ACM大一练习赛-第三场------C - 薛定谔的猫《打表枚举的方法》

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Description

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;

我感觉我的思路比较简单，就是直接枚举，然后将字符逐个输出！要注意的是：将数组开的大一点，否则会runtime errror!因为题上给的是100kb。

代码：

#include <stdio.h>
#include <string.h>
char a[300000];
char b[300000];
void init()
{
	a['A']='A';a['a']='a';a['B']='X';a['b']='x';a['C']='J';a['c']='j';
	a['D']='E';a['d']='e';a['E']='>';a['e']='.';a['F']='U';a['f']='u';
	a['G']='I';a['g']='i';a['H']='D';a['h']='d';a['I']='C';a['i']='c';
	a['J']='H';a['j']='h';a['K']='T';a['k']='t';a['L']='N';a['l']='n';
	a['M']='M';a['m']='m';a['N']='B';a['n']='b';a['O']='R';a['o']='r';
	a['P']='L';a['p']='l';a['Q']='\"';a['q']='\'';a['R']='P';a['r']='p';
	a['S']='O';a['s']='o';a['T']='Y';a['t']='y';a['U']='G';a['u']='g';
	a['V']='K';a['v']='k';a['W']='<';a['w']=',';a['X']='Q';a['x']='q';
	a['Y']='F';a['y']='f';a['Z']=':';a['z']=';';a['{']='?';a['[']='/';
	a['}']='+';a[']']='=';a[':']='S';a[';']='s';a['\"']='_';a['\'']='-';
	a['<']='W';a[',']='w';a['>']='V';a['.']='v';a['?']='Z';a['/']='z';
	a['~']='~';a['`']='`';a['!']='!';a['1']='1';a['@']='@';a['2']='2';
	a['#']='#';a['3']='3';a['$']='$';a['4']='4';a['%']='%';a['5']='5';
	a['^']='^';a['6']='6';a['&']='&';a['7']='7';a['*']='*';a['(']='(';
	a[')']=')';a['8']='8';a['9']='9';a['0']='0';a[32]=' '; a['|']='|';a['\\']='\\';
	a['+']='}';a['=']=']';a['_']='{';a['-']='[';
}
int main()
{
	int len;
	init();
	while(gets(b))
	{
		len=strlen(b);
		for(int i=0;i<len;i++)
		{
			printf("%c",a[b[i]]);
		}
		printf("\n");
	}
	return 0;
}

ACM大一练习赛-第三场------C - 薛定谔的猫《打表枚举的方法》

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原文地址：http://blog.csdn.net/dxx_111/article/details/47839647