Codeforces 474 A

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                                        ***A. Keyboard***Our good friend Mole is trying to code a big message. He is typing on an unusual keyboard with characters arranged in following way:qwertyuiopasdfghjkl;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.InputFirst 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.OutputPrint a line that contains the original message.Sample test(s)inputRs;;upimrrfod;pbroutputallyouneedislove

题目大意：就是给你一个字符串，让你向右或向左移动一位

直接上代码吧，太水了。。。

/*2015 - 8 - 29 中午Author: ITAK今日的我要超越昨日的我，明日的我要胜过今日的我，以创作出更好的代码为目标，不断地超越自己。*/#include <iostream>#include <cstring>using namespace std;char str[105];char st[105];char s[31] = {"qwertyuiopasdfghjkl;zxcvbnm,./"};int main(){    char c;    while(cin>>c)    {        cin>>str;        int len = strlen(str);        if(c == 'R')        {            for(int i=0; i<len; i++)            {                for(int j=0; j<30; j++)                {                    if(str[i] == s[j])                    {                        cout<<s[j-1];                        break;                    }                }            }        }        else        {            for(int i=0; i<len; i++)            {                for(int j=0; j<30; j++)                {                    if(str[i] == s[j])                    {                        cout<<s[j+1];                        break;                    }                }            }        }    }    return 0;}