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Transformation PicoCTF reverse engineering challenge
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| enc_dec = [28777, 25455, 17236, 18043, 12598, 24418, 26996, 29535, 26990, 29556, 13108, 25695, 28518, 24376, 24421, 14128, 13154, 13368, 13949] | |
| # Up to 0 to 126 because of pairs of characters. | |
| for val in enc_dec: | |
| for i in range(127): | |
| if 0 <= (val - (i << 8)) <= 126: | |
| print(chr(i) + chr(val - (i * 256)), end='', flush=True) | |
| # OR | |
| decoded_chars = ''.join([ | |
| chr(i) + chr(val - (i << 8)) | |
| for val in enc_dec | |
| for i in range(127) | |
| if 0 <= (val - (i << 8)) <= 126 | |
| ]) |
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| enc = "灩捯䍔䙻ㄶ形楴獟楮獴㌴摟潦弸彥ㄴㅡて㝽" | |
| output, output2 = "", "" | |
| for i in range(len(enc)): | |
| output += chr(ord(enc[i])>>8) | |
| output += chr(ord(enc[i]) - ((ord(enc[i])>>8)<<8)) | |
| two_byte_char_code = ord(enc[i]) | |
| upper_byte_char_code = ord(enc[i])>>8 | |
| upper_byte_isolated_char_code = (ord(enc[i])>>8)<<8 # The shifting to the left zeros out the lower byte. | |
| lower_byte_char_code = ord(enc[i]) - ((ord(enc[i])>>8)<<8) | |
| # Alternatively convert to string and then split the string binary representation. | |
| two_byte_char_code_in_bin = bin(two_byte_char_code)[2:].zfill(16) | |
| upper_byte_char_code1 = two_byte_char_code_in_bin[0:8] | |
| lower_byte_char_code1 = two_byte_char_code_in_bin[8:16] | |
| #print(int('01110000', 2)) | |
| #print(upper_byte_char_code1, lower_byte_char_code1) | |
| #print(chr(int(upper_byte_char_code1, 2)) , chr(int(lower_byte_char_code1, 2))) | |
| output2 += chr(int(upper_byte_char_code1, 2)) + chr(int(lower_byte_char_code1, 2)) | |
| print(bin(two_byte_char_code), bin(upper_byte_char_code), bin(upper_byte_isolated_char_code), bin(lower_byte_char_code)) | |
| #print(two_byte_char_code, upper_byte_char_code, upper_byte_isolated_char_code, lower_byte_char_code) | |
| #print(chr(upper_byte_char_code), chr(lower_byte_char_code)) | |
| #output += chr(ord(enc[i]) - ((ord(enc[i])//256)*256)) | |
| #output += chr(ord(enc[i]) & 0xFF) | |
| print(output) | |
| print(output2) |
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| decoded = ''.join([chr((ord(enc[i]) >> 8) & 0xFF) + chr(ord(enc[i]) & 0xFF) for i in range(0, len(enc), 2)]) |
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