[BUUOJ] reverse3

[潘一辰]

32位IDA打开，F5一下main函数

 __int64 __cdecl main_0()
 {
  size_t v0; // eax
  const char *v1; // eax
  size_t v2; // eax
  int v3; // edx
  __int64 v4; // ST08_8
  signed int j; // [esp+DCh] [ebp-ACh]
  signed int i; // [esp+E8h] [ebp-A0h]
  signed int v8; // [esp+E8h] [ebp-A0h]
  char Dest[108]; // [esp+F4h] [ebp-94h]
  char Str; // [esp+160h] [ebp-28h]
  char v11; // [esp+17Ch] [ebp-Ch]

  for ( i = 0; i < 100; ++i )
  {
    if ( (unsigned int)i >= 0x64 )
      j____report_rangecheckfailure();
    Dest[i] = 0;
  }
  sub_41132F("please enter the flag:");
  sub_411375("%20s", &Str);
  v0 = j_strlen(&Str);
  v1 = (const char *)sub_4110BE(&Str, v0, &v11);
  strncpy(Dest, v1, 0x28u);
  v8 = j_strlen(Dest);
  for ( j = 0; j < v8; ++j )
    Dest[j] += j;
  v2 = j_strlen(Dest);
  if ( !strncmp(Dest, Str2, v2) )
    sub_41132F("rigth flag!\n");
  else
    sub_41132F("wrong flag!\n");
  HIDWORD(v4) = v3;
  LODWORD(v4) = 0;
  return v4;
}

可以看出输入的flag经过sub_4110BE函数的处理，再经过for循环最后与Str2比较。
打开Str2

.data:0041A034 Str2            db '[email protected]@dH',0 ; DATA XREF: _main_0+142↑o

把Str2反向经历for循环是容易的
再看sub_4110BE函数

void *__cdecl sub_411AB0(char *a1, unsigned int a2, int *a3)
{
  int v4; // STE0_4
  int v5; // STE0_4
  int v6; // STE0_4
  int v7; // [esp+D4h] [ebp-38h]
  signed int i; // [esp+E0h] [ebp-2Ch]
  unsigned int v9; // [esp+ECh] [ebp-20h]
  int v10; // [esp+ECh] [ebp-20h]
  signed int v11; // [esp+ECh] [ebp-20h]
  void *Dst; // [esp+F8h] [ebp-14h]
  char *v13; // [esp+104h] [ebp-8h]

  if ( !a1 || !a2 )
    return 0;
  v9 = a2 / 3;
  if ( (signed int)(a2 / 3) % 3 )
    ++v9;
  v10 = 4 * v9;
  *a3 = v10;
  Dst = malloc(v10 + 1);
  if ( !Dst )
    return 0;
  j_memset(Dst, 0, v10 + 1);
  v13 = a1;
  v11 = a2;
  v7 = 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 *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
        v4 = v7 + 1;
        *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
        *((_BYTE *)Dst + v4++) = aAbcdefghijklmn[64];
        *((_BYTE *)Dst + v4) = aAbcdefghijklmn[64];
        v7 = v4 + 1;
        break;
      case 2:
        *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
        v5 = v7 + 1;
        *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
        *((_BYTE *)Dst + v5++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)];
        *((_BYTE *)Dst + v5) = aAbcdefghijklmn[64];
        v7 = v5 + 1;
        break;
      case 3:
        *((_BYTE *)Dst + v7) = aAbcdefghijklmn[(signed int)(unsigned __int8)byte_41A144[0] >> 2];
        v6 = v7 + 1;
        *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[1] & 0xF0) >> 4) | 16 * (byte_41A144[0] & 3)];
        *((_BYTE *)Dst + v6++) = aAbcdefghijklmn[((byte_41A144[2] & 0xC0) >> 6) | 4 * (byte_41A144[1] & 0xF)];
        *((_BYTE *)Dst + v6) = aAbcdefghijklmn[byte_41A144[2] & 0x3F];
        v7 = v6 + 1;
        break;
    }
  }
  *((_BYTE *)Dst + v7) = 0;
  return Dst;
}

很明显看到输入的字符串经过了数组aAbcdefghijklmn[]的变换，打开这个数组

.rdata:00417B30 aAbcdefghijklmn db 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='

从'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/='可以看出，这个函数应该是base64的加密函数，因此只需要解密即可。

脚本

import base64

a= '[email protected]@dH'
b = list(a)
flag = ''

for i in range(0,16):
    b[i] = chr(ord(b[i])-i)

c = ''.join(b)

flag = base64.b64decode(c)
flag = flag.decode('ASCII')
print(flag)

flag

{i_l0ve_you}