[C++知识库]buu-[羊城杯 2020]easyre

无壳64位exe

运行还是老样子

查看字符串，先放一个base64在这里

跟进主函数

int __cdecl main(int argc, const char **argv, const char **envp)
{
  int v3; // eax
  int v4; // eax
  int v5; // eax
  int result; // eax
  char Str; // [rsp+20h] [rbp-60h]
  char Str1; // [rsp+50h] [rbp-30h]
  char v9; // [rsp+90h] [rbp+10h]
  char v10; // [rsp+D0h] [rbp+50h]
  char Str2[8]; // [rsp+110h] [rbp+90h]
  int v12; // [rsp+14Ch] [rbp+CCh]

  _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
    && (v3 = strlen(&Str), (unsigned int)encode_one(&Str, v3, &v10, &v12) == 0)
    && (v4 = strlen(&v10), (unsigned int)encode_two(&v10, v4, &v9, &v12) == 0)
    && (v5 = strlen(&v9), (unsigned int)encode_three(&v9, v5, &Str1, &v12) == 0)
    && !strcmp(&Str1, Str2) )
  {
    puts("you are right!");
    result = 0;
  }
  else
  {
    printf("Something wrong. Keep going.");
    result = 0;
  }
  return result;
}

一个输入，三个加密函数，逐步分析

encode_one

__int64 __fastcall encode_one(const char *a1, int a2, char *a3, int *a4)
{
  int v5; // esi
  int v6; // esi
  int v7; // esi
  int v8; // [rsp+34h] [rbp-1Ch]
  int v9; // [rsp+38h] [rbp-18h]
  char *v10; // [rsp+40h] [rbp-10h]
  int v11; // [rsp+48h] [rbp-8h]
  int i; // [rsp+4Ch] [rbp-4h]
  unsigned __int8 *v13; // [rsp+70h] [rbp+20h]
  int v14; // [rsp+78h] [rbp+28h]
  int *v15; // [rsp+88h] [rbp+38h]

  v13 = (unsigned __int8 *)a1;
  v14 = a2;
  v15 = a4;
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  v11 = 0;
  if ( a2 % 3 )
    v11 = 3 - a2 % 3;
  v9 = a2 + v11;
  v8 = 8 * (a2 + v11) / 6;
  v10 = a3;
  for ( i = 0; i < v9; i += 3 )
  {
    *v10 = alphabet[(char)*v13 >> 2];
    if ( v14 + v11 - 3 == i && v11 )
    {
      if ( v11 == 1 )
      {
        v5 = (char)cmove_bits(*v13, 6u, 2u);
        v10[1] = alphabet[v5 + (char)cmove_bits(v13[1], 0, 4u)];
        v10[2] = alphabet[(char)cmove_bits(v13[1], 4u, 2u)];
        v10[3] = 61;
      }
      else if ( v11 == 2 )
      {
        v10[1] = alphabet[(char)cmove_bits(*v13, 6u, 2u)];
        v10[2] = 61;
        v10[3] = 61;
      }
    }
    else
    {
      v6 = (char)cmove_bits(*v13, 6u, 2u);
      v10[1] = alphabet[v6 + (char)cmove_bits(v13[1], 0, 4u)];
      v7 = (char)cmove_bits(v13[1], 4u, 2u);
      v10[2] = alphabet[v7 + (char)cmove_bits(v13[2], 0, 6u)];
      v10[3] = alphabet[v13[2] & 0x3F];
    }
    v10 += 4;
    v13 += 3;
  }
  if ( v15 )
    *v15 = v8;
  return 0i64;
}

alphabet就是密码表

encode_two
encode_one中v10 = a3;

__int64 __fastcall encode_two(const char *a1, int a2, char *a3, int *a4)
{
  char *Source; // [rsp+40h] [rbp+10h]
  char *v6; // [rsp+50h] [rbp+20h]

  Source = (char *)a1;
  v6 = a3;
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  strncpy(a3, a1 + 26, 0xDui64);
  strncpy(v6 + 13, Source, 0xDui64);
  strncpy(v6 + 26, Source + 39, 0xDui64);
  strncpy(v6 + 39, Source + 13, 0xDui64);
  return 0i64;
}

encode_three

__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]
  char *v7; // [rsp+18h] [rbp-8h]
  const char *v8; // [rsp+30h] [rbp+10h]

  v8 = a1;
  if ( !a1 || !a2 )
    return 0xFFFFFFFFi64;
  v7 = a3;
  for ( i = 0; i < a2; ++i )
  {
    v5 = *v8;
    if ( *v8 <= 64 || v5 > 90 )
    {
      if ( v5 <= 96 || v5 > 122 )
      {
        if ( v5 <= 47 || v5 > 57 )
          *v7 = v5;
        else
          *v7 = (v5 - 48 + 3) % 10 + 48;
      }
      else
      {
        *v7 = (v5 - 97 + 3) % 26 + 97;
      }
    }
    else
    {
      *v7 = (v5 - 65 + 3) % 26 + 65;
    }
    ++v7;
    ++v8;
  }
  return 0i64;
}

偏移为3的凯撒密码

凯撒解密不带数字玩，就自己写一个
（好像手动减三会比写脚本方便点）

import base64
Str1=list('BjYjM5Mjk7NzMR4dIVHs5NzJjY3MTEzM5VhMn3=zQ6NzhhMzhlOD')
for i in range(len(Str1)):
     if Str1[i]>='0' and Str1[i]<='9':
          Str1[i]=str(int(Str1[i])-3)
Str1=''.join(Str1)

flag = Str1[13:26] + Str1[39:] + Str1[:13] + Str1[26:39]
print(base64.b64decode(flag))

flag{672cc4778a38e80cb362987341133ea2}