buuctf刷题记录3 [2019红帽杯]easyRE

先查壳，没有壳，拖进ida，f12大法

发现敏感句子，进入，代码如下

signed __int64 sub_4009C6()
{
  const char *v0; // rdi
  signed __int64 result; // rax
  __int64 v2; // ST10_8
  __int64 v3; // ST18_8
  __int64 v4; // ST20_8
  __int64 v5; // ST28_8
  __int64 v6; // ST30_8
  __int64 v7; // ST38_8
  __int64 v8; // ST40_8
  __int64 v9; // ST48_8
  __int64 v10; // ST50_8
  int i; // [rsp+Ch] [rbp-114h]
  char v12; // [rsp+60h] [rbp-C0h]
  char v13; // [rsp+61h] [rbp-BFh]
  char v14; // [rsp+62h] [rbp-BEh]
  char v15; // [rsp+63h] [rbp-BDh]
  char v16; // [rsp+64h] [rbp-BCh]
  char v17; // [rsp+65h] [rbp-BBh]
  char v18; // [rsp+66h] [rbp-BAh]
  char v19; // [rsp+67h] [rbp-B9h]
  char v20; // [rsp+68h] [rbp-B8h]
  char v21; // [rsp+69h] [rbp-B7h]
  char v22; // [rsp+6Ah] [rbp-B6h]
  char v23; // [rsp+6Bh] [rbp-B5h]
  char v24; // [rsp+6Ch] [rbp-B4h]
  char v25; // [rsp+6Dh] [rbp-B3h]
  char v26; // [rsp+6Eh] [rbp-B2h]
  char v27; // [rsp+6Fh] [rbp-B1h]
  char v28; // [rsp+70h] [rbp-B0h]
  char v29; // [rsp+71h] [rbp-AFh]
  char v30; // [rsp+72h] [rbp-AEh]
  char v31; // [rsp+73h] [rbp-ADh]
  char v32; // [rsp+74h] [rbp-ACh]
  char v33; // [rsp+75h] [rbp-ABh]
  char v34; // [rsp+76h] [rbp-AAh]
  char v35; // [rsp+77h] [rbp-A9h]
  char v36; // [rsp+78h] [rbp-A8h]
  char v37; // [rsp+79h] [rbp-A7h]
  char v38; // [rsp+7Ah] [rbp-A6h]
  char v39; // [rsp+7Bh] [rbp-A5h]
  char v40; // [rsp+7Ch] [rbp-A4h]
  char v41; // [rsp+7Dh] [rbp-A3h]
  char v42; // [rsp+7Eh] [rbp-A2h]
  char v43; // [rsp+7Fh] [rbp-A1h]
  char v44; // [rsp+80h] [rbp-A0h]
  char v45; // [rsp+81h] [rbp-9Fh]
  char v46; // [rsp+82h] [rbp-9Eh]
  char v47; // [rsp+83h] [rbp-9Dh]
  char v48[32]; // [rsp+90h] [rbp-90h]
  int v49; // [rsp+B0h] [rbp-70h]
  char v50; // [rsp+B4h] [rbp-6Ch]
  char v51; // [rsp+C0h] [rbp-60h]
  char v52; // [rsp+E7h] [rbp-39h]
  char v53; // [rsp+100h] [rbp-20h]
  unsigned __int64 v54; // [rsp+108h] [rbp-18h]

