【RCTF2018】部分Writeup

作者: Kirin_say | 来源:发表于2018-05-25 20:09 被阅读14次

0x01 Sign

签到题
载入ida
Alt+T查找RCTF即得flag

0x02 git

签到题
查看commit信息：

Revert
# 请为您的变更输入提交说明。以 '#' 开始的行将被忽略，而一个空的提交
# 说明将会终止提交。
#
# 位于分支 rctf
# 要提交的变更：
#   删除：     flag.txt
#

需要利用git revert来恢复flag.txt
git bash 下打开:

git reflog

看到有一次合并：

f671986 HEAD@{5}: checkout: moving from master to develop

而后：

git revert f671986

即可恢复txt文件，得到flag

0x03 cpushop

hash长度扩展攻击
我们需要flag的价格:price<money
利用hashdump附加请求信息：&price=0
而后根据正常的sign值得到附加信息后的sign
发送新的请求即得flag

0x04 Number Game

第一关直接碰撞即可
而后可以看到number game是很早的一个猜数游戏
github上可以找到一种生成树策略：

https://github.com/AustinGuo/GessNumber

直接套入脚本，过关即得flag

0x05 babyre

调试后发现堆中几个参数不可控
直接运行发现每个字节依次运算，与位置无关
找到对应关系根据密文破解
或者直接pwntools按位爆破也可以:

key = [0xb8a6ee10,0xb986234f,0xf1b1602,0x2571a420,0x805feee3,0x958b0534,0x8a0ad7,0xf3a32248,0x45c66c5a,0xcda47499,0xee152dad,0x77591917,0xb118982f,0x7386bd6a,0xdf1708d2,0x67d38af2,0xc567b6b6,0x526ba076,0xae4c822d,0xa1e10430,0xc3f2de73,0x7f08e077,0x7240b7ba,0x20067e4b,0x28def57e,0x545c15ad,0xb02881cf,0x15d71041,0x9f9e33a3,0x6958d9df,0x51a1924f,0x299fa574,0x3e968078,0x6e7b9de9,0x3e5daef6,0x70573efe,0x94bec810,0xde26ea93,0x7f7a8193,0x7147adb7,0xe554cf78,0xf0dc5e15,0x730a451b,0x90347c6c,0x8699b5f,0x834c762a,0x97e8bb26,0xc1ca6398,0x6aed0d44,0xe498e228,0xb80c91fe,0x9935e5ee,0xbeed92ae,0x5067d2a7,0x239b768c,0x3b6eb4ce,0x354689e2,0x8028bcb,0x86447eef,0x86c95e28,0x8a0ad7,0xe887091c,0xa8f27e12,0x7d8aa463,0xc15dfd82,0x9c513266,0x30a3769a,0xaa8afbba,0x1edc3864,0xaa7a15df,0x6c1194bf,0xeade7bf1,0x115f3801,0x42bafa35,0xa5a30516,0x4e032f9f,0x121ee8ab,0xe75d55e4,0x7f543d62,0x153f1a32,0x6d076b18,0xf3857621,0x6d7df3e1,0xfbcb7aa0,0x36b897bd,0x82c974fc,0xf5650025,0x90f77ba1,0xf9b8d92b,0x9323bc07,0xa5c88e2,0x7390c17b,0x760e65f9,0xb13ad888,0x1667aba8]
crypto = [0xb80c91fe,0x70573efe,0xbeed92ae,0x7f7a8193,0x7390c17b,0x90347c6c,0xaa7a15df,0xaa7a15df,0x526ba076,0x153f1a32,0x545c15ad,0x7d8aa463,0x526ba076,0xfbcb7aa0,0x7d8aa463,0x9c513266,0x526ba076,0x6d7df3e1,0xaa7a15df,0x9c513266,0x1edc3864,0x9323bc07,0x7d8aa463,0xfbcb7aa0,0x153f1a32,0x526ba076,0xf5650025,0xaa7a15df,0x1edc3864,0xb13ad888]
flag=""
for i in crypto:
     j=0
     for k in range(32,127):
          if key[k]==i:
             flag+=chr(k)
             break
          k+=1
     print flag

