[BUUCTF]REVERSE——[V&N2020 公开赛]strangeCpp

[V&N2020 公开赛]strangeCpp

附加

步骤

1. 查壳，无壳，64位程序
   
2. 64位ida载入，没有main函数，根据程序里的字符串，去查看函数

__int64 __fastcall sub_140013AA0(__int64 a1, __int64 a2, __int64 *a3)
{
  char *v3; // rdi
  signed __int64 i; // rcx
  __int64 v5; // rax
  __int64 v6; // rax
  __int64 v7; // rax
  __int64 v8; // rax
  char v10; // [rsp+0h] [rbp-20h]
  struct _SYSTEM_INFO SystemInfo; // [rsp+28h] [rbp+8h]
  __int64 *j; // [rsp+78h] [rbp+58h]
  __int64 v13; // [rsp+98h] [rbp+78h]
  __int64 *v14; // [rsp+1A0h] [rbp+180h]

  v14 = a3;
  v3 = &v10;
  for ( i = 94i64; i; --i )
  {
    *(_DWORD *)v3 = 3435973836;
    v3 += 4;
  }
  sub_1400110AA((__int64)&unk_140027033);
  GetSystemInfo(&SystemInfo);
  putchar(byte_140021004);                      // w
  putchar(byte_140021005);                      // e
  putchar(byte_140021006);                      // l
  putchar(byte_140021007);                      // c
  putchar(byte_140021019);                      // o
  putchar(byte_14002101A);                      // m
  putchar(byte_140021005);                      // e
  putchar(10);
  puts("Let me have a look at your computer...");
  for ( j = v14; *j; ++j )
  {
    v13 = *j;
    sub_140011226((__int64)"%s\n", v13);   // 循环输出变量
  }
  std::basic_ostream<char,std::char_traits<char>>::operator<<(std::cout, sub_140011127);
  dword_140021190 = SystemInfo.dwNumberOfProcessors;  // CPU数量
  sub_140011226((__int64)"now system cpu num is %d\n", SystemInfo.dwNumberOfProcessors);
  if ( dword_140021190 < 8 )
  {
    puts("Are you in VM?");
    _exit(0);
  }
  if ( GetUserNameA(Str1, &pcbBuffer) )   // 获取用户名
  {
    v5 = sub_140011172(std::cout, "this is useful");
    std::basic_ostream<char,std::char_traits<char>>::operator<<(v5, sub_140011127);
  }
  v6 = std::basic_ostream<char,std::char_traits<char>>::operator<<(std::cout, sub_140011127);
  v7 = sub_140011172(v6, "ok,I am checking...");
  std::basic_ostream<char,std::char_traits<char>>::operator<<(v7, sub_140011127);
  if ( !j_strcmp(Str1, "cxx") )
  {
    v8 = sub_140011172(std::cout, "flag{where_is_my_true_flag?}");
    std::basic_ostream<char,std::char_traits<char>>::operator<<(v8, sub_140011127);
    _exit(0);
  }
  system("pause");
  sub_1400113E3(&v10, &unk_14001DE50);
  return 0i64;
}

以为flag{where_is_my_true_flag?}是答案，还是太天真了

1. 找了半天没找到关键点，想要动调但是报错，不清楚这是为什么。都没搞清楚这个程序要干嘛
2. 在查看了其他师傅的wp之后，发现了这题的关键点
   我们之前那个welcome的字符那边定义了一个数组
   
3. 找到交叉引用它的地方
   
   当满足21行的if条件的情况时，25行会根据异或算法，输出一个字符串，猜测是flag
4. result的值由sub_140011384函数得到，看一下sub_140011384函数
   
   根据20~23行的算法，加上上一张图里的dword_140021190 <= 14549743，可以穷举暴力破解出result里的值

result=0
for v8 in range(14549743):
    v7 = (((v8 << 8) ^ (v8 >> 12))*291)&0xFFFFFFFF
    if (v7 == 607052314):
        result = v8
        break

print(result)

v7的类型是unsigned int–0~0xFFFFFFFF,输出的值需要截断，python默认没把数据截断
算出了result的值，看一下这个putchar语句输出的值

result=0
for v8 in range(14549743):
    v7 = (((v8 << 8) ^ (v8 >> 12))*291)&0xFFFFFFFF # 原文是unsigned int--0~0xFFFFFFFF,输出的值需要截断
    if (v7 == 607052314):
        result = v8
        break

print(result)

a=[0x26, 0x2C, 0x21, 0x27, 0x3B, 0x0D, 0x04, 0x75, 0x68, 0x34,
  0x28, 0x25, 0x0E, 0x35, 0x2D, 0x69, 0x3D, 0x6F, 0x6D, 0x00]

for j in range(17):
    print(chr((v8^a[j])&0xff),end="")

终于看到了曙光，讲123456拿去md5加密一下

flag{e10adc3949ba59abbe56e057f20f883e}