SUSCTF2022 Misc

Misc is speechless

Misc

checkin

ra2

The download is Red Alert 2. Since it is not opened by exe, the map editor is useless. After opening the game, I found that there is a mission, which is
related to the flag, so I used notepad++ to search the contents of the mission globally, and found ctf-01.
The rules were found in ctf-01.lua, because the difficulty is only the last two, so I modified it. one time

Then I found that it was only 30s during the day, so I changed it to 120s (in fact, it should be changed to a larger number).
Then it is basically invincible, go to run

the map flag is there

SUSCTF{RED_ALERT_WINNER!!!}

Tanner

Get a picture, the end of the file is the format and rules of the flag.
Go to Baidu and check the above check nodes and bit nodes

See this: https://zhuanlan.zhihu.com/p/20258157

So the matrix can be recovered, but it is not important for the time being, and then look down and find:

table = [0,1]
for c1 in table:
    for c2 in table:
        for c3 in table:
            for c4 in table:
                for c5 in table:
                    for c6 in table:
                        for c7 in table:
                            for c8 in table:
                                for c9 in table:
                                    for c10 in table:
                                        if(c1^c2^c3^c4 == 0):
                                            if(c1^c5^c6^c7 == 0):
                                                if(c2^c5^c8^c9==0):
                                                    if(c3^c6^c8^c10==0):
                                                        if(c4^c7^c9^c10==0):
                                                            print(c1,c2,c3,c4,c5,c6,c7,c8,c9,c10)

Then use notepad to replace the spaces with empty, and write a script

='''0000000000
0000000111
0000011001
0000011110
0000101010
0000101101
0000110011
0000110100
0011000001
0011000110
0011011000
0011011111
0011101011
0011101100
0011110010
0011110101
0101000010
0101000101
0101011011
0101011100
0101101000
0101101111
0101110001
0101110110
0110000011
0110000100
0110011010
0110011101
0110101001
0110101110
0110110000
0110110111
1001001000
1001001111
1001010001
1001010110
1001100010
1001100101
1001111011
1001111100
1010001001
1010001110
1010010000
1010010111
1010100011
1010100100
1010111010
1010111101
1100001010
1100001101
1100010011
1100010100
1100100000
1100100111
1100111001
1100111110
1111001011
1111001100
1111010010
1111010101
1111100001
1111100110
1111111000
1111111111'''
s = s.split('\n')
tmp = 0
for i in s:
    tmp += int(i,2)
    print(tmp)
print(bin(tmp))
#111111111100000

Submit the result after wrapping SUSCTF{} with sha256

misound

The thief is speechless, the whole point is meaningful. The
spectrum can see the string, and the AnEWmuLTiPLyis_etimes_wiLLbEcomE_B
SSTV can see it

However, this question does not use SSTV at all,
and then silenteye solves the wav and gets a series of bases. decoded

207 359 220 224 352 315 359 374 290 310 277 507 391 513 423 392 508 383 440 322 420 427 503 460 295 318 245 302 407 414 410 130 369 317

According to this pulpy English and hint can know. He wants to express
e*_=B
e is plaintext, _ is a parameter, B is ciphertext,
just understand
e = 101 _=95
101*95 = 369*26 + 1
so you can get flag

s2 = 'AnEWmuLTiPLyis_etimes_wiLLbEcomE_B'
s3 = '207 359 220 224 352 315 359 374 290 310 277 507 391 513 423 392 508 383 440 322 420 427 503 460 295 318 245 302 407 414 410 130 369 317'
s3 = s3.split(' ')
for i in range(len(s2)):
    print(chr(round((int(s3[i])*26+1)/ord(s2[i]))),end='')

It doesn't matter if you don't understand it, I don't understand the purpose of this question either.

AUDIO

A simple text description
First of all, wav is a floating point type. When looking at the end of the wav file, it is found that the file is exported by AU.
Then found in the second half, wav has Morse code sound.
Then I found that the amplitude of the wav file and the mp3 file is exactly half the difference.
So use AU software to convert mp3 to 32-bit floating point (called wav2) like wav.
Then invert the wav file (up and down) and increase the amplitude by the same amplitude as wav2.

Then export the processed audio and put it into the AU to see the spectrum, and you can see very obvious Morse.
Of course, a more standard approach is to use libraries such as scipy to operate (because the wav is a 32-bit floating point, the wave library cannot operate on it.)

I don't want to write it down because I don't want to do it again (

Re

DigitalCircuits

First, use python-exe-unpacker-master+uncompyle6 to reverse python's exe, get the python file and search for
the variables
in it. There is 10011110001101110111100110111001 in f10.

Then you can directly find a tea algorithm to solve it.

SUSCTF{XBvfaEdQvbcrxPBh8AOcJ6gA}