This is my Solution in Java!!
import java.io.File;
import java.io.FileNotFoundException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class alien {
/**
* @param args
* @throws FileNotFoundException
*/
public static void main(String[] args) throws FileNotFoundException {
Scanner in = new Scanner(new File("aliens.in"));
PrintWriter out = new PrintWriter(new File("aliens.out"));
String[] data = in.nextLine().split(" ");
int l = Integer.parseInt(data[0]);
int d = Integer.parseInt(data[1]);
int cas = Integer.parseInt(data[2]);
String rawWord;
int count = 0;
String[] dico = new String[d];
for (int i = 0; i < dico.length; i++) {
dico[i] = in.nextLine();
}
for (int i = 1; i <= cas; i++) {
rawWord = in.nextLine();
List<String> cases = new ArrayList<String>();
String mot = "";
char currentDictionnaryChar;
String letters = "";
cases = getTokens(rawWord);
// on traite chaque mot du dico pour vérifier si oui ou non il
peut
// être construit avec la chaine reçue
for (int j = 0; j < dico.length; j++) {
mot = dico[j];
boolean found = true;
for (int k = 0; k < l; k++) {
currentDictionnaryChar = mot.charAt(k);
letters = cases.get(k);
if(letters.indexOf(""+currentDictionnaryChar) ==
-1){
found = false;
break;
}
}
if(found) count++;
} // tous les mots du dico sont traités
out.print("Case #"+ i +": "+count);
count = 0;
if(in.hasNext()) out.println();
}
out.close();
}
public static List<String> getTokens(String raw) {
List<String> temp = new ArrayList<String>();
StringBuffer rawTemp = new StringBuffer(raw);
int i = 0, j = 0;
while(rawTemp.capacity() != 0){
i = rawTemp.indexOf("(") ;
j = rawTemp.indexOf(")") ;
switch(i) {
case 0:
temp.add("" + rawTemp.substring(i+1, j));
rawTemp = rawTemp.delete(i, j+1);
break;
case -1 :
for (int j2 = 0; j2 < rawTemp.length(); j2++) {
temp.add("" + rawTemp.charAt(j2));
}
rawTemp = rawTemp.delete(0, rawTemp.length());
break;
default :
for (int j2 = 0; j2 < i; j2++) {
temp.add("" + rawTemp.charAt(j2));
}
rawTemp = rawTemp.delete(0, i);
break;
}
rawTemp.trimToSize();
}
return temp;
}
}
2011/5/6 이홍일 <[email protected]>
> You can get source code on scoreboard.
> That is not difficult problem.
> You can get solution if you think little more.
>
> 2011/5/6 Matty <[email protected]>
>
>> Does anybody have a solution code for this problem written in Java?
>> I'm stuck...
>>
>> Thanks in advance,
>>
>> Matt
>>
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>>
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