  v54 = __readfsqword(0x28u);
  v12 = 73;
  v13 = 111;
  v14 = 100;
  v15 = 108;
  v16 = 62;
  v17 = 81;
  v18 = 110;
  v19 = 98;
  v20 = 40;
  v21 = 111;
  v22 = 99;
  v23 = 121;
  v24 = 127;
  v25 = 121;
  v26 = 46;
  v27 = 105;
  v28 = 127;
  v29 = 100;
  v30 = 96;
  v31 = 51;
  v32 = 119;
  v33 = 125;
  v34 = 119;
  v35 = 101;
  v36 = 107;
  v37 = 57;
  v38 = 123;
  v39 = 105;
  v40 = 121;
  v41 = 61;
  v42 = 126;
  v43 = 121;
  v44 = 76;
  v45 = 64;
  v46 = 69;
  v47 = 67;
  memset(v48, 0, sizeof(v48));
  v49 = 0;
  v50 = 0;
  sub_4406E0(0LL, v48, 37LL);
  v50 = 0;
  v0 = v48;
  if ( sub_424BA0(v48) == 36 )
  {
    for ( i = 0; ; ++i )
    {
      v0 = v48;
      if ( i >= (unsigned __int64)sub_424BA0(v48) )
        break;
      if ( (unsigned __int8)(v48[i] ^ i) != *(&v12 + i) )
      {
        result = 4294967294LL;
        goto LABEL_13;
      }
    }
    sub_410CC0("continue!");
    memset(&v51, 0, 0x40uLL);
    v53 = 0;
    sub_4406E0(0LL, &v51, 64LL);
    v52 = 0;
    v0 = &v51;
    if ( sub_424BA0(&v51) == 39 )
    {
      v2 = sub_400E44(&v51);
      v3 = sub_400E44(v2);
      v4 = sub_400E44(v3);
      v5 = sub_400E44(v4);
      v6 = sub_400E44(v5);
      v7 = sub_400E44(v6);
      v8 = sub_400E44(v7);
      v9 = sub_400E44(v8);
      v10 = sub_400E44(v9);
      v0 = (const char *)sub_400E44(v10);
      if ( !(unsigned int)sub_400360(v0, off_6CC090) )
      {
        sub_410CC0("You found me!!!");
        v0 = "bye bye~";
        sub_410CC0("bye bye~");
      }
      result = 0LL;
    }
    else
    {
      result = 4294967293LL;
    }
  }
  else
  {
    result = 0xFFFFFFFFLL;
  }
LABEL_13:
  if ( __readfsqword(0x28u) != v54 )
    sub_444020(v0);
  return result;
}

里面有两处关键运算

这是里面第一个关键运算，逻辑很简单，构造脚本

num=[73,111,100,108,62,81,110,98,40,111,99,121,127,121,46,105,127,100,96,51,119,125,119,101,107,57,123,105,121,61,126,121,76,64,69,67]
text=''
for i in range(0,36):
    text=text+chr(num[i]^i)
print(text)

得到：Info:The first four chars are flag

前四个字母是flag

第二个关键运算：

点进去后发现是一个类似base64的运算（10次），而且得到的结果与off_6cc090比较，而6cc090是一串base64编码，于是就猜这个是base64加密了10次，解密得

https://bbs.pediy.com/thread-254172.htm

发现是个网址，但是没有卵用，应该是干扰（打死出题人！）

没有思路后，看了看别人的wp：

在off_6cc090下面还有一个常量，在sub_400D35函数中调用

进入到该函数中

通过分析，v6（或v3）在里面起到很重要的作用，在

if ( ((unsigned __int8)v3 ^ byte_6CC0A0[0]) == ‘f’ && (HIBYTE(v6) ^ (unsigned __int8)unk_6CC0A3) == ‘g’ )

里面，判断v3第一个和第四个是不是f和g，那就猜测这四个字符是flag。（v3-v6地址是连着的，可以将v3看作一个char数组）

然后循环24次做异或操作。

先求出来v3：这里key就是v3

num2=[0x40,0x35,0x20,0x56,0x5D,0x18,0x22,0x45,0x17,0x2F,0x24,0x6E,0x62,0x3C,0x27,0x54,0x48,0x6C,0x24,0x6E,0x72,0x3C,0x32,0x45,0x5B]
text2='flag'
key=''
for i in range(0,4):
    key=key+chr(ord(text2[i])^num2[i])
print（key）

得到v3： &YA1

再求出v0（24次异或）

num2=[0x40,0x35,0x20,0x56,0x5D,0x18,0x22,0x45,0x17,0x2F,0x24,0x6E,0x62,0x3C,0x27,0x54,0x48,0x6C,0x24,0x6E,0x72,0x3C,0x32,0x45,0x5B]
key='&YA1'
flag=''
for i in range(0,25):
    flag=flag+chr(num2[i]^ord(key[i%4]))
print(flag)

就得到flag了（曲折，脑壳痛）

flag{Act1ve_Defen5e_Test}