58 lines
2.2 KiB
Java
58 lines
2.2 KiB
Java
import java.util.*;
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class VaultDoor1 {
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public static void main(String args[]) {
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VaultDoor1 vaultDoor = new VaultDoor1();
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Scanner scanner = new Scanner(System.in);
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System.out.print("Enter vault password: ");
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String userInput = scanner.next();
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String input = userInput.substring("picoCTF{".length(),userInput.length()-1);
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if (vaultDoor.checkPassword(input)) {
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System.out.println("Access granted.");
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} else {
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System.out.println("Access denied!");
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}
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}
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// I came up with a more secure way to check the password without putting
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// the password itself in the source code. I think this is going to be
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// UNHACKABLE!! I hope Dr. Evil agrees...
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//
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// -Minion #8728
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public boolean checkPassword(String password) {
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return password.length() == 32 &&
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password.charAt(0) == 'd' &&
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password.charAt(29) == 'a' &&
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password.charAt(4) == 'r' &&
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password.charAt(2) == '5' &&
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password.charAt(23) == 'r' &&
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password.charAt(3) == 'c' &&
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password.charAt(17) == '4' &&
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password.charAt(1) == '3' &&
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password.charAt(7) == 'b' &&
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password.charAt(10) == '_' &&
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password.charAt(5) == '4' &&
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password.charAt(9) == '3' &&
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password.charAt(11) == 't' &&
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password.charAt(15) == 'c' &&
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password.charAt(8) == 'l' &&
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password.charAt(12) == 'H' &&
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password.charAt(20) == 'c' &&
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password.charAt(14) == '_' &&
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password.charAt(6) == 'm' &&
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password.charAt(24) == '5' &&
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password.charAt(18) == 'r' &&
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password.charAt(13) == '3' &&
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password.charAt(19) == '4' &&
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password.charAt(21) == 'T' &&
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password.charAt(16) == 'H' &&
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password.charAt(27) == '6' &&
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password.charAt(30) == 'f' &&
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password.charAt(25) == '_' &&
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password.charAt(22) == '3' &&
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password.charAt(28) == 'd' &&
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password.charAt(26) == 'f' &&
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password.charAt(31) == '4';
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}
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}d35cr4mbl3_tH3
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