KCTF2022-第二题-末日邀请

KCTF2022-第二题-末日邀请

字符串转换：

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int __fastcall str2int(int src, int len)
{
  int i; // esi
  char v4; // dl
  char v5; // al

  for ( i = 0; i < len; ++i )
  {
    v4 = *(_BYTE *)(i + src);
    v5 = '0';
    if ( v4 >= ':' )
      v5 = 0x37;
    *(_BYTE *)(i + src) = v4 - v5;
  }
  return 0;
}

image-20220511180010693

image-20220511180412661

image-20220511180602745

image-20220511180921771

暴力破解:

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def check(l):
  v37 = 0
  for i in range(9):
    v23 = l[i] + 10*v37
    v24 = v23 - 0x37373737
    if v23 <= 0x4B435445:
      v24 = v23
    v37 = v24
    if v24 %(i+1) == 0:
      continue
    else:
      return False
  return True

for i in itertools.permutations([1,2,3,4,5,6,7,8,9], 9):
  l = []
  for c in i:
    l.append(c)
  if check(l):
    print(l)
1
[3, 8, 1, 6, 5, 4, 7, 2, 9]

目前还有前key的前3位未知，需要满足条件：

image-20220511181049091

可以通过暴力尝试得到前三位为421。

最终key： 421KCTF381654729。

image-20220511181441556