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[原创]2021KCTF秋季赛 迷失丛林
2021-11-30 14:04 12352

[原创]2021KCTF秋季赛 迷失丛林

2021-11-30 14:04
12352
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do
{
  byte_404220[0] = *((_BYTE *)&dword_404000 + v2 - 1);// 取第一个hex
  byte_404221 = v2;
  v3 = byte_404220;
  v4 = &dword_404100;
  v5 = &byte_404220[dword_404100];
  do
  {
    v6 = *v4;
    if ( *v4 > 0 )
    {
      do
      {
        v7 = v5 + 1;
        *(v7 - 1) = *((_BYTE *)&dword_404000 + (unsigned __int8)*v3);
        *v7 = *v3 + 1;
        v5 = v7 + 1;
        ++v3;
        --v6;
      }
      while ( v6 );
    }
    ++v4;
  }
  while ( (int)v4 < (int)&unk_40411C );
  v8 = 256;
  do
  {
    ++v25[(unsigned __int8)*v3++];
    --v8;
  }
  while ( v8 );
  ++v2;
  v25 += 256;
}
while ( v2 - 1 < 256 );
v9 = &unk_404448;
v10 = 256;
do
{
  if ( *(v9 - 40) )
    ++v21;
  if ( *(v9 - 26) )
    ++v22;
  if ( *v9 )
    ++v23;
  if ( v9[39] )
    ++v24;
  v9 += 256;
  --v10;
}
while ( v10 );
if ( v21 == (char)0xA9 && v22 == (char)0xAC && v23 == (char)0xA7 && v24 > 0xC8u )

这里是验证第一段序列号的逻辑

 

可以发现其实前8个字节被放进了dword_404000里,而且还测试了这个其实是个s盒,没有重复的数字。

 

所以可以确定前个8个字节是什么,但是并不知道它们之间的顺序是什么。

 

0x1e,0x28,0x4b,0x6d,0x8c,0xa3,0xd2,0xfb

 

确定顺序之后其实就可以爆破了,大概10分钟可以爆出来。

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import sys
a=[ 0xA2, 0x9B,
  0xF4, 0xDF, 0xAC, 0x7C, 0xA1, 0xC6, 0x16, 0xD0, 0x0F, 0xDD,
  0xDC, 0x73, 0xC5, 0x6B, 0xD1, 0x96, 0x47, 0xC2, 0x26, 0x67,
  0x4E, 0x41, 0x82, 0x20, 0x56, 0x9A, 0x6E, 0x33, 0x92, 0x88,
  0x29, 0xB5, 0xB4, 0x71, 0xA9, 0xCE, 0xC3, 0x34, 0x50, 0x59,
  0xBF, 0x2D, 0x57, 0x22, 0xA6, 0x30, 0x04, 0xB2, 0xCD, 0x36,
  0xD5, 0x68, 0x4D, 0x5B, 0x45, 0x9E, 0x85, 0xCF, 0x9D, 0xCC,
  0x61, 0x78, 0x32, 0x76, 0x31, 0xE3, 0x80, 0xAD, 0x39, 0x4F,
  0xFA, 0x72, 0x83, 0x4C, 0x86, 0x60, 0xB7, 0xD7, 0x63, 0x0C,
  0x44, 0x35, 0xB3, 0x7B, 0x19, 0xD4, 0x69, 0x08, 0x0B, 0x1F,
  0x3D, 0x11, 0x79, 0xD3, 0xEE, 0x93, 0x42, 0xDE, 0x23, 0x3B,
  0x5D, 0x8D, 0xA5, 0x77, 0x5F, 0x58, 0xDB, 0x97, 0xF6, 0x7A,
  0x18, 0x52, 0x15, 0x74, 0x25, 0x62, 0x2C, 0x05, 0xE8, 0x0D,
  0x98, 0x2A, 0x43, 0xE2, 0xEF, 0x48, 0x87, 0x49, 0x1C, 0xCA,
  0x2B, 0xA7, 0x8A, 0x09, 0x81, 0xE7, 0x53, 0xAA, 0xFF, 0x6F,
  0x8E, 0x91, 0xF1, 0xF0, 0xA4, 0x46, 0x3A, 0x7D, 0x54, 0xEB,
  0x2F, 0xC1, 0xC0, 0x0E, 0xBD, 0xE1, 0x6C, 0x64, 0xBE, 0xE4,
  0x02, 0x3C, 0x5A, 0xA8, 0x9F, 0x37, 0xAF, 0xA0, 0x13, 0xED,
  0x1B, 0xEC, 0x8B, 0x3E, 0x7E, 0x27, 0x99, 0x75, 0xAB, 0xFE,
  0xD9, 0x3F, 0xF3, 0xEA, 0x70, 0xF7, 0x95, 0xBA, 0x1D, 0x40,
  0xB0, 0xF9, 0xE5, 0xF8, 0x06, 0xBC, 0xB6, 0x03, 0xC9, 0x10,
  0x9C, 0x2E, 0x89, 0x5C, 0x7F, 0xB1, 0x1A, 0xD6, 0x90, 0xAE,
  0xDA, 0xE6, 0x5E, 0xB9, 0x84, 0xE9, 0x55, 0xBB, 0xC7, 0x0A,
  0xE0, 0x66, 0xF2, 0xD8, 0xCB, 0x00, 0x12, 0xB8, 0x17, 0x94,
  0x6A, 0x4A, 0x01, 0x24, 0x14, 0x51, 0x07, 0x65, 0x21, 0xC8,
  0x38, 0xFD, 0x8F, 0xC4, 0xF5, 0xFC]
 
