http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5502
The 12th Zhejiang Provincial Collegiate Programming Contest - J
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;
分析:
直接打表模拟对照输出即可。
AC代码:
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 #include <stack> 5 #include <queue> 6 #include <map> 7 #include <set> 8 #include <vector> 9 #include <math.h> 10 #include <algorithm> 11 using namespace std; 12 #define ls 2*i 13 #define rs 2*i+1 14 #define up(i,x,y) for(i=x;i<=y;i++) 15 #define down(i,x,y) for(i=x;i>=y;i--) 16 #define mem(a,x) memset(a,x,sizeof(a)) 17 #define w(a) while(a) 18 #define LL long long 19 const double pi = acos(-1.0); 20 #define Len 20005 21 #define mod 19999997 22 const int INF = 0x3f3f3f3f; 23 24 char s1[]= {"-=_+qwertyuiop[]QWERTYUIOP{}asdfghjkl;'ASDFGHJKL:"zxcvbnm,./ZXCVBNM<>?"}; 25 char s2[]= {"[]{}',.pyfgcrl/="<>PYFGCRL?+aoeuidhtns-AOEUIDHTNS_;qjkxbmwvz:QJKXBMWVZ"}; 26 char c; 27 28 char print(char c) 29 { 30 for(int i=0; s1[i]; i++) 31 if(s1[i]==c) 32 return s2[i]; 33 return c; 34 } 35 int main() 36 { 37 w(~scanf("%c",&c)) 38 printf("%c",print(c)); 39 40 return 0; 41 }