[原创]2021MTCTF-Inject

发表于: 2021-5-28 00:24 6496

[原创]2021MTCTF-Inject

SYJ-Re

2021-5-28 00:24

6496

Inject

这道题，我是真的无语，那有个点，卡了好久

没想到是猜出来的？？？？？

详情请往下看

分析过程

1. check_key1_key2函数分析

拿到题目之后，解压得到一个Inject.exe和notepad2.exe

直接将Inject.exe拖入工具进行查壳，然后发现无壳直接拖入IDA进行分析

int check_key1_key2()
{
  unsigned __int64 v0; // r11
  double v1; // xmm0_8
  float chushu; // xmm0_4 float占4个字节
  unsigned __int64 data; // r8
  __int64 count; // rcx
  unsigned __int64 v5; // r10
  __int64 count1; // rcx
  unsigned __int64 data1; // r9
  unsigned __int64 yushu; // rdx
  unsigned __int64 key2; // [rsp+20h] [rbp-18h] BYREF
 
  printf("Please input the key1 and key2:");
  scanf("%x %lld", &key1, &key2);
  v0 = key2;
  if ( (key2 & 0x8000000000000000ui64) != 0i64 )
    v1 = (double)(int)(key2 & 1 | (key2 >> 1)) + (double)(int)(key2 & 1 | (key2 >> 1));
  else
    v1 = (double)(int)key2;
  chushu = (double)(int)key1 / v1;              // 除法: key1 / key2
  HHHHHHHHHH = LODWORD(chushu);                 // 赋值给一个变量(这个变量之后会用来解密yyy文件的代码)
  if ( LODWORD(chushu) >= 0x75D05803ui64 )      // 必须<0x75D05803
    goto LABEL_14;
  data = 1i64;
  count = 8i64;
  v5 = LODWORD(chushu) % 0x75D05803ui64;        // v5 = (key1 / key2) % 0x75D05803
  do
  {
    data = v5 * data % 0x75D05803;              // 循环乘法,不就是次方吗,比赛脑子秀逗了?!
    --count;
  }
  while ( count );                              // data = pow(v5, 8) % 0x75D05803
  count1 = 2i64;
  data1 = 1i64;
  do
  {
    data1 = v5 * data1 % 0x75D05803;
    --count1;
  }
  while ( count1 );                             // data1 = pow(v5, 2) % 0x75D05803
  if ( (2 * (v5 % 0x75D05803) - data1 - 42 + data) % 0x75D05803 != 1 )// 关键的验证式子
    goto LABEL_14;
  yushu = key1 % key2;
  if ( yushu )
    v0 = sub_7FF6C0891280(key2, yushu);
  if ( v0 != 1 )
LABEL_14:
    exit(1);
  printf("You are right, let's to the next step.\n");
  return system("pause");
}

然后，稀里糊涂的就开始了爆破之旅（错误步骤啊，稍微看一下即可，也不是说错误，就只能爆出来一半，剩下一半爆不出来）

int main()
{
        unsigned int i = 0;
        for(i=0;i<0xFFFFFFFF;i++)
    {
        unsigned int data = 1;   // = (int)pow(i, 8) % 0x75D05803;
        //unsigned int data1;  // =(int)pow(i, 2) % 0x75D05803;
        int count = 8;
        do
        {
            data = i * data % 0x75D05803;              // 循环乘法,不就是次方吗,比赛脑子秀逗了?!
            --count;
        }while ( count );                              // data = pow(v5, 8) % 0x75D05803
 
        int count1 = 2;
        unsigned int data1 = 1;
        do
        {
            data1 = i * data1 % 0x75D05803;
            --count1;
        }while ( count1 );                             // data1 = pow(v5, 2) % 0x75D05803
 
        if((2 * (i % 0x75D05803) - data1 - 42 + data) % 0x75D05803 == 1)
        {
            printf("%x\n",i);
        }
    }
}

