Reverse｜Buuctf reverse3

查壳为32位程序，使用32位ida打开

查询到rright flag字符

查看伪代码

__int64 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;
}

输入一个字符串经过sub_4110BE函数进行加密，再通过一个for循环进行变换，然后与str2进行比较

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

sub_4110BE函数将3个字符变成4个字符，结合aAbcdefghijklmn的值，可以判断是base64
脚本

import base64
 
str = 'e3nifIH9b_C@n@dH'
flag = ''

for i in range(len(str)):
	flag += chr(ord(str[i])-i)

print(base64.b64decode(flag))

flag{i_l0ve_you}