署spark 1.0的流程 一.环境描写叙述 本实验在一台Windows7-64下安装Vmware.在Vmware里安装两虚拟机分别例如以下 主机名spark1(19 ...

  9. UltraEdit Companion Utility

    UltraEdit Companion Utility 配色组件 http://www.danielwmoore.com/extras/index.php?action=downloads;sa=vi ...

  10. Web测试要点 做移动端的测试,也做web端的测试,甚至后面桌面端的测试和后台的测试也做了,基本上把我们产品各个端都玩了一轮

    Web测试要点 一.功能测试 1.链接测试 (1).测试所有链接是否按指示的那样确实链接到了该链接的页面:  (2).测试所链接的页面是否存在:  (3).保证Web应用系统上没有孤立的页面(所谓孤立 ...