爆破出来两个满足的：

1 2	`37e3e317` `adb43b1a`

然后重新动调将内存中那个HHHHHHHHHH变量和变量v5都设为37e3e317

然后验证只过了一个

最后的这里
图片描述

就会exit

2. create_file函数分析

图片描述

就是一些创建文件等操作：

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_BYTE *create_file()
{
  size_t v0; // r8
  void **ptr; // rbx
  .....
  memset(Filename, 0, 0x104ui64);
  GetModuleFileNameA(0i64, Filename, 0x104u);   // 第一个参数为NULL, 该函数返回该应用程序全路径
  v37 = 0i64;
  v38 = 15i64;
  LOBYTE(exename[0]) = 0;
  v0 = -1i64;
  do
    ++v0;
  while ( Filename[v0] );
  strncpy(exename, Filename, v0);
  ptr = exename;
  if ( v38 >= 16 )
    ptr = (void **)exename[0];
  if ( v37 )
  {
    v39[0] = 0i64;
    v39[1] = 0i64;
    v39[2] = 0i64;
    v39[3] = 0i64;
    v39[4] = 0i64;
    v40 = 0i64;
    v41 = 0;
    memset(v43, 0, sizeof(v43));
    v42 = 1;
    v2 = (void **)((char *)ptr + v37 - 1);
    if ( *((_BYTE *)v39 + *(unsigned __int8 *)v2) )
    {
LABEL_9:
      v3 = (_DWORD)v2 - (_DWORD)ptr;
      goto LABEL_11;
    }
    while ( v2 != ptr )
    {
      v2 = (void **)((char *)v2 - 1);
      if ( *((_BYTE *)v39 + *(unsigned __int8 *)v2) )
        goto LABEL_9;
    }
  }
  v3 = -1;
LABEL_11:
  Size[0] = 0i64;
  Size[1] = 15i64;
  LOBYTE(Block[0]) = 0;
  v4 = v3;
  v5 = v3;
  if ( v37 < v3 )
    v5 = v37;
  v6 = exename;                                 // E:\SYJ\COMPETITIONS\2021美团\inject
  if ( v38 >= 16 )
    v6 = (void **)exename[0];
  strncpy(Block, v6, v5);
  v7 = strcat_0(Block, "\\flag.txt", 9ui64);    // 拼接字符串
                                                // v7 = E:\SYJ\COMPETITIONS\2021美团\inject\Inject.exe\flag.txt
  txt_path = 0ui64;
  *(_OWORD *)FileName = *v7;
  txt_path = v7[1];
  *((_QWORD *)v7 + 2) = 0i64;
  *((_QWORD *)v7 + 3) = 15i64;
  *(_BYTE *)v7 = 0;
  if ( Size[1] >= 0x10 )
  {
    v8 = Block[0];
    if ( Size[1] + 1 >= 0x1000 )
    {
      v8 = (void *)*((_QWORD *)Block[0] - 1);
      if ( (unsigned __int64)(Block[0] - v8 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v8);
  }
  Size[0] = 0i64;
  Size[1] = 15i64;
  LOBYTE(Block[0]) = 0;
  Src[2] = 0i64;
  v31 = 15i64;
  LOBYTE(Src[0]) = 0;
  if ( v37 < v4 )
    v4 = v37;
  v9 = exename;
  if ( v38 >= 0x10 )
    v9 = (void **)exename[0];
  strncpy(Src, v9, v4);
  notepad_path = strcat_0(Src, "\\notepad2.exe ", 0xEui64);// 拼接字符串
                                                // E:\SYJ\COMPETITIONS\2021美团\inject\notepad2.exe
  v29 = 0ui64;
  *(_OWORD *)v28 = *notepad_path;
  v29 = notepad_path[1];
  *((_QWORD *)notepad_path + 2) = 0i64;
  *((_QWORD *)notepad_path + 3) = 15i64;
  *(_BYTE *)notepad_path = 0;
  v11 = FileName;
