vault-door-1
This vault uses some complicated arrays! I hope you can make sense of it, special agent. The source code for this vault is here: VaultDoor1.java
Download the Java file.
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import java.util.*;
class VaultDoor1 {
public static void main(String args[]) {
VaultDoor1 vaultDoor = new VaultDoor1();
Scanner scanner = new Scanner(System.in);
System.out.print("Enter vault password: ");
String userInput = scanner.next();
String input = userInput.substring("picoCTF{".length(),userInput.length()-1);
if (vaultDoor.checkPassword(input)) {
System.out.println("Access granted.");
} else {
System.out.println("Access denied!");
}
}
// I came up with a more secure way to check the password without putting
// the password itself in the source code. I think this is going to be
// UNHACKABLE!! I hope Dr. Evil agrees...
//
// -Minion #8728
public boolean checkPassword(String password) {
return password.length() == 32 &&
password.charAt(0) == 'd' &&
password.charAt(29) == 'c' &&
password.charAt(4) == 'r' &&
password.charAt(2) == '5' &&
password.charAt(23) == 'r' &&
password.charAt(3) == 'c' &&
password.charAt(17) == '4' &&
password.charAt(1) == '3' &&
password.charAt(7) == 'b' &&
password.charAt(10) == '_' &&
password.charAt(5) == '4' &&
password.charAt(9) == '3' &&
password.charAt(11) == 't' &&
password.charAt(15) == 'c' &&
password.charAt(8) == 'l' &&
password.charAt(12) == 'H' &&
password.charAt(20) == 'c' &&
password.charAt(14) == '_' &&
password.charAt(6) == 'm' &&
password.charAt(24) == '5' &&
password.charAt(18) == 'r' &&
password.charAt(13) == '3' &&
password.charAt(19) == '4' &&
password.charAt(21) == 'T' &&
password.charAt(16) == 'H' &&
password.charAt(27) == '7' &&
password.charAt(30) == '3' &&
password.charAt(25) == '_' &&
password.charAt(22) == '3' &&
password.charAt(28) == '9' &&
password.charAt(26) == 'e' &&
password.charAt(31) == '8';
}
}
The code already specified the correct character at each password position.
The password had to be 32 characters long. The first character was d, the second was 3, and so on for all 32 positions.
d35cr4mbl3_tH3_cH4r4cT3r5_e79c38
Reconstruct the characters in order to recover the password!
So, the flag = picoCTF{d35cr4mbl3_tH3_cH4r4cT3r5_e79c38}
This post is licensed under CC BY 4.0 by the author.