BUUCTF RE [ACTF新生赛2020]usualCrypt

1.查壳 无壳
2.拖进IDA

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // esi
  int v5[3]; // [esp+8h] [ebp-74h] BYREF
  __int16 v6; // [esp+14h] [ebp-68h]
  char v7; // [esp+16h] [ebp-66h]
  char v8[100]; // [esp+18h] [ebp-64h] BYREF

  sub_403CF8(&unk_40E140);
  scanf("%s", v8);
  memset(v5, 0, sizeof(v5));
  v6 = 0;
  v7 = 0;
  sub_401080(v8, strlen(v8), v5);
  v3 = 0;
  while ( *((_BYTE *)v5 + v3) == byte_40E0E4[v3] )
  {
    if ( ++v3 > strlen((const char *)v5) )
      goto LABEL_6;
  }
  sub_403CF8(aError);
LABEL_6:
  if ( v3 - 1 == strlen(byte_40E0E4) )
    return sub_403CF8(aAreYouHappyYes);
  else
    return sub_403CF8(aAreYouHappyNo);
}

3.进入函数 sub_401080

int __cdecl sub_401080(int a1, int a2, int a3)
{
  int v3; // edi
  int v4; // esi
  int v5; // edx
  int v6; // eax
  int v7; // ecx
  int v8; // esi
  int v9; // esi
  int v10; // esi
  int v11; // esi
  _BYTE *v12; // ecx
  int v13; // esi
  int v15; // [esp+18h] [ebp+8h]

  v3 = 0;
  v4 = 0;
  sub_401000();
  v5 = a2 % 3;
  v6 = a1;
  v7 = a2 - a2 % 3;
  v15 = a2 % 3;
  if ( v7 > 0 )
  {
    do
    {
      LOBYTE(v5) = *(_BYTE *)(a1 + v3);
      v3 += 3;
      v8 = v4 + 1;
      *(_BYTE *)(v8 + a3 - 1) = byte_40E0A0[(v5 >> 2) & 0x3F];
      *(_BYTE *)(++v8 + a3 - 1) = byte_40E0A0[16 * (*(_BYTE *)(a1 + v3 - 3) & 3)
                                            + (((int)*(unsigned __int8 *)(a1 + v3 - 2) >> 4) & 0xF)];
      *(_BYTE *)(++v8 + a3 - 1) = byte_40E0A0[4 * (*(_BYTE *)(a1 + v3 - 2) & 0xF)
                                            + (((int)*(unsigned __int8 *)(a1 + v3 - 1) >> 6) & 3)];
      v5 = *(_BYTE *)(a1 + v3 - 1) & 0x3F;
      v4 = v8 + 1;
      *(_BYTE *)(v4 + a3 - 1) = byte_40E0A0[v5];
    }
    while ( v3 < v7 );
    v5 = v15;
  }
  if ( v5 == 1 )
  {
    LOBYTE(v7) = *(_BYTE *)(v3 + a1);
    v9 = v4 + 1;
    *(_BYTE *)(v9 + a3 - 1) = byte_40E0A0[(v7 >> 2) & 0x3F];
    v10 = v9 + 1;
    *(_BYTE *)(v10 + a3 - 1) = byte_40E0A0[16 * (*(_BYTE *)(v3 + a1) & 3)];
    *(_BYTE *)(v10 + a3) = 61;
LABEL_8:
    v13 = v10 + 1;
    *(_BYTE *)(v13 + a3) = 61;
    v4 = v13 + 1;
    goto LABEL_9;
  }
  if ( v5 == 2 )
  {
    v11 = v4 + 1;
    *(_BYTE *)(v11 + a3 - 1) = byte_40E0A0[((int)*(unsigned __int8 *)(v3 + a1) >> 2) & 0x3F];
    v12 = (_BYTE *)(v3 + a1 + 1);
    LOBYTE(v6) = *v12;
    v10 = v11 + 1;
    *(_BYTE *)(v10 + a3 - 1) = byte_40E0A0[16 * (*(_BYTE *)(v3 + a1) & 3) + ((v6 >> 4) & 0xF)];
    *(_BYTE *)(v10 + a3) = byte_40E0A0[4 * (*v12 & 0xF)];
    goto LABEL_8;
  }
LABEL_9:
  *(_BYTE *)(v4 + a3) = 0;
  return sub_401030(a3);

通过查看各位大佬博客得知 此处是BASE64的特征码 我没有看出来
4.这个函数就可以分为三个部分

1.开头sub_401000函数
2.中间base64加密
3.结尾sub_401030函数

sub_401000函数
int sub_401000()
{
  int result; // eax
  char v1; // cl

  for ( result = 6; result < 15; ++result )
  {
    v1 = byte_40E0AA[result];
    byte_40E0AA[result] = byte_40E0A0[result];
    byte_40E0A0[result] = v1;
  }
  return result;
}

看出来是在进行换位操作
这里就是一个对base64的变表处理

sub_401030函数
int __cdecl sub_401030(const char *a1)
{
  __int64 v1; // rax
  char v2; // al

  v1 = 0i64;
  if ( strlen(a1) )
  {
    do
    {
      v2 = a1[HIDWORD(v1)];
      if ( v2 < 97 || v2 > 122 )
      {
        if ( v2 < 65 || v2 > 90 )
          goto LABEL_9;
        LOBYTE(v1) = v2 + 32;
      }
      else
      {
        LOBYTE(v1) = v2 - 32;
      }
      a1[HIDWORD(v1)] = v1;
LABEL_9:
      LODWORD(v1) = 0;
      ++HIDWORD(v1);
    }
    while ( HIDWORD(v1) < strlen(a1) );
  }
  return v1;
}

将经过base64变表加密的结果,大小写互换。

因此,我们只需要反过来,先将byte_40E0E4大小写互换,构造base64变表,再利用变表将byte_40E0E4转换为正常的base64解密就行。

import base64

Str = list("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/")
model = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"
for i in range(6,15):
    Str[i],Str[i+10] = Str[i+10],Str[i]
Str = ''.join(Str)
enc = "zMXHz3TIgnxLxJhFAdtZn2fFk3lYCrtPC2l9".swapcase()
dec = ""
for i in range(len(enc)):
    dec += model[Str.find(enc[i])]
print (dec)
print (Str)
print (base64.b64decode(dec))

flag{bAse64_h2s_a_Surprise}

posted @ 2021-12-15 12:24  social_biao  阅读(53)  评论(0编辑  收藏  举报