  if ( *((_QWORD *)&txt_path + 1) >= 16ui64 )
    v11 = (char **)FileName[0];
  v12 = strcat_0(v28, v11, txt_path);
  *(_OWORD *)Block = *v12;
  *(_OWORD *)Size = v12[1];
  *((_QWORD *)v12 + 2) = 0i64;
  *((_QWORD *)v12 + 3) = 15i64;
  *(_BYTE *)v12 = 0;
  if ( *((_QWORD *)&v29 + 1) >= 0x10ui64 )
  {
    v13 = v28[0];
    if ( (unsigned __int64)(*((_QWORD *)&v29 + 1) + 1i64) >= 0x1000 )
    {
      v13 = (void *)*((_QWORD *)v28[0] - 1);
      if ( (unsigned __int64)(v28[0] - v13 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v13);
  }
  *(_QWORD *)&v29 = 0i64;
  *((_QWORD *)&v29 + 1) = 15i64;
  LOBYTE(v28[0]) = 0;
  if ( v31 >= 0x10 )
  {
    v14 = Src[0];
    if ( v31 + 1 >= 0x1000 )
    {
      v14 = (void *)*((_QWORD *)Src[0] - 1);
      if ( (unsigned __int64)(Src[0] - v14 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v14);
  }
  v15 = (const char *)FileName;
  if ( *((_QWORD *)&txt_path + 1) >= 16ui64 )
    v15 = FileName[0];
  fp = fopen(v15, "wb");
  fwrite("key3:", 1ui64, 5ui64, fp);            // E:\SYJ\COMPETITIONS\2021美团\inject\Inject.exe\flag.txt里面写入key3:这4个字符
  fclose(fp);
  v17 = Size[0];
  v18 = Size[0] + 1;
  if ( Size[0] == -1i64 )
    v18 = -1i64;
  v19 = malloc(v18);
  v20 = v19;
  v21 = Block;
  v22 = (char *)Block[0];
  v23 = Size[1];
  if ( Size[1] >= 0x10 )
    v21 = (void **)Block[0];
  memcpy(v19, v21, v17);
  v20[v17] = 0;
  if ( v23 >= 0x10 )
  {
    v24 = v22;
    if ( v23 + 1 >= 0x1000 )
    {
      v22 = (char *)*((_QWORD *)v22 - 1);
      if ( (unsigned __int64)(v24 - v22 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v22);
  }
  if ( *((_QWORD *)&txt_path + 1) >= 0x10ui64 )
  {
    v25 = FileName[0];
    if ( (unsigned __int64)(*((_QWORD *)&txt_path + 1) + 1i64) >= 0x1000 )
    {
      v25 = (char *)*((_QWORD *)FileName[0] - 1);
      if ( (unsigned __int64)(FileName[0] - v25 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v25);
  }
  *(_QWORD *)&txt_path = 0i64;
  *((_QWORD *)&txt_path + 1) = 15i64;
  LOBYTE(FileName[0]) = 0;
  if ( v38 >= 0x10 )
  {
    v26 = exename[0];
    if ( v38 + 1 >= 0x1000 )
    {
      v26 = (void *)*((_QWORD *)exename[0] - 1);
      if ( (unsigned __int64)(exename[0] - v26 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v26);
  }
  return v20;                    // 逐渐查看赋值链,得到这个返回的是notepad2.exe的路径
                                 // v20 = v19
                                 // memcpy(v19, v21, v17)
                                 // v21 = Block
                                 // *(_OWORD *)Block = *v12
                                 // v12 = ...
}

