• [2022安恒夏令营] 5个小题

    这个比赛的人比较少，中间又有强网杯，作了5个小题

    目录

    crypto

    misc 2

    misc 3

    re1

    re2

    crypto

    题目

    1. from secret import flag
    2. from Crypto.Util.number import *
    3.  
    4. seed = bytes_to_long(flag)
    5. bits = seed.bit_length()
    6.  
    7. while True:
    8.     p = getPrime(bits + 1)
    9.     if p > seed:
    10.         break
    11. print(p)
    12.  
    13. a = getRandomRange(1, p)
    14. b = getRandomRange(1, p)
    15.  
    16. for _ in range(3):
    17.     seed = (a * seed + b) % p
    18.     print(seed)
    19.  
    20. # 31893593182018727625473530765941216190921866039118147474754069955393226712079257707838327486268599271803
    21. # 25820280412859586557218124484272275594433027771091486422152141535682739897353623931875432576083022273940
    22. # 24295465524789348024814588142969609603624462580932512051939198335014954252359986260009296537423802567677
    23. # 14963686422550871447791815183480974143372785034397446416396172429864269108509521776424254168481536292904

    flag是第一个seed,然后每次用seed = (a*seed +b)%p得到下一个，可以用手工得到两个方程

    1. s3-s2 = a(s2-s1) mod p #求a
    2. s3 = a*s2 + b mod p    #求b

     得到a,b以后就能直接得到flag

    1. p = 31893593182018727625473530765941216190921866039118147474754069955393226712079257707838327486268599271803
    2. s1= 25820280412859586557218124484272275594433027771091486422152141535682739897353623931875432576083022273940
    3. s2= 24295465524789348024814588142969609603624462580932512051939198335014954252359986260009296537423802567677
    4. s3= 14963686422550871447791815183480974143372785034397446416396172429864269108509521776424254168481536292904
    5.  
    6. from gmpy2 import *
    7. # s3-s2 = a(s2-s1) mod p 
    8. t = invert(s2-s1, p)
    9. a = (t*(s3-s2))%p
    10. a = 25714049524051800625758372225598770398614533519595256157902456884372427008002817231382013589153912429484
    11.  
    12. # s3 = a*s2 + b mod p 
    13. b = (s3 - a*s2 )%p
    14. b = 10170297191068284996927402215696494489341479919496467043387273609973505047115652911086290426648313876559
    15. #s1 = a*s0 + b mod p 
    16. a1 = invert(a,p)
    17. s0 = ((s1 - b)*a1 )%p
    18. s0 = 14337636555117933152506165016723944787939761429733562849369091223517166614830298165864272285381681301117
    19. bytes.fromhex(hex(s0)[2:])
    20. #flag{e4syRsa1snotdifficult5996642D0A7415EF}
    21. #e4syRsa1snotdifficult5996642D0A7415EF

    misc 2

    题目程序

    1. # from flag import FLAG
    2. import random
    3. from PIL import Image
    4. from hashlib import md5
    5. from Crypto.Util.number import long_to_bytes as n2b
    6.  
    7. FLAG = b'A'*28
    8.  
    9. random.seed(793211)
    10.  
    11.  
    12. def pbl(bits):
    13.     num = random.getrandbits(bits)
    14.     bins = []
    15.     while num:
    16.         bins.append(num & 1)
    17.         num >>= 1
    18.     while len(bins) != bits:
    19.         bins.append(0)
    20.     return bins
    21.  
    22.  
    23. assert len(FLAG) == 28
    24. cat = Image.open('1.png')
    25. cat1 = Image.new('L', cat.size)
    26. cat2 = Image.new('L', cat.size)
    27. cat3 = Image.new('L', cat.size)
    28.  
    29. x, y = cat.size
    30. bits = x * y
    31. r1, r2 = pbl(bits), pbl(bits)
    32. r3 = FLAG + n2b(random.getrandbits((bits - len(FLAG)) * 8))
    33. r3 = list(r3)
    34. random.shuffle(r3)
    35.  
    36. for i in range(x):
    37.     for j in range(y):
    38.         pix = cat.getpixel((i, j))
    39.         cat1.putpixel((i, j), pix[0] ^ r1[i * y + j])
    40.         cat2.putpixel((i, j), pix[1] ^ r2[i * y + j])
    41.         cat3.putpixel((i, j), pix[2] ^ r3[i * y + j])
    42.  
    43. img = Image.new('RGB', cat.size)
    44.  
    45.  
    46. for p1, p3 in zip(cat1.getdata(), cat3.getdata()):
    47.     print(p1, p3)
    48.     break
    49. #img.putdata([(p1, 0, p3) for p1, p3 in zip(cat1.getdata(), cat3.getdata())])
    50. #img.save('xx.png')