0x06 babyre2

载入ida调试,看到这里：

mov     rax, 20656D6F636C6557h
mov     qword ptr [rsp+198h+s], rax
mov     rax, 2046544352206F74h
mov     [rsp+198h+var_D0], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_110], rax
mov     rax, 6548202138313032h
mov     [rsp+198h+var_C8], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_108], rax
mov     rax, 2061207369206572h
mov     [rsp+198h+var_C0], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_100], rax
mov     rax, 6320455279626142h
mov     [rsp+198h+var_B8], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_B0], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_A8], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_A0], 0FFFFFFFFFFFFFFFFh
mov     [rsp+198h+var_F8], rax
mov     rax, 65676E656C6C6168h
mov     [rsp+198h+var_F0], rax
mov     rax, 756F7920726F6620h
mov     [rsp+198h+var_E8], rax

这些数据依照顺序依次参与：

mov     rax, qword ptr [rsp+198h+s]
xor     ecx, ecx
mul     [rsp+198h+var_198]
mov     rsi, rdx
mov     rdi, rax
mov     rdx, 0FFFFFFFFFFFFFFC5h
call    sub_400BA0

这里可以看到几个参数：

我们输入字符与上文数据的乘积
乘积高16位存于rsi寄存器
乘积低16位存于rdi寄存器
参数0FFFFFFFFFFFFFFC5h存于rdx中

看一下sub_400BA0的伪代码：
很显然进行运算的只有：

else
  {
    if ( a2 <= *((_QWORD *)&a1 + 1) )
    {
      if ( !a2 )
        v3 = 1 / 0uLL;
      *(_QWORD *)&v17 = a1;
      *((_QWORD *)&v17 + 1) = *((_QWORD *)&a1 + 1) % v3;
      v5 = v17 % v3;
    }
    else
    {
      v5 = a1 % a2;
    }
    result = v5;
  }
  return result;

很显然，这里的高16位与低16位都需要对0FFFFFFFFFFFFFFC5h取模，而后return
继续看数次调用sub_400BA0并retn后的运算：

movdqa  xmm1, [rsp+198h+var_98]
mov     qword ptr [rsp+198h+var_28+8], rax
movdqa  xmm0, cs:xmmword_602070
pxor    xmm1, cs:xmmword_602060
movdqa  xmm4, [rsp+198h+var_78]
pxor    xmm0, [rsp+198h+var_88]
movdqa  xmm3, [rsp+198h+var_68]
pxor    xmm4, cs:xmmword_602080
movdqa  xmm2, [rsp+198h+var_58]
pxor    xmm3, cs:xmmword_602090
por     xmm1, xmm0
pxor    xmm2, cs:xmmword_6020A0
movdqa  xmm0, [rsp+198h+var_38]
por     xmm4, xmm1
movdqa  xmm1, [rsp+198h+var_48]
pxor    xmm0, cs:xmmword_6020C0
por     xmm3, xmm4
pxor    xmm1, cs:xmmword_6020B0
por     xmm2, xmm3
movdqa  xmm3, xmm2
movdqa  xmm2, xmm1
movdqa  xmm1, xmm0
movdqa  xmm0, cs:xmmword_6020D0
por     xmm2, xmm3
pxor    xmm0, xmmword ptr [rsp+198h+var_28]
por     xmm1, xmm2
por     xmm0, xmm1
movdqa  xmm1, xmm0
psrldq  xmm1, 8
por     xmm0, xmm1
movq    rax, xmm0
test    rax, rax
jz      short loc_400A86
jz      short loc_400A86
mov     edi, offset s   ; "Incorrect."
call    _puts
loc_400A68:                             ; CODE XREF: main+520↓j
xor     eax, eax
mov     rcx, [rsp+198h+var_10]
xor     rcx, fs:28h
jnz     short loc_400A92
add     rsp, 190h
pop     rbp
 loc_400A86:                             ; CODE XREF: main+4EC↑j
mov     edi, offset aCorrectCongrat ; "Correct. Congratulations!"
call    _puts
jmp     short loc_400A68

