Codeforces Round #271 (Div. 2) --A Keyboard （暴力）

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

题目链接:http://codeforces.com/contest/474/problem/A

大致题意：就是键盘上的三行字符。每行十个，如今给你一个串，告诉你它的偏移方式（左偏一位或者右偏一位），让你输出原串。

解题思路：非常easy。数据给的不大，那就直接暴力了，直接把三十个字符开个string或数组存一下。能够开三个，也能够开一个。可是我建议还是开一个好。看得更清楚。

直接扫一遍度入的串，要是左偏，就输出它的后面一个字符。否则。输出它的左边一个字符就可以。

AC代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
#define INF 0x7fffffff
string a = "qwertyuiop";
string b = "asdfghjkl;";
string c = "zxcvbnm,./";

int main()
{
    #ifdef sxk
        freopen("in.txt","r",stdin);
    #endif
    string n, s;
    while(cin>>n>>s)
    {
        int len = s.size();
        if(n[0] == 'R'){                  
            for(int i=0; i<len; i++){
                for(int j=0; j<10; j++){
                    if(s[i] == a[j]){
                       printf("%c",a[j-1]);
                       break;
                    }
                }
                for(int j=0; j<10; j++){
                    if(s[i] == b[j]){
                       printf("%c",b[j-1]);
                       break;
                    }
                }
                for(int j=0; j<10; j++){
                    if(s[i] == c[j]){
                       printf("%c",c[j-1]);
                       break;
                    }
                }
            }
        }
        else{
            for(int i=0; i<len; i++){
                for(int j=0; j<10; j++){
                    if(s[i] == a[j]){
                       printf("%c",a[j+1]);
                       break;
                    }
                }
                for(int j=0; j<10; j++){
                    if(s[i] == b[j]){
                       printf("%c",b[j+1]);
                       break;
                    }
                }
                for(int j=0; j<10; j++){
                    if(s[i] == c[j]){
                       printf("%c",c[j+1]);
                       break;
                    }
                }
            }
        }
        printf("\n");
    }
    return 0;
}