    附件给了两个图片，然后给定加密程序，可以看到 第2个图的B通道是第1个图片B通过与flag异或的结果，异或序列是一个随机数序列，seed已经给出，通过seed可以得到这个序列。

    1. import random
    2. from PIL import Image
    3. from hashlib import md5
    4. from Crypto.Util.number import long_to_bytes as n2b
    5.  
    6. def pbl(bits):
    7.     num = random.getrandbits(bits)
    8.     bins = []
    9.     while num:
    10.         bins.append(num & 1)
    11.         num >>= 1
    12.     while len(bins) != bits:
    13.         bins.append(0)
    14.     return bins
    15.  
    16.  
    17. cat = Image.open('1.png')
    18. xx  = Image.open('xx.png')
    19.  
    20. x, y = cat.size
    21. bits = x * y
    22.  
    23. r3 = [0]*bits 
    24. for i in range(x):
    25.     for j in range(y):
    26.         pix = cat.getpixel((i, j))
    27.         pxx = xx.getpixel((i, j))
    28.         r3[i * y + j] = pix[2]^pxx[2]
    29.  
    30. #print(r3)
    31.  
    32. r_ord = [i for i in range(bits)]
    33. random.seed(793211)
    34. r1, r2 = pbl(bits), pbl(bits)
    35. random.getrandbits((bits - 28) * 8)
    36.  
    37. random.shuffle(r_ord)
    38. #print(r_ord)
    39.  
    40. flag = [0]*28
    41. for i in range(bits):
    42.     if r_ord[i]<28:
    43.         flag[r_ord[i]] = r3[i]
    44.  
    45. print(bytes(flag))
    46. #flag{lovely_cat_with_random}

    misc 3

    这个给了一个pyc的文件，无法反编译，先用dis打印出来可视字节码再手工翻译

    1. import marshal
    2. import dis
    3. import numpy as np
    4.  
    5. def a1():
    6.     code = open('whatisthis.cpython-38.pyc', 'rb').read()[16:]
    7.  
    8.     code = marshal.loads(code)
    9.     dis.dis(code)

