ZOJ3878:Convert QWERTY to Dvorak(浙江赛2015)

ZOJ3878:Convert QWERTY to Dvorak(浙江省赛2015)

The 12th Zhejiang Provincial Collegiate Programming Contest 

Edward, a poor copy typist, is a user of the Dvorak Layout. But now he has only a QWERTY Keyboard with a broken Caps Lock key, so Edward never presses the broken Caps Lock key. Luckily, all the other keys on the QWERTY keyboard work well. Every day, he has a lot of documents to type. Thus he needs a converter to translate QWERTY into Dvorak. Can you help him?

The QWERTY Layout and the Dvorak Layout are in the following:

The QWERTY Layout

The Dvorak Layout

Input

A QWERTY document Edward typed. The document has no more than 100 kibibytes. And there are no invalid characters in the document.

Output

The Dvorak document.

Sample Input

Jgw Gqm Andpw a H.soav Patsfk f;doe
Nfk Gq.d slpt a X,dokt vdtnsaohe
Kjd yspps,glu pgld; aod yso kd;kgluZ
1234567890
`~!@#$%^&*()}"']_+-=ZQqWEwe{[\|
ANIHDYf.,bt/
ABCDEFuvwxyz

Sample Output

Hi, I'm Abel, a Dvorak Layout user.
But I've only a Qwerty keyboard.
The following lines are for testing:
1234567890
`~!@#$%^&*()+_-={}[]:"'<>,.?/\|
ABCDEFuvwxyz
AXJE>Ugk,qf;

题意:这道题题目都不用看就能理解意思了，就是按照键盘对应位置去模拟即可

思路：直接打表即可

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stack>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <math.h>
#include <algorithm>
using namespace std;
#define ls 2*i
#define rs 2*i+1
#define up(i,x,y) for(i=x;i<=y;i++)
#define down(i,x,y) for(i=x;i>=y;i--)
#define mem(a,x) memset(a,x,sizeof(a))
#define w(a) while(a)
#define LL long long
const double pi = acos(-1.0);
#define Len 20005
#define mod 19999997
const int INF = 0x3f3f3f3f;

char s1[]= {"-=_+qwertyuiop[]QWERTYUIOP{}asdfghjkl;'ASDFGHJKL:\"zxcvbnm,./ZXCVBNM<>?"};
char s2[]= {"[]{}',.pyfgcrl/=\"<>PYFGCRL?+aoeuidhtns-AOEUIDHTNS_;qjkxbmwvz:QJKXBMWVZ"};
char c;

char print(char c)
{
    for(int i=0; s1[i]; i++)
        if(s1[i]==c)
            return s2[i];
    return c;
}
int main()
{
    w(~scanf("%c",&c))
    printf("%c",print(c));

    return 0;
}