首页
社区
课程
招聘
[看雪CTF2016]第七题分析
发表于: 2016-11-16 04:08 6281

[看雪CTF2016]第七题分析

2016-11-16 04:08
6281

1. 先定位关键点

只有第一个线程是有用的, 另外两个直接patch掉

.text:00403048                 push    esi
.text:00403049                 mov     esi, ds:CreateThread
.text:0040304F                 push    edi
.text:00403050                 xor     edi, edi
.text:00403052                 push    edi             ; lpThreadId
.text:00403053                 push    edi             ; dwCreationFlags
.text:00403054                 push    edi             ; lpParameter
.text:00403055                 push    offset thread1  ; lpStartAddress
.text:0040305A                 push    edi             ; dwStackSize
.text:0040305B                 push    edi             ; lpThreadAttributes
.text:0040305C                 call    esi ; CreateThread
.text:0040305E                 push    edi             ; lpThreadId
.text:0040305F                 push    edi             ; dwCreationFlags
.text:00403060                 push    edi             ; lpParameter
.text:00403061                 push    offset thread2  ; lpStartAddress
.text:00403066                 push    edi             ; dwStackSize
.text:00403067                 push    edi             ; lpThreadAttributes
.text:00403068                 mov     g_hThread1, eax
.text:0040306D                 call    esi ; CreateThread
.text:0040306F                 push    edi             ; lpThreadId
.text:00403070                 push    edi             ; dwCreationFlags
.text:00403071                 push    edi             ; lpParameter
.text:00403072                 push    offset thread3  ; lpStartAddress
.text:00403077                 push    edi             ; dwStackSize
.text:00403078                 push    edi             ; lpThreadAttributes
.text:00403079                 mov     g_hThread2, eax
.text:0040307E                 call    esi ; CreateThread
.text:00403080                 push    0FFFFFFFFh      ; dwMilliseconds
.text:00403082                 push    g_hThread1      ; hHandle
.text:00403088                 mov     g_hThread3, eax
.text:0040308D                 call    ds:WaitForSingleObject
.text:004030D5                 call    verify_all_402B93
.text:004030DA                 test    al, al
.text:004030DC                 jz      short loc_403106
.text:004030DE                 lea     eax, [ebp+var_28]
.text:004030E1                 push    eax
.text:004030E2                 call    decrypt_string
DWORD g_ary[0x3F2];
g_ary[0]=0x6C35B49D;
g_ary[1]=0xA645500D;
g_ary[2]=0xCB9E682E;
g_ary[n]=g_ary[n-1]+g_ary[n-2]+g_ary[n-3];

.text:004026CD                 push    esi
.text:004026CE                 push    4               ; flProtect
.text:004026D0                 push    3000h           ; flAllocationType
.text:004026D5                 mov     esi, 0FC8h
.text:004026DA                 push    esi             ; dwSize
.text:004026DB                 push    0               ; lpAddress
.text:004026DD                 call    ds:VirtualAlloc
.text:004026E3                 mov     g_ary, eax
.text:004026E8                 push    0Ch
.text:004026EA                 pop     edx
.text:004026EB                 mov     dword ptr [eax], 6C35B49Dh
.text:004026F1                 mov     eax, g_ary
.text:004026F6                 mov     dword ptr [eax+4], 0A645500Dh
.text:004026FD                 mov     eax, g_ary
.text:00402702                 mov     dword ptr [eax+8], 0CB9E682Eh
.text:00402709                 mov     ecx, g_ary
.text:0040270F                 mov     eax, [edx+ecx-8]
.text:00402713                 imul    eax, [edx+ecx-4]
.text:00402718                 add     eax, [edx+ecx-0Ch]
.text:0040271C                 mov     [edx+ecx], eax
.text:0040271F                 add     edx, 4
.text:00402722                 cmp     edx, esi
.text:00402724                 jl      short loc_402709
计算两个g_sum1及g_sum2
BOOL flag[0x3F2] = 01011010101010101010101010101010101010101010101010101010110101010101010101010101010101010101010101010101010101010101010101010101010101010110101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010110101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010100101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101010101001010101010101010101010101010101010101010101010010101010101010101010101010101010101010101010101010101010101010101010100101010101;
g_sum1=sum(g_ary中flag为1的项])
g_sum2=sum(g_ary中flag为0的项])=sum(g_ary)-g_sum1;

.text:00402C5B                 mov     esi, offset flNewProtect
.text:00402C60                 mov     ecx, esi
.text:00402C62                 push    edi
.text:00402C63                 mov     edi, 3F1h
.text:00402C68                 push    edi
.text:00402C69                 push    0
.text:00402C6B                 call    sum_array_what
.text:00402C70                 push    ebx
.text:00402C71                 push    edi
.text:00402C72                 push    0
.text:00402C74                 mov     ecx, esi
.text:00402C76                 mov     dword ptr g_sum1, eax
.text:00402C7B                 mov     dword ptr g_sum1+4, edx
.text:00402C81                 call    sum_array_all
.text:00402C86                 sub     eax, dword ptr g_sum1
.text:00402C8C                 pop     edi
.text:00402C8D                 sbb     edx, dword ptr g_sum1+4
.text:00402C93                 pop     esi
.text:00402C94                 mov     dword ptr g_sum2, eax
.text:00402C99                 mov     dword ptr g_sum2+4, edx
.text:00402850                 mov     ecx, [ebp+arg_0]
.text:00402853                 call    std_string_length
.text:00402858                 cmp     eax, 55h
.text:0040285B                 jnz     short loc_402871
.text:0040285D                 mov     ecx, [ebp+arg_0]
.text:00402860                 push    9
.text:00402862                 call    std_string_get_at
.text:00402867                 cmp     byte ptr [eax], 'F'
.text:0040286A                 jnz     short loc_402871

[培训]内核驱动高级班,冲击BAT一流互联网大厂工作,每周日13:00-18:00直播授课

收藏
免费 2
支持
分享
最新回复 (6)
雪    币: 10298
活跃值: (3157)
能力值: (RANK:520 )
在线值:
发帖
回帖
粉丝
2
试试,可以提交了

方便的话加比赛群:8601428
可以及时沟通
2016-11-16 10:27
0
雪    币: 29118
活跃值: (7589)
能力值: ( LV15,RANK:3306 )
在线值:
发帖
回帖
粉丝
3
好了,thx, 加了
2016-11-16 11:45
0
雪    币: 3053
活跃值: (891)
能力值: ( LV13,RANK:1300 )
在线值:
发帖
回帖
粉丝
4
验证g_sum1 ^ g_sum2 == (v[0] ^ v[1]);

不错,学习了
2016-11-16 17:36
0
雪    币: 449
活跃值: (233)
能力值: ( LV2,RANK:10 )
在线值:
发帖
回帖
粉丝
5
如何path反调式,谢谢楼主了。
2016-11-29 10:56
0
雪    币: 4357
活跃值: (974)
能力值: ( LV8,RANK:142 )
在线值:
发帖
回帖
粉丝
6
@qqsunqiang,直接nop掉后两个createthread即可
2017-4-20 13:56
0
雪    币: 21332
活跃值: (6251)
能力值: (RANK:445 )
在线值:
发帖
回帖
粉丝
7
"这样的话,  只需要保证根据flag计算v时,  序列一直是10即可"  这句话中,
‘序列一直是10即可’是什么意思?  怎么得到的?
2017-12-4 22:44
0
游客
登录 | 注册 方可回帖
返回
//