    输出

    1.   2             0 LOAD_CONST               0 (0)
    2.                 2 LOAD_CONST               1 (None)
    3.                 4 IMPORT_NAME              0 (marshal)
    4.                 6 STORE_NAME               0 (marshal)
    5.                 8 LOAD_CONST               0 (0)
    6.                10 LOAD_CONST               1 (None)
    7.                12 IMPORT_NAME              1 (numpy)
    8.                14 STORE_NAME               2 (np)
    9.                16 LOAD_CONST               2 (235)
    10.                18 LOAD_CONST               3 (26)
    11.                20 LOAD_CONST               4 (15)
    12.                22 LOAD_CONST               5 (8)
    13. ...同上都是数据，此处略掉十万行... 
    14.            259458 LOAD_CONST              19 (18)
    15.            259460 LOAD_CONST             123 (7)
    16.            259462 EXTENDED_ARG             1
    17.            259464 EXTENDED_ARG           506
    18.            259466 BUILD_LIST           129723
    19.            259468 STORE_NAME               3 (LKmDJNIyYcklUJMlklygvZUBQUjRuhRGVrWiKDrdMsRjAshBxlXysnzMLXrrIHpovIRyoQWZQjvPbolMpJBkxXFMJtrfuQmXoNUXPfHZwdAYdYwDkuGOtJZLcYUnzCnm)
    20.  
    21.   4        259470 LOAD_CONST             133 ("Don't try to reverse this python script. You will be disappointed about it ")
    22.            259472 STORE_NAME               4 (SrvcwKKBZzolvVvpsMTzdEwhhsZYRBHaCjDkbjjaeTtrvMAYGjFHGGJofdCqlwsVlHZardbOZULepuWWTLSbZwjixarkVXPHOeMpPdArcybkUWoaMvoZrxGVzKPsUUAn)
    23.  
    24.   5        259474 LOAD_CONST             134 ('475')
    25.            259476 STORE_NAME               5 (EPwKMzdvDfYeFVrZDDSRPDqULNPDfaOYUKVpGuxabUwVBfovYMvOjkRbvwTxzTpgUXJOKvlUFGGJbfEEvIlyQBdsCEoTcmMWEqLKTrMwWLdjVsYnfNJsXtRKefaSastz)
    26.  
    27.   6        259478 LOAD_NAME                2 (np)
    28.            259480 LOAD_ATTR                6 (random)
    29.            259482 LOAD_METHOD              7 (seed)
    30.            259484 LOAD_NAME                2 (np)
    31.            259486 LOAD_METHOD              8 (sum)
    32.            259488 LOAD_CONST             135 (object  at 0x0000022FAE5EA6B0, file "/root/share/vm_share/³ö̢/notDefined/new/NotDefined_obfu.py", line 5>)
    33.            259490 LOAD_CONST             136 ('')
    34.            259492 MAKE_FUNCTION            0
    35.            259494 LOAD_NAME                5 (EPwKMzdvDfYeFVrZDDSRPDqULNPDfaOYUKVpGuxabUwVBfovYMvOjkRbvwTxzTpgUXJOKvlUFGGJbfEEvIlyQBdsCEoTcmMWEqLKTrMwWLdjVsYnfNJsXtRKefaSastz)
    36.            259496 GET_ITER
    37.            259498 CALL_FUNCTION            1
    38.            259500 CALL_METHOD              1
    39.            259502 CALL_METHOD              1
    40.            259504 POP_TOP
    41.  
    42.   7        259506 LOAD_NAME                9 (range)
    43.            259508 LOAD_NAME               10 (len)
    44.            259510 LOAD_NAME                3 (LKmDJNIyYcklUJMlklygvZUBQUjRuhRGVrWiKDrdMsRjAshBxlXysnzMLXrrIHpovIRyoQWZQjvPbolMpJBkxXFMJtrfuQmXoNUXPfHZwdAYdYwDkuGOtJZLcYUnzCnm)
    45.            259512 CALL_FUNCTION            1
    46.            259514 CALL_FUNCTION            1
    47.            259516 GET_ITER
    48.            259518 FOR_ITER                32 (to 259584)
    49.            259520 STORE_NAME              11 (EFAAANzoMBMeiWmpiPLHzCedWRoczlJwzpxHuBDPSWIlfFxpOqFiIZmWbvpOgrVNCsngbdYnwnxskvnFKLFfwMYNRmzfDisiWwXruLUffUyerjtrsmPoNzMUMpAatNWH)
    50.  
    51.   8        259522 LOAD_NAME                3 (LKmDJNIyYcklUJMlklygvZUBQUjRuhRGVrWiKDrdMsRjAshBxlXysnzMLXrrIHpovIRyoQWZQjvPbolMpJBkxXFMJtrfuQmXoNUXPfHZwdAYdYwDkuGOtJZLcYUnzCnm)
    52.            259524 LOAD_NAME               11 (EFAAANzoMBMeiWmpiPLHzCedWRoczlJwzpxHuBDPSWIlfFxpOqFiIZmWbvpOgrVNCsngbdYnwnxskvnFKLFfwMYNRmzfDisiWwXruLUffUyerjtrsmPoNzMUMpAatNWH)
    53.            259526 DUP_TOP_TWO
    54.            259528 BINARY_SUBSCR
    55.            259530 LOAD_NAME                2 (np)
    56.            259532 LOAD_ATTR                6 (random)
    57.            259534 LOAD_METHOD             12 (randint)
    58.            259536 LOAD_CONST             137 (27)
    59.            259538 CALL_METHOD              1
    60.            259540 INPLACE_XOR
    61.            259542 ROT_THREE
    62.            259544 STORE_SUBSCR
    63.            259546 EXTENDED_ARG             3
    64.            259548 EXTENDED_ARG          1013
    65.            259550 JUMP_ABSOLUTE        259518 (to 519036)
    66.            259552 LOAD_NAME               13 (exec)
    67.            259554 LOAD_NAME                0 (marshal)
    68.            259556 LOAD_METHOD             14 (loads)
    69.            259558 LOAD_NAME               15 (bytearray)
    70.            259560 LOAD_NAME                3 (LKmDJNIyYcklUJMlklygvZUBQUjRuhRGVrWiKDrdMsRjAshBxlXysnzMLXrrIHpovIRyoQWZQjvPbolMpJBkxXFMJtrfuQmXoNUXPfHZwdAYdYwDkuGOtJZLcYUnzCnm)
    71.            259562 CALL_FUNCTION            1
    72.            259564 CALL_METHOD              1
    73.            259566 CALL_FUNCTION            1
    74.            259568 POP_TOP
    75.            259570 LOAD_CONST               1 (None)
    76.            259572 RETURN_VALUE
    77.  
    78. Disassembly of object  at 0x0000022FAE5EA6B0, file "/root/share/vm_share/³ö̢/notDefined/new/NotDefined_obfu.py", line 5>:
    79.   5           0 BUILD_LIST               0
    80.               2 LOAD_FAST                0 (.0)
    81.               4 FOR_ITER                12 (to 30)
    82.               6 STORE_FAST               1 (EFAAANzoMBMeiWmpiPLHzCedWRoczlJwzpxHuBDPSWIlfFxpOqFiIZmWbvpOgrVNCsngbdYnwnxskvnFKLFfwMYNRmzfDisiWwXruLUffUyerjtrsmPoNzMUMpAatNWH)
    83.         >>    8 LOAD_GLOBAL              0 (ord)
    84.              10 LOAD_FAST                1 (EFAAANzoMBMeiWmpiPLHzCedWRoczlJwzpxHuBDPSWIlfFxpOqFiIZmWbvpOgrVNCsngbdYnwnxskvnFKLFfwMYNRmzfDisiWwXruLUffUyerjtrsmPoNzMUMpAatNWH)
    85.              12 CALL_FUNCTION            1
    86.              14 LIST_APPEND              2
    87.              16 JUMP_ABSOLUTE            4 (to 8)
    88.              18 RETURN_VALUE