最后的返回值就是notepad2.exe的全路径

3. write_code_to_yyy函数分析

图片描述

这里面其实就是像yyy中写入代码

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_BYTE *write_code_to_yyy()
{
 ....
  memset(Filename, 0, 0x104ui64);
  GetModuleFileNameA(0i64, Filename, 0x104u);   // 获取当前程序全路径
  v39 = 0i64;
  v40 = 15i64;
  LOBYTE(exename[0]) = 0;
  v0 = -1i64;
  do
    ++v0;
  while ( Filename[v0] );
  strncpy(exename, Filename, v0);
  v1 = exename;
  if ( v40 >= 16 )
    v1 = (void **)exename[0];
  if ( v39 )
  {
    v41[0] = 0i64;
    v41[1] = 0i64;
    v41[2] = 0i64;
    v41[3] = 0i64;
    v41[4] = 0i64;
    v42 = 0i64;
    v43 = 0;
    memset(v45, 0, sizeof(v45));
    v44 = 1;
    v2 = (void **)((char *)v1 + v39 - 1);
    if ( *((_BYTE *)v41 + *(unsigned __int8 *)v2) )
    {
LABEL_9:
      v3 = (_DWORD)v2 - (_DWORD)v1;
      goto LABEL_11;
    }
    while ( v2 != v1 )
    {
      v2 = (void **)((char *)v2 - 1);
      if ( *((_BYTE *)v41 + *(unsigned __int8 *)v2) )
        goto LABEL_9;
    }
  }
  v3 = -1;
LABEL_11:
  yyy_name[2] = 0i64;
  v35 = 15i64;
  LOBYTE(yyy_name[0]) = 0;
  v4 = v3;
  if ( v39 < v3 )
    v4 = v39;
  v5 = exename;
  if ( v40 >= 0x10 )
    v5 = (void **)exename[0];
  strncpy(yyy_name, v5, v4);
  v6 = strcat_0(yyy_name, "\\yyy.a", 6ui64);    // 拼接字符串之后：
                                                // E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
  *(_OWORD *)FileName = *v6;
  *(_OWORD *)Size = v6[1];
  *((_QWORD *)v6 + 2) = 0i64;
  *((_QWORD *)v6 + 3) = 15i64;
  *(_BYTE *)v6 = 0;
  if ( v35 >= 0x10 )
  {
    v7 = yyy_name[0];
    if ( v35 + 1 >= 0x1000 )
    {
      v7 = (void *)*((_QWORD *)yyy_name[0] - 1);
      if ( (unsigned __int64)(yyy_name[0] - v7 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v7);
  }
  v8 = (const char *)FileName;
  if ( Size[1] >= 0x10 )
    v8 = FileName[0];
  fp = fopen(v8, "wb");                         // 打开文件E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
  v10 = FindResourceA(0i64, (LPCSTR)101, "code");// 查找CODE类型的资源
  v11 = SizeofResource(0i64, v10);
  v12 = LoadResource(0i64, v10);
  v13 = LockResource(v12);
  fwrite(v13, 1ui64, v11, fp);                  // 查找资源查找到之后(resource(101))直接写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
  v14 = FindResourceA(0i64, (LPCSTR)102, "code");
  v15 = SizeofResource(0i64, v14);
  v16 = LoadResource(0i64, v14);                // 装载指定资源到全局储存器
  ptr = LockResource(v16);                      // 锁定资源并得到资源在内存中的第一个字节的指针
  v18 = malloc((unsigned int)v15);
  count = 0;
  if ( (_DWORD)v15 )                            // 使用查找到的资源和之前我们在第一个函数里面key1 / key2得到的数据
  {                                             // 解密资源后存储到v18中
    v20 = v18;
    v21 = ptr - v18;                            // (resource(102)) ^ (key的8字节)
                                                // (以为key之前是8个字节的double)
    do
    {
      *v20 = v20[v21] ^ *((_BYTE *)&HHHHHHHHHH + (count++ & 7));// 异或8次进行解密
      ++v20;
    }
    while ( count < (unsigned int)v15 );
  }
  fwrite(v18, 1ui64, v15, fp);                  // 将v18中解密得到的数据,写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
  v22 = FindResourceA(0i64, (LPCSTR)103, "code");
  v23 = SizeofResource(0i64, v22);              // 看见这种FindResource函数，直接使用ResourceHacker工具可以查看
  v24 = LoadResource(0i64, v22);
  v25 = LockResource(v24);
  fwrite(v25, 1ui64, v23, fp);                  // 查找资源查找到之后又直接将(resource(103))写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
  fclose(fp);
  v26 = Size[0] + 1;
  if ( Size[0] == -1i64 )
    v26 = -1i64;
  v27 = malloc(v26);
  v28 = v27;
  v29 = FileName;
  if ( Size[1] >= 0x10 )
    v29 = (char **)FileName[0];
  v30 = Size[0];
  memcpy(v27, v29, Size[0]);
  v28[v30] = 0;
  if ( Size[1] >= 0x10 )
  {
    v31 = FileName[0];
    if ( Size[1] + 1 >= 0x1000 )
    {
      v31 = (char *)*((_QWORD *)FileName[0] - 1);
      if ( (unsigned __int64)(FileName[0] - v31 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v31);
  }
  Size[0] = 0i64;
  Size[1] = 15i64;
  LOBYTE(FileName[0]) = 0;
  if ( v40 >= 0x10 )
  {
    v32 = exename[0];
    if ( v40 + 1 >= 0x1000 )
    {
      v32 = (void *)*((_QWORD *)exename[0] - 1);
      if ( (unsigned __int64)(exename[0] - v32 - 8) > 0x1F )
        invalid_parameter_noinfo_noreturn();
    }
    j_j_free(v32);
  }
  return v28;
}

