Oracle

The problem provides you with a single Oracle.class file. Throwing it into an online Java decompiler gives this code:

public class Oracle
{
    private static final int FLAG_LENGTH = 42;
    private static final byte[] CHECK;
    private static byte[] numbers;
    
    private static void firstPass() {
        for (int i = 0; i < 42; ++i) {
            final byte[] numbers = Oracle.numbers;
            final int n = i;
            numbers[n] ^= (byte)(3 * i * i + 5 * i + 101 + i % 2);
        }
    }
    
    private static void secondPass() {
        final byte[] numbers = new byte[42];
        for (int i = 0; i < 42; ++i) {
            numbers[i] = (byte)(Oracle.numbers[(i + 42 - 1) % 42] << 4 | (Oracle.numbers[i] & 0xFF) >> 4);
        }
        Oracle.numbers = numbers;
    }
    
    private static void thirdPass() {
        for (int i = 0; i < 42; ++i) {
            final byte[] numbers = Oracle.numbers;
            final int n = i;
            numbers[n] += (byte)(7 * i * i + 31 * i + 127 + i % 2);
        }
    }
    
    private static void fail() {
        System.out.println("That's not the flag. Try again.");
        System.exit(1);
    }

    public static void main(final String[] array) {
        Oracle.numbers = System.console().readLine("Enter flag: ", new Object[0]).getBytes();
        if (Oracle.numbers.length != 42) {
            fail();
        }
        firstPass();
        secondPass();
        thirdPass();
        int n = 0;
        for (int i = 0; i < 42; ++i) {
            n |= (Oracle.CHECK[i] ^ Oracle.numbers[i]);
        }
        if (n != 0) {
            fail();
        }
        System.out.println("Good job. You found the flag!");
    }
    
    static {
        CHECK = new byte[] { 48, 6, 122, -86, -73, -59, 78, 84, 105, -119, -36, -118, 70, 17, 101, -85, 55, -38, -91, 32, -18, -107, 53, 99, -74, 67, 89, 120, -41, 122, -100, -70, 34, -111, 21, -128, 78, 27, 123, -103, 36, 87 };
    }
}

The program takes in user input, hashes it and then compares it to the CHECK array. To produce the correct flag the three hashing functions are applied in reverse order to retrieve the flag.

# check array
f = [48, 6, 122, -86, -73, -59, 78, 84, 105, -119, -36, -118, 70, 17, 101, -85, 55, -38, -91, 32, -18, -107, 53, 99, -74, 67, 89, 120, -41, 122, -100, -70, 34, -111, 21, -128, 78, 27, 123, -103, 36, 87];

# reverse the third hash function
for i in range(42):
	f[i] -= (7 * i * i + 31 * i + 127 + i % 2) % 256;

# reverse the second hash function
n = [0 for _ in range(42)];
for i in range(42):
	n[(i + 41) % 42] |= ((f[i] & 0xf0) >> 4);
	n[i] |= ((f[i] & 0xf) << 4);

# reverse and output the first hash function
print(''.join([chr(n[i] ^ (3 * i * i + 5 * i + 101 + i % 2) % 256) for i in range(42)]));

Flag: sdctf{u_f0und_th3_LANGu4ge_0f_th1s_0r4cl3}