    手工翻译结果

    1. import  marshal
    2. import numpy as np
    3.  
    4. LKmD = [...]
    5. Srvc = "Don't try to reverse this python script. You will be disappointed about it "
    6. EPwK = '475'
    7.  
    8. np.random.seed(np.sum([ord(i) for i in EPwK]))
    9.  
    10. for EFAA in range(len(LKmD)):
    11.     LKmD[EFAA] ^= np.random.randint(27)
    12. exec(marshal.loads(bytearray(LKmD)))

     大意是前边那个数组与已知seed的随机序列异或得到一个串再执行，写函数处理一下，再用dis输出

    1. def a2():
    2.     fp = open('aaa.txt')
    3.     [fp.readline() for i in range(8)]
    4.     v = [int(fp.readline().split('(')[1].split(')')[0]) for i in range(129723)]
    5.     
    6.     EPwK = '475'
    7.     np.random.seed(np.sum([ord(i) for i in EPwK]))
    8.  
    9.     for i in range(len(v)):
    10.         v[i] ^= np.random.randint(27)
    11.  
    12.     code = marshal.loads(bytes(v))
    13.     dis.dis(code)

     后边的结果就更简单了，就是直接给一个字符串的原码，然后执行，翻译后

    1. import marshal
    2.  
    3. sv1 = [...略掉十成个数字...]
    4. def sv2(key):
    5.     sv3 = list(range(256))
    6.     sv4 = 0
    7.     for sv5 in range(256):
    8.         sv4 = (sv4 + sv3[sv5] + ord(key[sv5 % len(key)])) % 256
    9.         sv3[sv5], sv3[sv4] = sv3[sv4], sv3[sv5]
    10.     return sv3
    11. def sv6(p):
    12.     sv7 = sv2('h0lyduck')
    13.     sv8 = []
    14.     sv5 = sv4 = 0
    15.     for sv9 in p:
    16.         sv5 = (sv5 + 1) % 256
    17.         sv4 = (sv4 + sv7[sv5]) % 256
    18.         sv7[sv5], sv7[sv4] = sv7[sv4], sv7[sv5]
    19.         sva = (sv7[sv5] + sv7[sv4]) % 256
    20.         svb = sv7[sva]
    21.         sv8.append(sv9 ^ svb)
    22.     return sv8
    23. #exec(marshal.loads(bytearray(sv6(sv1))))
    24.  
    25. import dis
    26.  
    27. code = marshal.loads(bytearray(sv6(sv1)))
    28. dis.dis(code)