用Resource Hacker直接查看他调用FindResource这个API获取的资源

它其实关键的加密就那一段：

v8 = (const char *)FileName;
 if ( Size[1] >= 0x10 )
   v8 = FileName[0];
 fp = fopen(v8, "wb");                         // 打开文件E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
 v10 = FindResourceA(0i64, (LPCSTR)101, "code");// 查找CODE类型的资源
 v11 = SizeofResource(0i64, v10);
 v12 = LoadResource(0i64, v10);
 v13 = LockResource(v12);
 fwrite(v13, 1ui64, v11, fp);                  // 查找资源查找到之后(resource(101))直接写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
 v14 = FindResourceA(0i64, (LPCSTR)102, "code");
 v15 = SizeofResource(0i64, v14);
 v16 = LoadResource(0i64, v14);                // 装载指定资源到全局储存器
 ptr = LockResource(v16);                      // 锁定资源并得到资源在内存中的第一个字节的指针
 v18 = malloc((unsigned int)v15);
 count = 0;
 if ( (_DWORD)v15 )                            // 使用查找到的资源和之前我们在第一个函数里面key1 / key2得到的数据
 {                                             // 解密资源后存储到v18中
   v20 = v18;
   v21 = ptr - v18;                            // (resource(102)) ^ (key的8字节)
                                               // (以为key之前是8个字节的double)
   do
   {
     *v20 = v20[v21] ^ *((_BYTE *)&HHHHHHHHHH + (count++ & 7));// 异或8次进行解密
     ++v20;
   }
   while ( count < (unsigned int)v15 );
 }
 fwrite(v18, 1ui64, v15, fp);                  // 将v18中解密得到的数据,写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
 v22 = FindResourceA(0i64, (LPCSTR)103, "code");
 v23 = SizeofResource(0i64, v22);              // 看见这种FindResource函数，直接使用ResourceHacker工具可以查看
 v24 = LoadResource(0i64, v22);
 v25 = LockResource(v24);
 fwrite(v25, 1ui64, v23, fp);                  // 查找资源查找到之后又直接将(resource(103))写入E:\SYJ\COMPETITIONS\2021美团\inject\yyy.a
 fclose(fp);

先向yyy.a写入resource101，然后将resource102和我们那个HHHHHHHH变量进行异或，总共8次，每次取HHHHHHHH变量的一个字节进行异或，然后向yyy.a写入异或之后的数据，然后再将resource103资源写入yyy.a
图片描述

然后，离谱的地方就来了，"因为resource103是直接写入yyy.a的，说明可以合理的猜测resouce102和我们的那个HHH....单字节异或之后的值也是0，就说明那个HHH...变量，也就是我们key1 / key2得到的那个变量，就是resource102的最后8字节"
图片描述

