re每日一题（20200530）（补）

re每日一题（20200530）（补）

GKCTF2020

0x03 EzMachine

载入IDA，main函数不能F5，需要将0x1591的后三个字节nop

int __cdecl main(int argc, const char **argv, const char **envp)
{
  unsigned int i; // [esp+10h] [ebp-8h]

  sub_6014F0();
  while ( 1 )
  {
LABEL_2:
    for ( i = 0; ; i += 2 )
    {
      if ( i >= 0x58 )
        goto LABEL_2;
      if ( code[4 * i] == byte_6449A0[E_ip] )
        break;
    }
    call[i]();
  }
}

查看code这个数组，可以发现是VM。

分析过后，整理字节码为指令，指令为三个字节长度

0x1 mov reg,const
0x2 push
0x3 mov [mem],reg
0x4 pop
0x5 print
0x6 add
0x7 sub
0x8 mul
0x9 idiv
0xa xor
0xb jmp
0xc cmp
0xd  jz  
0xf  ja
0x10 jb
0x11 gets
0x12 memset 
0x13 mov_reg_esp
0x14 movzx
0xff exit

大致整理后，加密算法为

cipher = []
for i in range(len(input)):
	tmp = ord(input[i])
	sum = 0
	if input[i] <= 'z' and input[i] >= 'a':
		sum = (tmp^0x47)+1
	elif input[i] <= 'Z' and input[i] >= 'A':
		sum = (tmp^0x4b)-1
	cipher.append(sum%0x10)
	cipher.append(sum/0x10)

通过加密算法可以写出解密算法。（由于在VM的栈中，需要反序）

cipher = [0x7,0xd,0x0,0x5,0x1,0xc,0x1,0x0,0x0,0xd,0x5,0xf,0x0,0x9,0x5,0xf,0x3,0x0,0x2,0x5,0x3,0x3,0x1,0x7,0x7,0xb,0x2,0x1,0x2,0x7,0x2,0xc,0x2,0x2]
cipher = cipher[::-1]
flag = ''
for i in range(0,len(cipher),2):
	tmp = cipher[i+1]*0x10+cipher[i]
	ch = chr((tmp-1)^0x47)
	if ch <= 'z' and ch >= 'a':
		flag += ch
		continue
	ch = chr((tmp+1)^0x4b)
	if ch <= 'Z' and ch >= 'A':
		flag += ch
		continue
	flag += chr(tmp)

print flag