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[原创]签到题分析
发表于: 2021-11-16 01:18 1771

[原创]签到题分析

2021-11-16 01:18
1771

没壳直接上IDA,因为程序是C++写的直接看DialogFunc然后F5
sub_401340(hWnd);为按钮执行函数。 以下直接在他函数上备注了。

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int __cdecl sub_401340(HWND hDlg)
{
  int v1; // ebx
  int v2; // ebx
  int v3; // ecx
  int v4; // eax
  int result; // eax
  signed int v6; // [esp+Ch] [ebp-260h]
  char v7; // [esp+10h] [ebp-25Ch]
  CHAR s1; // [esp+D8h] [ebp-194h]
  CHAR s2; // [esp+1A0h] [ebp-CCh]
 
  memset(&s1, 0, 0xC8u);
  memset(&v7, 0, 0xC8u);
  memset(&s2, 0, 0xC8u);
  v1 = GetDlgItemTextA(hDlg, 1000, &s1, 201); //获取Name
  if ( v1 //判断Name是否输入
    && (v6 = GetDlgItemTextA(hDlg, 1001, &s2, 201), v2 = sub_401260(&s1, v1), strspn(&s2, "0123456789") == strlen(&s2)) //判断SN是否输入以及判断SN是否为纯数字 sub_401260 下面贴上
    && v6 <= 10 //判断SN位数
    && (v4 = sub_40307F(v3, (int)&s2)) != 0 //sub_40307F下面标注
    && (unknown_libname_13(v2 ^ v4, &v7, 16), sub_401260(&v7, 8) == 330861687) ) //unknown_libname_13下面标注
  {
    SetDlgItemTextA(hDlg, 1001, "Success!");
    result = 1;
  }
  else
  {
    SetDlgItemTextA(hDlg, 1001, "Wrong Serial!");
    result = 0;
  }
  return result;
}
int __cdecl sub_401260(char *a1, int a2) //明显计算CRC32。。。
{
  signed int v2; // ecx
  unsigned int v3; // eax
  unsigned int v4; // eax
  unsigned int v5; // eax
  unsigned int v6; // eax
  unsigned int v7; // eax
  unsigned int v8; // eax
  unsigned int v9; // eax
  unsigned int v10; // eax
  int v11; // edx
  unsigned int v12; // ecx
  char *v13; // esi
  char v14; // al
  int v16[256]; // [esp+0h] [ebp-404h]
 
  v2 = 0;
  do
  {
    v3 = (unsigned int)v2 >> 1;
    if ( v2 & 1 )
      v3 ^= 0xEDB88320;
    if ( v3 & 1 )
      v4 = (v3 >> 1) ^ 0xEDB88320;
    else
      v4 = v3 >> 1;
    if ( v4 & 1 )
      v5 = (v4 >> 1) ^ 0xEDB88320;
    else
      v5 = v4 >> 1;
    if ( v5 & 1 )
      v6 = (v5 >> 1) ^ 0xEDB88320;
    else
      v6 = v5 >> 1;
    if ( v6 & 1 )
      v7 = (v6 >> 1) ^ 0xEDB88320;
    else
      v7 = v6 >> 1;
    if ( v7 & 1 )
      v8 = (v7 >> 1) ^ 0xEDB88320;
    else
      v8 = v7 >> 1;
    if ( v8 & 1 )
      v9 = (v8 >> 1) ^ 0xEDB88320;
    else
      v9 = v8 >> 1;
    if ( v9 & 1 )
      v10 = (v9 >> 1) ^ 0xEDB88320;
    else
      v10 = v9 >> 1;
    v16[v2++] = v10;
  }
  while ( v2 < 256 );
  v11 = a2;
  v12 = -1;
  if ( a2 )
  {
    v13 = a1;
    do
    {
      v14 = *v13++;
      v12 = v16[(unsigned __int8)(v12 ^ v14)] ^ (v12 >> 8);
      --v11;
    }
    while ( v11 );
  }
  return ~v12;
}
// Microsoft VisualC universal runtime
int __cdecl unknown_libname_13(int a1, int a2, int a3)
{
  // 虽然未识别,,,但是调用common_xtox肯定是itoa了
  char v4; // [esp+0h] [ebp-4h]
 
  if ( a3 != 10 || (v4 = 1, a1 >= 0) )
    v4 = 0;
  common_xtox<unsigned long,char>(a1, a2, -1, a3, v4);
  return a2;
}
int __cdecl sub_40307F(int a1)
{
  //字符串转为数字不备注了
  int v1; // ecx
  int v3; // [esp-14h] [ebp-14h]
  int v4; // [esp-10h] [ebp-10h]
  int v5; // [esp-Ch] [ebp-Ch]
  signed int v6; // [esp-8h] [ebp-8h]
  int v7; // [esp-4h] [ebp-4h]
 
  v7 = v1;
  v6 = 1;
  v5 = 10;
  v4 = v1;
  v3 = v1;
  unknown_libname_6(&v3, a1, 0);
  return __crt_strtox::parse_integer<unsigned long,__crt_strtox::c_string_character_source<char>>(0, v3, v4, v5, v6);
}

整个过程代码都有了,下面简单白话下他的过程
1.取Name计算CRC32
2.SN为纯数字小于等于10
3.将SN转为数字
4.将计算的crc32 ^ SN 的CRC32等于330861687即为验证成功.

 

下面为SN生成器
动态分析1386343770(52a1ed5a)的CRC32为330861687,所以直接用户名的crc32 ^ 1386343770 即可得出SN。

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uint32_t crc32(void const* data, int n) {
    uint32_t r = 0xFFFFFFFF;
 
    for (int i = 0; i < n; ++i) {
        r ^= ((char const*)data)[i];
 
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
        if ((r & 0x1) != 0) { r = (r >> 1) ^ 0xEDB88320; }
        else { r = r >> 1; }
    }
    return r ^ 0xFFFFFFFF;
}
uint32_t c32 = crc32((char*)"KCTF", strlen((char*)"KCTF"));
printf("SN:%d\n", c32 ^ 1386343770);
输出结果:SN:205824534

文章编写能力差。。。语句可能有些不通,望理解。。


[招生]科锐逆向工程师培训(2024年11月15日实地,远程教学同时开班, 第51期)

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