 
def check(d):
    for i in d:
        if d.count(i)>1:
            return 1
    return 0
burst_num =[0x1e,0x28,0x4b,0x6d,0x8c,0xa3,0xd2,0xfb]
def enc_fun(d):
    # print(hex(d[0]),hex(d[1]))
    e=[0 for i in range(256*256)]
    f=[0 for i in range(512)]
    v2 = 1
    m = 0
    for i in range(256):
        f[0] = d[v2-1]
        f[1] = v2&0xff
        l = 0
 
        for k in range(254):
            f[2+k*2]   = d[f[l]]
            f[2+k*2+1] = (f[l] +1)&0xff
            l+=1
        for j in range(256):
            e[m+f[l]] +=1
            l+=1
        v2+=1
        m += 256
    # print(e)
    n=0
    v21=0
    v22=0
    v23=0
    v24=0
    for i in range(256):
        if e[n]:
            v21+=1
        if e[n+14]:
            v22+=1
        if e[n+40]:
            v23+=1
        if e[n+79]:
            v24+=1
        n+=256
    if v24>0xc8:
        print(hex(v21),hex(v22),hex(v23),hex(v24))
    if v21 ==0xA9 and v22 ==0xAC and v23 ==0xA7 and v24 >0xc8:
        return 1
    return 0
 
def brust():
    flag = 0
    for i1 in burst_num:
        for i2 in burst_num:
            for i3 in burst_num:
                for i4 in burst_num:
                    for i5 in burst_num:
                        for i6 in burst_num:
                            for i7 in burst_num:
                                for i8 in burst_num:
                                    = [i1,i2,i3,i4,i5,i6,i7,i8]
                                    if check(d):
                                        continue
                                    += a
                                    if enc_fun(d):
                                        print(i1,i2,i3,i4,i5,i6,i7,i8)
                                        sys.exit()
if __name__ == '__main__':
    brust()
#0x4b,0x6d,0x28,0x8c,0xfb,0xd2,0x1e,0xa3

B4D682C8BF2DE13A 前16个序列号

 

后8个字节序列号的逻辑代码

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if ( v21 == (char)0xA9 && v22 == (char)0xAC && v23 == (char)0xA7 && v24 > 0xC8u )
  {
    v11 = 0;
    while ( 2 )
    {
      v12 = 0;
      while ( *((unsigned __int8 *)&dword_404000 + v12) != v11 )
      {
        if ( ++v12 >= 256 )
          goto LABEL_32;
      }
      v13 = *((_BYTE *)&dword_404000 + v11);
      v14 = 0;
      v15 = v11;
      if ( v11 != v13 )
      {
        do
        {
          ++v14;
          v16 = *((unsigned __int8 *)&dword_404000 + v15);
          *((_BYTE *)&dword_404000 + v15) = *((_BYTE *)&dword_404000 + v16);
          v15 = v16;
          if ( v14 >= 256 )
            return 0;
        }
        while ( v11 != *((unsigned __int8 *)&dword_404000 + v16) );
      }
      *((_BYTE *)&dword_404000 + v15) = v13;
LABEL_32:
      if ( ++v11 < 256 )
        continue;
      break;
    }
    v17 = 0;
    Str1 = dword_404000;
    dword_414424 = dword_404004;
    do
    {
      for ( i = 0; i < 8; ++i )
      {
        if ( v17 >= 8 )
        {
          if ( !i || i == 7 )
            --*((_BYTE *)&Str1 + i);
        }
        else
        {
          if ( (*(_BYTE *)(a2 + i) & 1) != 0 )
            v19 = *((_BYTE *)&Str1 + i) + 1;
          else
            v19 = *((_BYTE *)&dword_404000 + *((unsigned __int8 *)&Str1 + i));
          *((_BYTE *)&Str1 + i) = v19;
          *(_BYTE *)(a2 + i) >>= 1;
        }
      }
      ++v17;
    }
    while ( v17 < 9 );
    if ( !strncmp(&Str1, Str2, 8) )
    {
      MessageBoxA(this[1], Str2, Caption, 0x40u);// 输出正确
      return 1;
    }
  }