    第3得到都是直接的串，同样方法处理

    1. import marshal
    2. import struct
    3. from typing import List
    4. sv1 = [...略掉数据...]
    5. class sv2:
    6.     def __init__(self, val: int):
    7.         assert isinstance(val, int)
    8.         self.val = val
    9.     def __add__(self, other):
    10.         return sv2((self.val + other.val) & 0xffffffff)
    11.     def __xor__(self, other):
    12.         return sv2(self.val ^ other.val)
    13.     def __lshift__(self, nbit: int):
    14.         sv3 = (self.val << nbit % 32) & 0xffffffff
    15.         sv4 = (self.val & 0xffffffff) >> (32 - (nbit % 32))
    16.         return sv2(sv3 | sv4)
    17.     def __repr__(self):
    18.         return hex(self.val)
    19.     def __int__(self):
    20.         return int(self.val)
    21. def sv5(sv6: sv2, b: sv2, sv7: sv2, d: sv2):
    22.     sv6 += b
    23.     d ^= sv6
    24.     d <<= 16
    25.     sv7 += d
    26.     b ^= sv7
    27.     b <<= 12
    28.     sv6 += b
    29.     d ^= sv6
    30.     d <<= 8
    31.     sv7 += d
    32.     b ^= sv7
    33.     b <<= 7
    34.     return sv6, b, sv7, d
    35. def sv8(sv9: List[sv2], idx1, idx2, idx3, idx4):
    36.     sv9[idx1], sv9[idx2], sv9[idx3], sv9[idx4] = \
    37.     sv5(sv9[idx1], sv9[idx2], sv9[idx3], sv9[idx4])
    38. def sva(sv9: List[sv2]):
    39.     sv8(sv9, 0, 4, 8, 12)
    40.     sv8(sv9, 1, 5, 9, 13)
    41.     sv8(sv9, 2, 6, 10, 14)
    42.     sv8(sv9, 3, 7, 11, 15)
    43.     sv8(sv9, 0, 5, 10, 15)
    44.     sv8(sv9, 1, 6, 11, 12)
    45.     sv8(sv9, 2, 7, 8, 13)
    46.     sv8(sv9, 3, 4, 9, 14)
    47.     return sv9
    48. def svb(sv9: List[sv2]) -> List[bytes]:
    49.     return b''.join([struct.pack(', int(svc)) for svc in sv9])
    50. def svd(sve: bytes, svf: int, svg: bytes) -> bytes:
    51.     svh = [sv2(svj) for svj in struct.unpack(', b'expand 32-byte k')]
    52.     sve = [sv2(svj) for svj in struct.unpack(', sve)]
    53.     svf = [sv2(svf)]
    54.     svg = [sv2(svj) for svj in struct.unpack(', svg)]
    55.     sv9 = svh + sve + svf + svg
    56.     svk = sv9[:]
    57.     for svm in range(10):
    58.         sv9 = sva(sv9)
    59.     sv9 = [svc + svn for svc, svn in zip(sv9, svk)]
    60.     return svb(sv9)
    61. def svp(svj: bytes, y: bytes):
    62.     return bytes(sv6 ^ b for sv6, b in zip(svj, y))
    63. def svq(sve: bytes, svf: int, svg: bytes, plaintext: bytes):
    64.     sv7 = bytearray(0)
    65.     for svr in range(len(plaintext) // 64):
    66.         svs = svd(sve, svf + svr, svg)
    67.         svt = plaintext[svr * 64:(svr + 1) * 64]
    68.         sv7 += svp(svt, svs)
    69.     if len(plaintext) % 64 != 0:
    70.         svr = len(plaintext) // 64
    71.         svs = svd(sve, svf + svr, svg)
    72.         svt = plaintext[svr * 64:]
    73.         sv7 += svp(svt, svs)
    74.     return sv7
    75. sve = bytes.fromhex(
    76.     '000102030405060708090a0b0c0d0e0f101112131415161718191a1b1c1d1e1f')
    77. svf = 0x00000001
    78. svg = bytes.fromhex('000000000000004a00000000')
    79. sv6 = svq(sve, svf, svg, bytearray(sv1))
    80. code = marshal.loads(sv6)
    81. import dis 
    82. dis.dis(code)
    83.

