BUUCTF-RE-[羊城杯 2020]easyre

进入main，简单分析之后进入encode_one

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  int v4; // eax
  int v5; // eax
  char Str[48]; // [rsp+20h] [rbp-60h] BYREF
  char Str1[64]; // [rsp+50h] [rbp-30h] BYREF
  char v9[64]; // [rsp+90h] [rbp+10h] BYREF
  char v10[64]; // [rsp+D0h] [rbp+50h] BYREF
  char Str2[60]; // [rsp+110h] [rbp+90h] BYREF
  int v12; // [rsp+14Ch] [rbp+CCh] BYREF

  _main();
  strcpy(Str2, "EmBmP5Pmn7QcPU4gLYKv5QcMmB3PWHcP5YkPq3=cT6QckkPckoRG");
  puts("Hello, please input your flag and I will tell you whether it is right or not.");
  scanf("%38s", Str);
  if ( strlen(Str) != 38                        // 输入的flag长度为38
    || (v3 = strlen(Str), (unsigned int)encode_one(Str, v3, v10, &v12))
    || (v4 = strlen(v10), (unsigned int)encode_two(v10, v4, v9, &v12))
    || (v5 = strlen(v9), (unsigned int)encode_three(v9, v5, Str1, &v12))
    || strcmp(Str1, Str2) )                     // 添加比较条件，相同为0，输出正确
  {
    printf("Something wrong. Keep going.");
    return 0;
  }
  else
  {
    puts("you are right!");
    return 0;
  }
}

encode_one,本来还在认真分析代码的，然后发现
alphabet = ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/
大概率为base64加密，直接换到下一个函数

__int64 __fastcall encode_one(char *a1, int a2, char *a3, int *a4)
					// a1 = flag  a2 = 38   a3:char v10[64]    a4: int v12
{
  int v5; // esi
  int v6; // esi
  int v7; // esi
  int v8; // [rsp+34h] [rbp-1Ch]
  int v9; // [rsp+38h] [rbp-18h]
  int v11; // [rsp+48h] [rbp-8h]
  int i; // [rsp+4Ch] [rbp-4h]
  unsigned __int8 *v13; // [rsp+70h] [rbp+20h]

  v13 = (unsigned __int8 *)a1;
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  v11 = 0;
  if ( a2 % 3 )
    v11 = 3 - a2 % 3;
  v9 = a2 + v11;
  v8 = 8 * (a2 + v11) / 6;
  for ( i = 0; i < v9; i += 3 )
  {
    *a3 = alphabet[(char)*v13 >> 2];
    if ( a2 + v11 - 3 == i && v11 )
    {
      if ( v11 == 1 )
      {
        v5 = (char)cmove_bits(*v13, 6u, 2u);
        a3[1] = alphabet[v5 + (char)cmove_bits(v13[1], 0, 4u)];
        a3[2] = alphabet[(char)cmove_bits(v13[1], 4u, 2u)];
        a3[3] = 61;
      }
      else if ( v11 == 2 )
      {
        a3[1] = alphabet[(char)cmove_bits(*v13, 6u, 2u)];
        a3[2] = 61;
        a3[3] = 61;
      }
    }
    else
    {
      v6 = (char)cmove_bits(*v13, 6u, 2u);
      a3[1] = alphabet[v6 + (char)cmove_bits(v13[1], 0, 4u)];
      v7 = (char)cmove_bits(v13[1], 4u, 2u);
      a3[2] = alphabet[v7 + (char)cmove_bits(v13[2], 0, 6u)];
      a3[3] = alphabet[v13[2] & 0x3F];
    }
    a3 += 4;
    v13 += 3;
  }
  if ( a4 )
    *a4 = v8;
  return 0i64;
}

encode_two结构简单

__int64 __fastcall encode_two(const char *a1, int a2, char *a3, int *a4)
{
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  strncpy(a3, a1 + 26, 0xDui64);
  strncpy(a3 + 13, a1, 0xDui64);
  strncpy(a3 + 26, a1 + 39, 0xDui64);
  strncpy(a3 + 39, a1 + 13, 0xDui64);
  return 0i64;
}

encode_three，网上检索发现是凯撒密码（特征为他只位移字母），位移为3

__int64 __fastcall encode_three(const char *a1, int a2, char *a3, int *a4)
{
  char v5; // [rsp+Fh] [rbp-11h]
  int i; // [rsp+14h] [rbp-Ch]
  const char *v8; // [rsp+30h] [rbp+10h]

  v8 = a1;
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  for ( i = 0; i < a2; ++i )
  {
    v5 = *v8;
    if ( *v8 <= 64 || v5 > 90 )
    {
      if ( v5 <= 96 || v5 > 122 )
      {
        if ( v5 <= 47 || v5 > 57 )
          *a3 = v5;
        else
          *a3 = (v5 - 48 + 3) % 10 + 48;
      }
      else
      {
        *a3 = (v5 - 97 + 3) % 26 + 97;
      }
    }
    else
    {
      *a3 = (v5 - 65 + 3) % 26 + 65;
    }
    ++a3;
    ++v8;
  }
  return 0i64;
}

最后加密完成的字符串为EmBmP5Pmn7QcPU4gLYKv5QcMmB3PWHcP5YkPq3=cT6QckkPckoRG

第一次揭秘后：BjYjM5Mjk7NzMR4dIVHs5NzJjY3MTEzM5VhMn3=zQ6NzhhMzhlOD

第二次解密为：是位置调换，要注意这里使用指针变换，所以最后为R1dIVHs2NzJjYzQ3NzhhMzhlODBjYjM2Mjk4NzM0MTEzM2VhMn0=

第三次base64解密：GWHT{672cc4778a38e80cb362987341133ea2}