SICTF2023

发表于 更新于 分类于 阅读次数： Waline： 本文字数： 1.3k 阅读时长 ≈ 5 分钟

SICTF2023 #Round2 Crypto方向

古典大杂烩

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🐩👃🐪🐼👅🐯🐩👈👇👭👟👝🐺🐭👉👙👤👋👚🐪🐫👍👢👮👱🐼👢👨👠👭🐽🐰🐻👚👂👧👠👥👛👮👯👮👬🐾👐👛👌👚👞🐨👏👉👆🐿👆👘👇🐺👦🐸👃🐭👟👑👪👃👁🐻🐻👜🐧👇👊🐧🐾🐼👇🐫🐺👐👆👪🐼👋👌👧🐻👐🐩🐺👥🐽👋👉🐰👎👠👠👣🐧🐫👧🐭👢🐯👑👑🐮👂👏🐻👥👚🐮👋👬👌👥👁👣👅👧👯👦👌👌👍👠👌🐽👉👃👊🐫👉🐨🐮👩👆🐪🐯👘👏👏🐼👩👍👊👍👡👀👰👋👣👨👧👍👜👐👛🐮👘👅👠🐿👂👰👄👈👝👠👤👃👛👘🐭👅👱👆👬👫👥👆🐽👁👐👥👊👇👉👊👩👌👭🐫🐫👬👱🐯👇🐺👁👞👑👙🐮👜👋👘👪👩👚👦👨👀👩👐👉👃🐾👥👀🐫👝👍🐩🐧👰👆👇👨🐪👃🐭👦🐫👱

base家族

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base100 --> base62 --> base64 --> base58 --> base32 --> base62
#SICTF{fe853b49-8730-462e-86f5-fc8e9789f077}

Radio

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from Crypto.Util.number import *
from flag import flag
m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
n1 = p * q
p = getPrime(1024)
q = getPrime(1024)
n2 = p * q
p = getPrime(1024)
q = getPrime(1024)
n3 = p * q
e = 17
c1 = pow(m,e,n1)
c2 = pow(m,e,n2)
c3 = pow(m,e,n3)
print("n1 =",n1)
print("n2 =",n2)
print("n3 =",n3)
print("c1 =",c1)
print("c2 =",c2)
print("c3 =",c3)
'''
n1 = 14628911682936716611458501697007036859460044243525290515096052103585430459755335375005202100114469571371360084664887335211277585652711111523095037589648375630146039444071400098427638768750755153219974194380355807078158427824557754939604018020265955042573660474772006646525311705184431094905718137297923127124517126579859336516891364853724635334011666814712424599592662398013241607855160919361308195967978220182785816761656927836373944699635667244275310680450562446433724968942835275279255823144471582249379035668825437133182865600026935116686574740844588839352146024513673500770611055698030333734066230166111140083923 
n2 = 16756694748293603983474688536179571665757862433174984877308316444468003022266277794769268134195205510197588585566270416339902269736376811449830775290335951504698137924773942880807921752691668522662285163130340474205633998154849689387759453003838730282756734975490180702422176361373516245372635401939755527017589503572550811648345570775428936487145892225736625411540461653083957762795820510109891180906709827194217045059033312564525916136573856999724346161896146703174418039344166251503310869772735585554127509732135494936119159784702673291794381095696332128950979288440758815310482211285712819274848744478643590996499 
n3 = 12023158079717019193506148537498877243668782424904061914991928068483879707115315968983829360560644394409575645736275352836086080024994045582242629571839276759393418303915955798990522990081795218822313146157773272844272865701134880180795342597049645358985187689813369428579614193015028249821853347208001645148169449968882591709833452960545988520048722323580338213590245476892223967673180144525106292453573842357322398199104132677638909964034937501684668442732786408572501007756270725934445316827054687741612177409932320532825182104820899546084015733164816993674100635828218335112393003462442685677115798304835391938681 
c1 = 786426913645332991929803636719878643130489430090701482974255190570111407517277263761161970232982615374753982050075781017755721714929721429185828101898786972242994012456972241276851428750970754773002966788642795040933520662931514953660571657013642671173456750800960592586345219252277575624120271330470724245201080094330964145796872211627254805407394764183615099525852600855622089361965086460279057625205099471122036599934609091062009161119885692567925924978687256063116915630947838112126347748759078024890458539541208153526564434483654508834147071166870006117573542198238493913144419569943131642262575848786399020602    
c2 = 14269311999815379511888097227418748728398011595172649708273598243317106830139061994801598925448165045032084910971094414749744701731066555194159863759072739031915833091715422787808666326235589236328864675164322734119047182014621724868200908222400504845559290620275973427127376594365043386362821355037781568524903149101953873768462097165128186788759111090267131443645126715520994688945363059795513931799317608292977574376954729552861360597103229877031117089231816770880909815561950691603994439997197261395452797893557057320175747162837857668062550646101714062365530246698404923128445182100334335447738834779014705114350  
c3 = 3204718091370324153305164801961074660508922478706979436653573192321723216725523523538914956544950802616295043619768261075799875855502834749045520466140056621489305006966280527055668378303630674311102581232313032585389907028715671091914904062961720585667564982641321454541632782484415075257140508738041786400512095949826279576159569786734978545737717138115729502475357594151593143140355121154223614868465202149338507796306863351134218879326031985027900678671697876083351974546516576983143592764763925335805465720148057651958521255276602933604064541840892578409973858867533575728482926007556060584654853884046046420855 
'''