    最后一次得到的文件，可以看到flag

    1.   3           0 LOAD_CONST               0 (0)
    2.               2 LOAD_CONST               1 (None)
    3.               4 IMPORT_NAME              0 (builtins)
    4.               6 STORE_NAME               0 (builtins)
    5.  
    6.   5           8 LOAD_CONST               2 ('flag{Wow_Y0o0U_@re_MaSt3r_0F_Not_d3f1neD}')
    7.              10 STORE_NAME               1 (flag)
    8.  
    9.  19          12 LOAD_CONST               3 (object enc at 0x000002567BCAD370, file "", line 5>)
    10.              14 LOAD_CONST               4 ('enc')
    11.              16 MAKE_FUNCTION            0
    12.              18 STORE_NAME               2 (enc)

    re1

    main里边直接调用加密程序，然后与密文核对

    1. int __cdecl main(int argc, const char **argv, const char **envp)
    2. {
    3.   __int64 v3; // rcx
    4.   __int64 v4; // rdx
    5.   __int128 v6[2]; // [rsp+20h] [rbp-38h] BYREF
    6.   char v7; // [rsp+40h] [rbp-18h]
    7.  
    8.   memset(v6, 0, sizeof(v6));
    9.   v7 = 0;
    10.   sub_140001020("I prefer TeaProPlus! can you give me this?\n");
    11.   sub_140001020("TeaProPlus code:");
    12.   sub_140001080("%32s", (const char *)v6);
    13.   sub_1400010E0(v3, (unsigned int *)v6);
    14.   v4 = 0i64;
    15.   while ( *((_BYTE *)v6 + v4) == byte_1400032F0[v4] )
    16.   {
    17.     if ( (unsigned __int64)++v4 >= 0x20 )
    18.     {
    19.       sub_140001020("Oh yeah! Good code! flag{(your input)}\n");
    20.       return 0;
    21.     }
    22.   }
    23.   puts("oh NO!I want to drink TeaProPlus");
    24.   return -1;
    25. }

     加密程序是一个变种的tea

    1. __int64 __fastcall sub_1400010E0(__int64 a1, unsigned int *a2)
    2. {
    3.   unsigned int v7; // r9d
    4.   unsigned int key; // edi
    5.   unsigned int v1; // ebp
    6.   unsigned int v0; // esi
    7.   unsigned int v2; // r14d
    8.   unsigned int v3; // r15d
    9.   unsigned int v4; // r12d
    10.   unsigned int v5; // r13d
    11.   _DWORD *key2; // r11
    12.   _DWORD *v11; // r10
    13.   __int64 v12; // rbx
    14.   _DWORD *v13; // r9
    15.   __int64 result; // rax
    16.   bool v15; // zf
    17.   unsigned int v6; // [rsp+40h] [rbp+8h]
    18.   int round; // [rsp+48h] [rbp+10h]
    19.   unsigned int v18; // [rsp+50h] [rbp+18h]
    20.  
    21.   v7 = a2[7];
    22.   key = 0;
    23.   v1 = a2[1];
    24.   v0 = *a2;
    25.   v2 = a2[2];
    26.   v3 = a2[3];
    27.   v4 = a2[4];
    28.   v5 = a2[5];
    29.   v6 = a2[6];
    30.   v18 = v7;
    31.   round = 12;
    32.   do
    33.   {
    34.     key -= 0x61C88647;
    35.     key2 = &key3[(key >> 2) & 3];
    36.     v0 += ((key ^ v1) + (*key2 ^ v7)) ^ (((16 * v7) ^ (v1 >> 3)) + ((v7 >> 5) ^ (4 * v1)));
    37.     *a2 = v0;
    38.     v11 = &key3[(key >> 2) & 3 ^ 1i64];
    39.     v1 += ((*v11 ^ v0) + (key ^ v2)) ^ (((16 * v0) ^ (v2 >> 3)) + ((v0 >> 5) ^ (4 * v2)));
    40.     a2[1] = v1;
    41.     v12 = (key >> 2) & 3 ^ 3i64;
    42.     v13 = &key3[(key >> 2) & 3 ^ 2i64];
    43.     v2 += ((key ^ v3) + (*v13 ^ v1)) ^ (((16 * v1) ^ (v3 >> 3)) + ((v1 >> 5) ^ (4 * v3)));
    44.     a2[2] = v2;
    45.     v3 += ((key ^ v4) + (key3[v12] ^ v2)) ^ (((16 * v2) ^ (v4 >> 3)) + ((v2 >> 5) ^ (4 * v4)));
    46.     a2[3] = v3;
    47.     v4 += ((key ^ v5) + (*key2 ^ v3)) ^ (((16 * v3) ^ (v5 >> 3)) + ((v3 >> 5) ^ (4 * v5)));
    48.     a2[4] = v4;
    49.     v5 += ((*v11 ^ v4) + (key ^ v6)) ^ (((16 * v4) ^ (v6 >> 3)) + ((v4 >> 5) ^ (4 * v6)));
    50.     a2[5] = v5;
    51.     LODWORD(key2) = (((key ^ v18) + (*v13 ^ v5)) ^ (((16 * v5) ^ (v18 >> 3)) + ((v5 >> 5) ^ (4 * v18)))) + v6;
    52.     a2[6] = (unsigned int)key2;
    53.     v6 = (unsigned int)key2;
    54.     result = key ^ v0;
    55.     v18 += (result + (key3[v12] ^ (unsigned int)key2)) ^ (((16 * (_DWORD)key2) ^ (v0 >> 3))
    56.                                                         + (((unsigned int)key2 >> 5) ^ (4 * v0)));
    57.     v15 = round-- == 1;
    58.     v7 = v18;
    59.     a2[7] = v18;
    60.   }
    61.   while ( !v15 );
    62.   return result;
    63. }

