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

题解:水题,打两个表即可了。

#include <iostream>
#include <cstdio>
#include <cstring> using namespace std; char c,s[150];
int r[30]={97,'v','x','s','w','d','f','g','u','h','j','k','n','b','i','o',97,'e','a','r','y','c','q','z','t',97},
l[30]={'s','n','v','f','r','g','h','j','o','k','l',';',',','m','p','[','w','t','d','y','i','b','e','c','u','x'}; int main()
{
scanf("%c",&c);
if (c=='R')
{
scanf("%s",s);
for (int i=0;i<strlen(s);i++) if (s[i]=='[') cout <<"p";
else if (s[i]==';') cout <<"l";
else if (s[i]==',') cout <<"m";
else if (s[i]=='.') cout <<",";
else if (s[i]=='/') cout <<".";
else printf("%c",r[s[i]-97]);
}
if (c=='L')
{
scanf("%s",s);
for (int i=0;i<strlen(s);i++) if (s[i]=='.') cout <<"/";
else if (s[i]==',') cout <<".";
else printf("%c",l[s[i]-97]);
}
return 0;
}

【CODEFORCES】 A. Keyboard的更多相关文章

  1. 【Codeforces】Round #491 (Div. 2) 总结

    [Codeforces]Round #491 (Div. 2) 总结 这次尴尬了,D题fst,E没有做出来.... 不过还好,rating只掉了30,总体来说比较不稳,下次加油 A:If at fir ...

  2. 【Codeforces】Round #488 (Div. 2) 总结

    [Codeforces]Round #488 (Div. 2) 总结 比较僵硬的一场,还是手速不够,但是作为正式成为竞赛生的第一场比赛还是比较圆满的,起码没有FST,A掉ABCD,总排82,怒涨rat ...

  3. 【CodeForces】601 D. Acyclic Organic Compounds

    [题目]D. Acyclic Organic Compounds [题意]给定一棵带点权树,每个点有一个字符,定义一个结点的字符串数为往下延伸能得到的不重复字符串数,求min(点权+字符串数),n&l ...

  4. 【Codeforces】849D. Rooter's Song

    [算法]模拟 [题意]http://codeforces.com/contest/849/problem/D 给定n个点从x轴或y轴的位置p时间t出发,相遇后按对方路径走,问每个数字撞到墙的位置.(还 ...

  5. 【CodeForces】983 E. NN country 树上倍增+二维数点

    [题目]E. NN country [题意]给定n个点的树和m条链,q次询问一条链(a,b)最少被多少条给定的链覆盖.\(n,m,q \leq 2*10^5\). [算法]树上倍增+二维数点(树状数组 ...

  6. 【CodeForces】925 C.Big Secret 异或

    [题目]C.Big Secret [题意]给定数组b,求重排列b数组使其前缀异或和数组a单调递增.\(n \leq 10^5,1 \leq b_i \leq 2^{60}\). [算法]异或 为了拆位 ...

  7. 【CodeForces】700 D. Huffman Coding on Segment 哈夫曼树+莫队+分块

    [题目]D. Huffman Coding on Segment [题意]给定n个数字,m次询问区间[l,r]的数字的哈夫曼编码总长.1<=n,m,ai<=10^5. [算法]哈夫曼树+莫 ...

  8. 【CodeForces】906 D. Power Tower 扩展欧拉定理

    [题目]D. Power Tower [题意]给定长度为n的正整数序列和模数m,q次询问区间[l,r]累乘幂%m的答案.n,q<=10^5,m,ai<=10^9. [算法]扩展欧拉定理 [ ...

  9. 【CodeForces】741 D. Arpa’s letter-marked tree and Mehrdad’s Dokhtar-kosh paths(dsu on tree)

    [题意]给定n个点的树,每条边有一个小写字母a~v,求每棵子树内的最长回文路径,回文路径定义为路径上所有字母存在一种排列为回文串.n<=5*10^5. [算法]dsu on tree [题解]这 ...

随机推荐

  1. [BZOJ4012][HNOI2015]开店(动态点分治,树链剖分)

    4012: [HNOI2015]开店 Time Limit: 70 Sec  Memory Limit: 512 MBSubmit: 2168  Solved: 947[Submit][Status] ...

  2. Mac 配置几个环境变量

    终端    open -t ~/.bash_profile  打开.bash_profile export PATH=${PATH}:/Users/maxinliang/Android/sdk/pla ...

  3. 最封闭的开源系统,话说Android的八宗罪

    最封闭的开源系统,话说Android的八宗罪 http://www.ifanr.com/9833 Android 分裂的多面性 http://tech.cncms.com/shouji/android ...

  4. [POJ1205]Water Treatment Plants

    题目大意: 有一排n个格子,要在它们上面装管道. 每个格子上的管道都是T形的,但是可以有三种流动的方向. 每种都是把两个方向的水往另一个方向排出. 如果方向是向左或向右,就是排到相邻的格子里. 特别地 ...

  5. dubbo基础(初学习dubbo)

    1. 扩展   Soap是webService协议.是http+xml. Rest ful是http+json.相对于soap来说rest ful就是轻量的,因为==.   Rpc与soa区别? Rp ...

  6. [bzoj1016][JSOI2008]最小生成树计数 (Kruskal + Matrix Tree 定理)

    Description 现在给出了一个简单无向加权图.你不满足于求出这个图的最小生成树,而希望知道这个图中有多少个不同的最小生成树.(如果两颗最小生成树中至少有一条边不同,则这两个最小生成树就是不同的 ...

  7. JDK源码(1.7) -- java.util.ListIterator<E>

    java.util.ListIterator<E> 源码分析(JDK1.7) ------------------------------------------------------- ...

  8. extjs用iframe的问题

    项目中用extjs做前提系统的界面是左边用树做目录 右边用tabpanel做内容展示点击树节点的时候 在tabpanel添加新的tab JScript code var newTab = center ...

  9. How Basic Performance Analysis Saved Us Millions-------火焰图

    ENGINEERING How Basic Performance Analysis Saved Us Millions  Michael Malis May 19, 2017 9 min read ...

  10. 源码管理--llorch的Visual Studio基本教程(四)

    通用的演示样例说明: 本系列博客仅仅讨论工具的基础,不讨论不论什么语言. 甚至不讨论快捷键:-) 能够用鼠标就完毕本教程 IDE默认指代的是Visual Studio 2013 Community E ...