对应伪代码（实际汇编的运算逻辑更清晰一些）：

5 = _mm_or_si128(
         _mm_xor_si128(_mm_load_si128((const __m128i *)&xmmword_6020D0), v46),
         _mm_or_si128(
           _mm_xor_si128(_mm_load_si128((const __m128i *)&v45), (__m128i)xmmword_6020C0),
           _mm_or_si128(
             _mm_xor_si128(_mm_load_si128((const __m128i *)&v44), (__m128i)xmmword_6020B0),
             _mm_or_si128(
               _mm_xor_si128(_mm_load_si128((const __m128i *)&v43), (__m128i)xmmword_6020A0),
               _mm_or_si128(
                 _mm_xor_si128(_mm_load_si128((const __m128i *)&v42), (__m128i)xmmword_602090),
                 _mm_or_si128(
                   _mm_xor_si128(_mm_load_si128((const __m128i *)&v41), (__m128i)xmmword_602080),
                   _mm_or_si128(
                     _mm_xor_si128(v4, (__m128i)xmmword_602060),
                     _mm_xor_si128(_mm_load_si128((const __m128i *)&xmmword_602070), v40))))))));
  if ( (unsigned __int64)*(_OWORD *)&_mm_or_si128(v5, _mm_srli_si128(v5, 8)) )
    puts("Incorrect.");
  else
    puts("Correct. Congratulations!");
  return 0LL;

很显然，这里只是对应比较运算后的数据与内存中固定的数据：

.data:0000000000602060 xmmword_602060  xmmword 7BA58F82BD8980352B7192452905E8FBh
.data:0000000000602060                                         ; DATA XREF: main+438↑r
.data:0000000000602070 xmmword_602070  xmmword 163F756FCC221AB0A3112746582E1434h
.data:0000000000602070                                         ; DATA XREF: main+430↑r
.data:0000000000602080 xmmword_602080  xmmword 0DCDD8B49EA5D7E14ECC78E6FB9CBA1FEh
.data:0000000000602080                                         ; DATA XREF: main+45B↑r
.data:0000000000602090 xmmword_602090  xmmword 0AAAAAAAAAA975D1CA2845FE0B3096F8Eh
.data:0000000000602090                                         ; DATA XREF: main+46C↑r
.data:00000000006020A0 xmmword_6020A0  xmmword 55555555555559A355555555555559A3h
.data:00000000006020A0                                         ; DATA XREF: main+478↑r
.data:00000000006020B0 xmmword_6020B0  xmmword 55555555555559A355555555555559A3h
.data:00000000006020B0                                         ; DATA XREF: main+4A2↑r
.data:00000000006020C0 xmmword_6020C0  xmmword 55555555555559A355555555555559A3h
.data:00000000006020C0                                         ; DATA XREF: main+496↑r
.data:00000000006020D0 xmmword_6020D0  xmmword 55555555555559A355555555555559A3h
.data:00000000006020D0                                         ; DATA XREF: main+4BA↑r

shift+E提取数据
我们需要求出对应的x满足：

x:所要求的的flag
y:内存中的比较数据(已知)
k:最开始的参与乘积的数据(已知)
x需要满足：
kx ≡ y mod 0FFFFFFFFFFFFFFC5h

实际上，我们只需要求k关于0FFFFFFFFFFFFFFC5h的乘法逆元，即X满足：

kX  ≡ 1 mod 0FFFFFFFFFFFFFFC5h

而后kXy即满足：

kXy ≡ y mod 0FFFFFFFFFFFFFFC5h

故而：

所求：x=Xy

这里直接利用PyCrypto库中的inverse函数(求乘法逆元)即可：
逐位运算即得flag：

from pwn import *
from Crypto.Util.number import *

key1 =[0x20656D6F636C6557,0x2046544352206F74,0x6548202138313032,0x2061207369206572,0x6320455279626142,0x65676E656C6C6168,0x756F7920726F6620,0xFFFFFFFFFFFF002E,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF,0xFFFFFFFFFFFFFFFF]
key2 =[0x2B7192452905E8FB,0x7BA58F82BD898035,0xA3112746582E1434,0x163F756FCC221AB0,0xECC78E6FB9CBA1FE,0x0DCDD8B49EA5D7E14,0xA2845FE0B3096F8E,0x0AAAAAAAAAA975D1C,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3,0x55555555555559A3]
key3 = 0xFFFFFFFFFFFFFFC5
flag=""
for i in range(len(key1)):
   flag+=p64(key2[i] * inverse(key1[i],key3)%key3)
print flag

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本文标题：【RCTF2018】部分Writeup

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