中国剩余定理

exp:

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from Crypto.Util.number import long_to_bytes
from sympy.ntheory.modular import crt
import gmpy2

n1 =   
n2 =   
n3 =   
c1 =     
c2 =   
c3 =  
 
m_list = [n1,n2,n3]
a_list = [c1,c2,c3]

x = crt(m_list,a_list)[0]
e = 17
m = gmpy2.iroot(x,e)
if m[1]:
    print(long_to_bytes(m[0]))
#SICTF{fdc0afb5-1c81-46b9-a28a-241f5f64419d}

MingTianPao

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import binascii
from Crypto.Util.strxor import strxor
from secret import flag, message
# message is a Classic English Story

for i in range(10):
    tmp = (message[i*30:(i+1)*30].encode())
    print(binascii.hexlify(strxor(tmp,flag)).decode())

# 1f2037202a1e6d06353b61263d050a0538493b3018544e14171d2b1c4218
# 3769373b66142f31297f291126410e042b01162d59103a0c005221075013
# 37242c202e1e3f743c36371130410c1e2b491a31574406014505291a550e
# 7f6922742e1a213270372e01264105193004532b1f554e120c1e2a145618
# 7d69143c23156d18392b35183141310e3b49213613590003453a291a555d
# 36273731341e297424372454230e0c0f2c49127f005f020245112718545d
# 26396320295b2531227161273c04430f360d533118444e0f0b1d31554615
# 323d6335660c24373b3a2554350f0a063e05533712101905165e66145f19
# 733e222766152220703e27063508074b300f53371e5d40444735291a555d
# 37283a7432146d2d3f2a6d541808171f330c530d12544e360c162f1b565d

MTP

网上找个脚本

需要改动一点东西

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import Crypto.Util.strxor as xo
import libnum, codecs, numpy as np

def isChr(x):
    if ord('a') <= x and x <= ord('z'): return True
    if ord('A') <= x and x <= ord('Z'): return True
    return False


def infer(index, pos):
    if msg[index, pos] != 0:
        return
    msg[index, pos] = ord(' ')
    for x in range(len(c)):
        if x != index:
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(' ')

def know(index, pos, ch):
    msg[index, pos] = ord(ch)
    for x in range(len(c)):
        if x != index:
            msg[x][pos] = xo.strxor(c[x], c[index])[pos] ^ ord(ch)


dat = []

def getSpace():
    for index, x in enumerate(c):
        res = [xo.strxor(x, y) for y in c if x!=y]
        f = lambda pos: len(list(filter(isChr, [s[pos] for s in res])))
        cnt = [f(pos) for pos in range(len(x))]
        for pos in range(len(x)):
            dat.append((f(pos), index, pos))

