【Reverse】每日必逆0x00

附件：https://files.buuoj.cn/files/aa4f6c7e8d5171d520b95420ee570e79/a9d22a0e-928d-4bb4-8525-e38c9481469e.rar

题解

拿到程序先进行查壳，避免做过多无用功
确认是无壳，32位可执行文件，直接将其拖入到IDAPro中
直接f5大法，分析伪代码

int __cdecl main_0(int argc, const char **argv, const char **envp)

{

  size_t v3; // eax

  const char *v4; // eax

  size_t v5; // eax

  char v7; // [esp+0h] [ebp-188h]

  char v8; // [esp+0h] [ebp-188h]

  signed int j; // [esp+DCh] [ebp-ACh]

  int i; // [esp+E8h] [ebp-A0h]

  signed int v11; // [esp+E8h] [ebp-A0h]

  char Destination[108]; // [esp+F4h] [ebp-94h] BYREF

  char Str[28]; // [esp+160h] [ebp-28h] BYREF

  char v14[8]; // [esp+17Ch] [ebp-Ch] BYREF

  for ( i = 0; i < 100; ++i ) //貌似是初始化空间

  {

    if ( (unsigned int)i >= 0x64 )

      j____report_rangecheckfailure();

    Destination[i] = 0;

  }

  sub_41132F("please enter the flag:", v7);

  sub_411375("%20s", (char)Str);

  v3 = j_strlen(Str);         //v3->Str的长度

  v4 = (const char *)sub_4110BE(Str, v3, v14);

  strncpy(Destination, v4, 0x28u);  //将v4的前0x28个字符复制到Destination

  v11 = j_strlen(Destination);  //v11-》des的长度

  for ( j = 0; j < v11; ++j )   //des每个字符+i

    Destination[j] += j;

  v5 = j_strlen(Destination);   //v5-》des的长度

  if ( !strncmp(Destination, Str2, v5) )  //就是将des和Str2进行比较

    sub_41132F("rigth flag!\n", v8);

  else

    sub_41132F("wrong flag!\n", v8);

  return 0;

}

分析文件执行流向

读取字符串-》进行sub_4110BE()加密处理=》再将加密后的字符串的每个字符按顺序加所在顺序索引=》最后再将该字符串和Str2进行比较就可以了
sub_4110BE()是一个处理函数，跟进去看

void *__cdecl sub_411AB0(char *a1, unsigned int a2, int *a3)

{

  int v4; // [esp+D4h] [ebp-38h]

  int v5; // [esp+D4h] [ebp-38h]

  int v6; // [esp+D4h] [ebp-38h]

  int v7; // [esp+D4h] [ebp-38h]

  int i; // [esp+E0h] [ebp-2Ch]

  unsigned int v9; // [esp+ECh] [ebp-20h]

  int v10; // [esp+ECh] [ebp-20h]

  int v11; // [esp+ECh] [ebp-20h]

  void *v12; // [esp+F8h] [ebp-14h]

  char *v13; // [esp+104h] [ebp-8h]

  if ( !a1 || !a2 )

    return 0;

  v9 = a2 / 3;

  if ( (int)(a2 / 3) % 3 )

    ++v9;

  v10 = 4 * v9;

  *a3 = v10;

  v12 = malloc(v10 + 1);

  if ( !v12 )

    return 0;

  j_memset(v12, 0, v10 + 1);

  v13 = a1;

  v11 = a2;

  v4 = 0;

  while ( v11 > 0 )

  {

    byte_41A144[2] = 0;

    byte_41A144[1] = 0;

    byte_41A144[0] = 0;

    for ( i = 0; i < 3 && v11 >= 1; ++i )

    {

      byte_41A144[i] = *v13;

      --v11;

      ++v13;

    }

    if ( !i )

      break;

    switch ( i )

    {

      case 1:

        *((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];

        v5 = v4 + 1;

        *((_BYTE *)v12 + v5) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];

        *((_BYTE *)v12 + ++v5) = aAbcdefghijklmn[64];

        *((_BYTE *)v12 + ++v5) = aAbcdefghijklmn[64];

        v4 = v5 + 1;

        break;

      case 2:

        *((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];

        v6 = v4 + 1;

        *((_BYTE *)v12 + v6) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];

        *((_BYTE *)v12 + ++v6) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | (4 * (byte_41A144[1] & 0xF))];

        *((_BYTE *)v12 + ++v6) = aAbcdefghijklmn[64];

        v4 = v6 + 1;

        break;

      case 3:

        *((_BYTE *)v12 + v4) = aAbcdefghijklmn[(int)(unsigned __int8)byte_41A144[0] >> 2];

        v7 = v4 + 1;

        *((_BYTE *)v12 + v7) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | (16 * (byte_41A144[0] & 3))];

        *((_BYTE *)v12 + ++v7) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | (4 * (byte_41A144[1] & 0xF))];

        *((_BYTE *)v12 + ++v7) = aAbcdefghijklmn[byte_41A144[2] & 0x3F];

        v4 = v7 + 1;

        break;

    }

  }

  *((_BYTE *)v12 + v4) = 0;

  return v12;

}

这里我也是看wp的，才知道是个base64加密，还得积累

-点击Str2，进行查看，得到一个字符串

逆向分析

将字符串按顺序减掉数组索引，再进行base64解密，即可得到flag

巴特西

【Reverse】每日必逆0x00

题解

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