17 E3 E3 37 17 E3 E3 00

4. 重新调试

将HHHHHHH变量直接内存中修改为上面得到的那个变量就可以过掉所有的检测

在write_code_to_yyy函数下方直接下个断点，F9之后可以看到文件夹中生成了yyy.a和flag.txt

这些代码就是创建远程线程将我们的yyy.a(其实就是一个dll)注入进notepad2.exe

if ( !CreateProcessA(0i64, v3, 0i64, 0i64, 0, 0, 0i64, 0i64, &StartupInfo, &ProcessInformation) )
  {
    printf("Create Fali!\n");
    exit(1);
  }
  v4 = OpenProcess(0x1FFFFFu, 0, ProcessInformation.dwProcessId);
  v5 = write_code_to_yyy();
  v6 = -1i64;
  v7 = v5;
  v8 = -1i64;
  do
    ++v8;
  while ( v5[v8] );
  v9 = VirtualAllocEx(v4, 0i64, v8 + 1, 0x1000u, 4u);
  v10 = v9;
  while ( v7[++v6] != 0 )
    ;
  WriteProcessMemory(v4, v9, v7, v6 + 1, 0i64);
  v12 = GetModuleHandleA("kernel32.dll");
  LoadLibraryA = (HMODULE (__stdcall *)(LPCSTR))GetProcAddress(v12, "LoadLibraryA");
  v14 = CreateRemoteThread(v4, 0i64, 0i64, (LPTHREAD_START_ROUTINE)LoadLibraryA, v10, 0, 0i64);
  WaitForSingleObject(v14, 0xFFFFFFFF);
  CloseHandle(v14);
  CloseHandle(v4);

5.分析生成的yyy.a

查壳发现有upx壳，利用工具FUPX脱一下就行

然后拖入IDA进行分析

创建线程钩取了WriteFile函数

DLLMain：

BOOL __stdcall DllMain(HINSTANCE hinstDLL, DWORD fdwReason, LPVOID lpvReserved)
{
  HANDLE v3; // rax
 
  if ( fdwReason == 1 )                         // 如果调用原因是DLL_PROCESS_ATTACH,进程调用DLL
  {
    v3 = CreateThread(0i64, 0i64, (LPTHREAD_START_ROUTINE)ThreadProc, 0i64, 0, 0i64);// 创建线程
    CloseHandle(v3);
  }
  return 1;
}

进入ThreadProc

（这里就钩取的关键函数）

__int64 __fastcall ThreadProc(LPVOID lpThreadParameter)
{
  HMODULE kernel32_base_addr; // rax
  BOOL (__stdcall *WriteFile)(HANDLE, LPCVOID, DWORD, LPDWORD, LPOVERLAPPED); // rax
  char *v3; // rbx
  DWORD flOldProtect; // [rsp+20h] [rbp-18h] BYREF
 
  MessageBoxA(
    0i64,
    "Now input key3 in notepad like `key3:abcdef`.\nSave it.Reopen the file, you will get the flag.\n",
    "Message",
    0);
  kernel32_base_addr = GetModuleHandleA("kernel32.dll");
  WriteFile = (BOOL (__stdcall *)(HANDLE, LPCVOID, DWORD, LPDWORD, LPOVERLAPPED))GetProcAddress(
                                                                                   kernel32_base_addr,
                                                                                   "WriteFile");
  v3 = (char *)WriteFile + *(unsigned int *)((char *)WriteFile + 2);
  fucn = *(__int64 (__fastcall **)(_QWORD, _QWORD, _QWORD, _QWORD, _QWORD))(v3 + 6);
  VirtualProtect(v3 + 6, 8ui64, 4u, &flOldProtect);
  check_key3_addr = (__int64)check_key3;
  *(_QWORD *)(v3 + 6) = check_key3;             // 赋值成我们的check_key3的函数地址
  VirtualProtect(v3 + 6, 8ui64, flOldProtect, &flOldProtect);
  return 0i64;
}