最后要等于“GoodJob~”

 

得到前8个序列号后我们就可以进入这里面了,然后可以dump出dword_40400的数据,这里是对后8个字节进行加密。

 

这里也是采用爆破,这里可以单字节爆破,我是一个个字节来爆破的。

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    data=[0xC1, 0x9B, 0x7F, 0x58, 0x64, 0xD5, 0x77, 0x21, 0x74, 0xEB,
  0x14, 0xBF, 0xDF, 0x25, 0x5A, 0x37, 0x85, 0x2C, 0xAF, 0x8C,
  0xDA, 0x26, 0xE2, 0x7A, 0x87, 0x4C, 0x60, 0x99, 0x54, 0x3C,
  0x95, 0xC0, 0xB9, 0x0C, 0xBC, 0x0E, 0xE7, 0x2D, 0x86, 0xBE,
  0x67, 0xD3, 0xD8, 0xFC, 0x30, 0xB6, 0xC8, 0x57, 0x1E, 0x62,
  0x3E, 0xCE, 0xA0, 0xCD, 0xF5, 0xEE, 0xA7, 0xCF, 0x45, 0xFE,
  0xD0, 0x80, 0x05, 0xAD, 0x13, 0xF3, 0xB7, 0x6B, 0x22, 0x2B,
  0xBD, 0x69, 0x42, 0x4B, 0xA5, 0xEA, 0xA6, 0xD2, 0x6F, 0x4F,
  0x4E, 0x07, 0xE1, 0x36, 0x01, 0xB5, 0xAA, 0xB1, 0x94, 0x0B,
  0x35, 0x3A, 0xC7, 0x49, 0x53, 0x82, 0xC3, 0x7B, 0x32, 0xFF,
  0x19, 0xC4, 0xF1, 0xC9, 0xE8, 0xF7, 0x56, 0x15, 0xA3, 0x46,
  0x89, 0x43, 0x9D, 0x8F, 0x20, 0xEF, 0xBB, 0x2A, 0xCB, 0x09,
  0x93, 0x4A, 0x1C, 0xE3, 0x33, 0xD1, 0xE0, 0x1D, 0x72, 0x7C,
  0x27, 0xE9, 0x17, 0x28, 0x6D, 0x6A, 0xD9, 0x00, 0x9A, 0xE5,
  0x63, 0xDE, 0x23, 0x9F, 0x0D, 0x47, 0x3B, 0x65, 0x08, 0x84,
  0x6C, 0x1A, 0x88, 0x12, 0xA1, 0xA4, 0xB3, 0x18, 0x24, 0x1B,
  0xD7, 0x44, 0xDB, 0xAC, 0x6E, 0x7D, 0x51, 0x5E, 0xED, 0x50,
  0xD6, 0x11, 0x5B, 0x9C, 0xB4, 0x68, 0x3D, 0x2F, 0x03, 0x40,
  0xBA, 0x2E, 0xCA, 0x02, 0xE6, 0xA8, 0xEC, 0x83, 0x06, 0x5D,
  0xB8, 0x4D, 0x97, 0x66, 0xF0, 0xFB, 0x8A, 0x55, 0xAB, 0xB2,
  0x04, 0xFA, 0x0A, 0x31, 0x71, 0xCC, 0x8B, 0x73, 0xA9, 0x48,
  0x5C, 0xF9, 0x98, 0xE4, 0xC6, 0x34, 0xC5, 0x7E, 0x81, 0x75,
  0x90, 0x1F, 0x92, 0x3F, 0x9E, 0x10, 0x29, 0x52, 0x39, 0xF4,
  0x41, 0x78, 0x5F, 0x16, 0x79, 0xC2, 0xB0, 0xDD, 0xF2, 0x61,
  0x0F, 0x70, 0xD4, 0x91, 0xDC, 0xF6, 0xF8, 0xFD, 0x59, 0x38,
  0x8D, 0x96, 0xAE, 0x8E, 0x76, 0xA2]
    # print(chr(0x58))
    print("start:")
    for i in range(0xff):
        c=[0xc1,0x9b,0x7f,0x58,0x64,0xd5,0x77,0x21]
        tmp = i
        for j in range(9):
            if j>=8:
                c[0]-=1
                c[7]-=1
            if j<8:
                if tmp&1 !=0:
                    v19 = c[7]+1
                else:
                    v19 = data[c[7]]
                c[7] = v19&0xff
                tmp>>=1
 
        if c[7] == ord('~'):
            print(hex(i))
#0x9D 0x6b 0xea 0x2f 0xa4 0x8 0xbc 0x22

得到序列号:B4D682C8BF2DE13AD9B6AEF24A80CB22


【看雪培训】目录重大更新!《安卓高级研修班》2022年春季班开始招生!

最后于 2021-11-30 16:47 被n3ko编辑 ,原因:
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