输入s二进制数据
TYPE_HEX = 0 TYPE_ASCII = 1 TYPE_BINARY = 2 TYPE_BRAILLE = 3 with open('lookup_table.txt', 'r') as f: rows = f.read().splitlines() table = list(map(lambda x: x.split('\t'), rows)) def convert(text_array, in_type=TYPE_BINARY, out_type=TYPE_ASCII, sep=''): output = [] input_table = list(map(lambda x: x[in_type], table)) output_table = list(map(lambda x: x[out_type], table)) for item in text_array: index = input_table.index(item.upper()) output.append(output_table[index]) return sep.join(output) s='100100011100010100100100011110110100000111110001101001100100110000111010001111100110010100111000111000001111100100000101011111' x = [] for i in range(0,len(s),6): x.append(s[i:i+6]) y=convert(x, TYPE_BINARY, TYPE_BRAILLE) print(y)
#⠉⠎⠊⠉⠞⠋⠸⠣⠥⠉⠃⠗⠼⠙⠊⠇⠇⠼⠉⠨⠾ z=convert(y, TYPE_BRAILLE, TYPE_HEX) print(z)
#4353494354465F3C554342522344494C4C23432E29 print(bytes.fromhex(z).decode())
#CSICTF_<UCBR#DILL#C.)
输入结果为CSICTF_<UCBR#DILL#C.)，实际上需要转换，将所有的大写转成小写，'_>'转成'{'，'.)'转成'}'，#C对应3，#D对应4，如此类推
即结果为csictf{ucbr4ill3}
lookup_table.txt
Hex ASCII Binary Braille 20 000000 ⠀ 21 ! 011101 ⠮ 22 " 000010 ⠐ 23 # 001111 ⠼ 24 $ 110101 ⠫ 25 % 100101 ⠩ 26 & 111101 ⠯ 27 ' 001000 ⠄ 28 ( 111011 ⠷ 29 ) 011111 ⠾ 2A * 100001 ⠡ 2B + 001101 ⠬ 2C , 000001 ⠠ 2D - 001001 ⠤ 2E . 000101 ⠨ 2F / 001100 ⠌ 30 0 001011 ⠴ 31 1 010000 ⠂ 32 2 011000 ⠆ 33 3 010010 ⠒ 34 4 010011 ⠲ 35 5 010001 ⠢ 36 6 011010 ⠖ 37 7 011011 ⠶ 38 8 011001 ⠦ 39 9 001010 ⠔ 3A : 100011 ⠱ 3B ; 000011 ⠰ 3C < 110001 ⠣ 3D = 111111 ⠿ 3E > 001110 ⠜ 3F ? 100111 ⠹ 40 @ 000100 ⠈ 41 A 100000 ⠁ 42 B 110000 ⠃ 43 C 100100 ⠉ 44 D 100110 ⠙ 45 E 100010 ⠑ 46 F 110100 ⠋ 47 G 110110 ⠛ 48 H 110010 ⠓ 49 I 010100 ⠊ 4A J 010110 ⠚ 4B K 101000 ⠅ 4C L 111000 ⠇ 4D M 101100 ⠍ 4E N 101110 ⠝ 4F O 101010 ⠕ 50 P 111100 ⠏ 51 Q 111110 ⠟ 52 R 111010 ⠗ 53 S 011100 ⠎ 54 T 011110 ⠞ 55 U 101001 ⠥ 56 V 111001 ⠧ 57 W 010111 ⠺ 58 X 101101 ⠭ 59 Y 101111 ⠽ 5A Z 101011 ⠵ 5B [ 010101 ⠪ 5C \ 110011 ⠳ 5D ] 110111 ⠻ 5E ^ 000110 ⠘ 5F _ 000111 ⠸