将WriteFile函数直接改成了我们的check_key3函数

而且这里还有提示，"Now input key3 in notepad like key3:abcdef.\nSave it.Reopen the file, you will get the flag.\n",

6. check_key3函数分析

__int64 __fastcall check_key3(__int64 a1, char *real_key3, __int64 a3, __int64 a4, __int64 a5)
{
  unsigned int v6; // edi
  char *flag; // rax
  int v10; // er9
  char *adr; // r11
  const char *flag_; // r10
  int i; // er8
  unsigned __int64 j; // rdx
  __int64 v15; // rbx
  char temp_char; // al
  int v17; // eax
  char v18; // cl
  __int64 v19; // r8
  const char *v20; // rdx
 
  v6 = a3;
  if ( a5 )
  {
    fucn(a1, real_key3, a3, a4, a5);
  }
  else
  {
    if ( strstr(real_key3, "key3:") == real_key3 )// 加个验证
    {
      sscanf(real_key3, "key3:%x", &save_area);
      *(float *)&data = (double)save_area * 1.818989403545856e-12 * 1.818989403545856e-12 * 0.00000002980232238769531;
      if ( check() )                            // 检测key3之后，才会运行生成flag的代码
      {
        flag = (char *)malloc(0x2Bui64);
        v10 = 5;
        adr = &byte_180003280;
        flag_ = flag;
        i = 0;
        j = 5i64;
        qmemcpy(flag, "flag{", 5);
        v15 = 0x10842000i64;
        do
        {
          if ( j <= 0x1C && _bittest64(&v15, j) )// 分隔符
          {
            temp_char = '-';
          }
          else
          {
            v17 = i % 8;
            v18 = *adr;
            ++i;
            ++adr;
            temp_char = v18 ^ *((_BYTE *)&data + v17);
          }
          flag_[j] = temp_char;
          ++v10;
          ++j;
        }
        while ( i < 32 );
        v19 = 42i64;
        v20 = flag_;
        *(_WORD *)&flag_[v10] = '}';
      }
      else
      {
        v19 = 0x17i64;
        v20 = "your key isn't correct.";
      }
      fucn(a1, v20, v19, a4, 0i64);
      exit(0);
    }
    fucn(a1, real_key3, v6, a4, 0i64);
  }
  return 1i64;
}

7. 爆破key3

bool check(unsigned input)
{
  unsigned long long v0; // r8
  unsigned long long v1; // r9
  unsigned long long v2; // rcx
  unsigned long long v3; // r11
  unsigned long long v4; // r10
  unsigned long long v5; // rcx
  unsigned long long v6; // rbx
 
  v0 = 1;
  v1 = 1;
  v2 = 4;
  v3 = input % 0x6440DB83u;
  do
  {
    v1 = v3 * v1 % 0x6440DB83;
    --v2;
  }
  while ( v2 );
  v4 = 1;
  v5 = 3;
  do
  {
    v4 = v3 * v4 % 0x6440DB83;
    --v5;
  }
  while ( v5 );
  v6 = 2;
  do
  {
    v0 = v3 * v0 % 0x6440DB83;
    --v6;
  }
  while ( v6 );
  return (2 * v0 - v4 + v1 - 32) % 0x6440DB83 == 1;
}
 
int main()
{
    for(unsigned int i=0;i<0xffffffff;i++)
    {
        if(check(i))
        {
            printf("%u\n",i);
            break;
        }
    }
    //爆破得到值为386499290
    unsigned int input_=0;
    for(unsigned int i=0;i<0xffffffff;i++)
    {
        *(float *)&input_ = (double)i
                       * 1.818989403545856e-12
                       * 1.818989403545856e-12
                       * 0.00000002980232238769531;
        if((input_)==386499290)
        {
            printf("爆破得到我们的key3为:%x",i);
            break;
        }
    }
    //爆破得到我们的key3为: 44c16d
    return 0;
}