uvaoj-10082:WERTYU

A common typing error is to place the hands on the keyboard one row to the right of the correct position. So "Q" is typed as "W" and "J" is typed as "K" and so on. You are to decode a message typed
in this manner.

Input consists of several lines of text. Each line may contain digits, spaces, upper case letters (except Q, A, Z), or punctuation shown above [except back-quote (`)]. Keys labelled with words [Tab,
BackSp, Control, etc.] are not represented in the input. You are to replace each letter or punction symbol by the one immediately
to its left on the QWERTY keyboard shown above. Spaces in the input should be echoed in the output.

Sample Input

O S, GOMR YPFSU/

Output for Sample Input

I AM FINE TODAY.

题解：lrj的白书上第三章的题目，他用的getchar读的，我用的gets，经证实gets也不会导致wa。但是还是推荐用getline，用gets纯粹是因为懒。。//个人认为这题命题不严，并没有说AZQ~这四个字符输入以后会输出什么，实际实验里，输入A输出是]，输入Z输出是'。题目说\t之类的转义字符都不考虑，所以这个AZQ之类的情况就可以不考虑了。；

code：

#include <iostream>

#include <cstring>

#include <cstdio>

using namespace std;

char s[]="`1234567890-=QWERTYUIOP[]\ASDFGHJKL;'ZXCVBNM,./";//用字符串保存来存储各个字符，建议定义长度，我不定义是因为懒；

int main()

{

char ans[1005];//其实订10005更保险一点；

while(gets(ans))

{

int lenans=strlen(ans);

int lens=strlen(s);

for(int i=0; i<lenans; i++)

{

for(int j=1; j<lens; j++)//注意是从1开始循环，不然在j=0的时候可能会访问数组外空间导致wa；

{

if(s[j]==ans[i])

{

ans[i]=s[j-1];

break;

}

}

}

puts(ans);

}

return 0;

}