    解密程序

    1. from pwn import u32,p32
    2.  
    3. s = bytes.fromhex('473BBD10F9E05561C5EBF76A5F43238D0516091AEF403DD4676AB1B4A978356B')
    4. c = [0]*8
    5. for i in range(8):
    6.     c[i] = u32(s[i*4: i*4+4])
    7.  
    8. key = [0]*13
    9. k = 0
    10. for i in range(13):
    11.     k = (k - 0x61C88647)%0x100000000
    12.     key[i] = k 
    13.  
    14. key3 = [0x1234, 0x2345, 0x4567, 0x6789]
    15. for i in range(11, -1, -1):
    16.     for j in range(7, -1, -1):
    17.         p_next = (j+1)%8
    18.         p_pre  = (j-1)%8
    19.         t = j%4
    20.         c[j] = ( c[j] - 
    21.                  (   ((key[i]^c[p_next]) + (key3[((key[i]>>2)&3) ^ t]^c[p_pre]) ) 
    22.                    ^ (((c[p_pre]<<4)^(c[p_next]>>3)) + ((c[p_pre]>>5)^(c[p_next]<<2)) )   ) 
    23.             ) %0x100000000
    24.     print(hex(key[i]))
    25.  
    26. flag = b''
    27. for i in range(8):
    28.     flag += p32(c[i])
    29.  
    30. print(flag)
    31. #7f943921724d63dc0ac9c6febf99fa88

     

    re2

    main里边啥都没有，发现有反调，调用了TlsCallback_0

    1. void __stdcall TlsCallback_0(int a1, DWORD flOldProtect, int a3)
    2. {
    3.   int i; // eax
    4.  
    5.   if ( flOldProtect == 1 )
    6.   {
    7.     VirtualProtect(StartAddress, 0x320u, 0x40u, &flOldProtect);
    8.     for ( i = 0; i < 0x320; ++i )
    9.       *((_BYTE *)StartAddress + i) ^= i;
    10.     CreateThread(0, 0, StartAddress, 0, 0, 0);
    11.   }
    12. }

    函数对程序块异或解密后起线程执行，先用程序给程序patch

    1. a1 = open('re2.exe', 'rb').read()
    2. a1 = list(a1)
    3. for i in range(800):
    4.     a1[0x440 + i ] ^= i&0xff 
    5.  
    6. open('re_patch.exe', 'wb').write(bytes(a1))

    然后用ida打开patch后的程序可以看到加密代码

    1. DWORD __stdcall StartAddress(LPVOID lpThreadParameter)
    2. {
    3.   int k; // ebx
    4.   int v4[65534]; // [esp+Ch] [ebp-50530h]
    5.   char v5[66816]; // [esp+40004h] [ebp-10538h] BYREF
    6.   int v6[6]; // [esp+50504h] [ebp-38h]
    7.   char *v7; // [esp+5051Ch] [ebp-20h]
    8.   int v8; // [esp+50520h] [ebp-1Ch]
    9.   int v9; // [esp+50524h] [ebp-18h]
    10.   int v10; // [esp+50528h] [ebp-14h]
    11.   int v11; // [esp+5052Ch] [ebp-10h]
    12.   int i; // [esp+50530h] [ebp-Ch]
    13.   int v13; // [esp+50534h] [ebp-8h]
    14.   int j; // [esp+50538h] [ebp-4h]
    15.  
    16.   while ( !dword_41B2E4 )
    17.     Sleep(0x3E8u);
    18.   for ( i = 0; i < 112; ++i )
    19.     byte_41B30A += *((_BYTE *)&loc_401450 + i);
    20.   for ( i = 0; byte_41B2EC[i]; ++i )
    21.     byte_41B2EC[i] ^= byte_41B30A;
    22.   v6[0] = 81;
    23.   v6[1] = 102;
    24.   v6[2] = 114;
    25.   v6[3] = 111;
    26.   v6[4] = 115;
    27.   v6[5] = 116;
    28.   for ( j = 0; j < 256; ++j )
    29.     *(_DWORD *)&v5[4 * j + 0x10100] = j;
    30.   for ( j = 0; j < 256; ++j )
    31.     v5[j + 0x10000] = v6[j % 6];                // key
    32.   v13 = 0;
    33.   for ( j = 0; j < 256; ++j )
    34.   {
    35.     v13 = (v5[j + 0x10000] + *(_DWORD *)&v5[4 * j + 65792] + v13) % 256;
    36.     v10 = *(_DWORD *)&v5[4 * j + 65792];
    37.     *(_DWORD *)&v5[4 * j + 65792] = *(_DWORD *)&v5[4 * v13 + 65792];
    38.     *(_DWORD *)&v5[4 * v13 + 65792] = v10;
    39.   }
    40.   v11 = dword_41B2E8;
    41.   v9 = 0;
    42.   v13 = 0;
    43.   j = 0;
    44.   while ( v11-- )
    45.   {
    46.     j = (j + 1) % 256;
    47.     v13 = (*(_DWORD *)&v5[4 * j + 65792] + v13) % 256;
    48.     v10 = *(_DWORD *)&v5[4 * j + 65792];
    49.     *(_DWORD *)&v5[4 * j + 65792] = *(_DWORD *)&v5[4 * v13 + 65792];
    50.     *(_DWORD *)&v5[4 * v13 + 65792] = v10;
    51.     v8 = (*(_DWORD *)&v5[4 * v13 + 65792] + *(_DWORD *)&v5[4 * j + 65792]) % 256;
    52.     v4[v9++] = *(_DWORD *)&v5[4 * v8 + 65792];
    53.   }
    54.   for ( j = 0; j < dword_41B2E8; ++j )
    55.     v5[j] = LOBYTE(v4[j]) ^ byte_41B2EC[j];
    56.   v7 = v5;
    57.   for ( k = 0; k < dword_41B2E8; ++k )
    58.   {
    59.     if ( v7[k] != *(_BYTE *)(k + 0x401307) )
    60.     {
    61.       puts(0x401345);
    62.       _loaddll(0);
    63.       break;
    64.     }
    65.   }
    66.   puts((int)"right\n");
    67.   j___fgetchar();
    68.   return 0;
    69. }

    感觉好像叫RC4还是啥来着，先生成加密流，再与明文异或。反正对应写程序就行了

    1. a1 = open('re_patch.exe', 'rb').read()[0x707: 0x707+ 0x100]
    2.  
    3. v6 = b'Qfrost'
    4. v51 = [i for i in range(256)]
    5. key = [v6[j%6] for j in range(256)]
    6. v13 = 0
    7. for j in range(256):
    8.     v13 = (key[j] + v51[j] + v13) % 256
    9.     v51[j], v51[v13] = v51[v13], v51[j]
    10.  
    11. flen = 50
    12. v13 = 0
    13. v4 = [0]*flen 
    14. j = 0
    15. for i in range(flen):
    16.     j = (j+1)%256
    17.     v13 = (v51[j] + v13)%256
    18.     v51[j], v51[v13] = v51[v13], v51[j]
    19.     v8 = (v51[j] + v51[v13])%256
    20.     v4[i] = v51[v8]^a1[i]
    21.  
    22. for i in range(256):
    23.     tmp = bytes([i^v for v in v4])
    24.     #if b'flag{' in tmp:
    25.     print(tmp)
    26.  
    27. #QDbg-Is_an_1ntereSting-Game!

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  • 原文地址：https://blog.csdn.net/weixin_52640415/article/details/126119424