#这里传密文
c = [codecs.decode(x.strip().encode(), 'hex') for x in open('0.txt', 'r').readlines()]

msg = np.zeros([len(c), len(c[0])], dtype=int)

getSpace()

dat = sorted(dat)[::-1]
for w, index, pos in dat:
    infer(index, pos)

know(0, 24, 'r')	#如果需要修正， 传入所在被改字母所在行和列，再输入替代的字母,例如alwa s 改为always
know(1, 10, 'h')
know(1, 12, 'r')
know(2, 16, 't')
know(2, 28, 'd')


print('\n'.join([''.join([chr(c) for c in x]) for x in msg]))

key = xo.strxor(c[0], ''.join([chr(c) for c in msg[0]]).encode())
print(key)

Easy_CopperSmith

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from Crypto.Util.number import *
from flag import flag
p = getPrime(512)
q = getPrime(512)
n = p * q
e = 65537
leak = p >> 230
m = bytes_to_long(flag)
c = pow(m,e,n)
print(n)
print(leak)
print(c)
'''
114007680041157617250208809154392208683967639953423906669116998085115503737001019559692895227927818755160444076128820965038044269092587109196557720941716578025622244634385547194563001079609897387390680250570961313174656874665690193604984942452581886657386063927035039087208310041149977622001887997061312418381
6833525680083767201563383553257365403889275861180069149272377788671845720921410137177
87627846271126693177889082381507430884663777705438987267317070845965070209704910716182088690758208915234427170455157948022843849997441546596567189456637997191173043345521331111329110083529853409188141263211030032553825858341099759209550785745319223409181813931086979471131074015406202979668575990074985441810
'''

一般情况下需要已知286位才能求解512的p

但是把small_roots(epsilon = 0.01)改动epsilon参数为0.01可以在已知264位的情况下求解p

题目给出高282，修改参数epsilon即可求解

exp:

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#sage
import gmpy2
from Crypto.Util.number import *

n = 
c = 
e = 65537
high_p = 6833525680083767201563383553257365403889275861180069149272377788671845720921410137177
high_p = high_p << 230

R.<x> = PolynomialRing(Zmod(n))
f = high_p + x
roots = f.small_roots(X = 2^230,beta = 0.4,epsilon=0.01)
if roots:
    p = high_p + int(roots[0])
    q = n // p
    d = gmpy2.invert(e,(p-1)*(q-1))
    m = pow(c,d,n)
    print(long_to_bytes(int(m)))

签到题来咯

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from secret import flag
from  Crypto.Util.number import *

m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
e = getPrime(10)
n = p*q
c1 = pow(114*m+2333,e,n)
c2 = pow(514*m+4555,e,n)
print(f'n = {n}')
print(f'c1 = {c1}')
print(f'c2 = {c2}')
'''
n = 18993579800590288733556762316465854395650778003397512624355925069287661487515652428099677335464809283955351330659278915073219733930542167360381688856732762552737791137784222098296804826261681852699742456526979985201331982720936091963830799430264680941164508709453794113576607749669278887105809727027129736803614327631979056934906547015919204770702496676692691248702461766117271815398943842909579917102217310779431999448597899109808086655029624478062317317442297276087073653945439820988375066353157221370129064423613949039895822016206336117081475698987326594199181180346821431242733826487765566154350269651592993856883
c1 = 3089900890429368903963127778258893993015616003863275300568951378177309984878857933740319974151823410060583527905656182419531008417050246901514691111335764182779077027419410717272164998075313101695833565450587029584857433998627248705518025411896438130004108810308599666206694770859843696952378804678690327442746359836105117371144846629293505396610982407985241783168161504309420302314102538231774470927864959064261347913286659384383565379900391857812482728653358741387072374314243068833590379370244368317200796927931678203916569721211768082289529948017340699194622234734381555103898784827642197721866114583358940604520
c2 = 6062491672599671503583327431533992487890060173533816222838721749216161789662841049274959778509684968479022417053571624473283543736981267659104310293237792925201009775193492423025040929132360886500863823523629213703533794348606076463773478200331006341206053010168741302440409050344170767489936681627020501853981450212305108039373119567034948781143698613084550376070802084805644270376620484786155554275798939105737707005991882264123315436368611647275530607811665999620394422672764116158492214128572456571553281799359243174598812137554860109807481900330449364878168308833006964726761878461761560543284533578701661413931
'''

相关消息攻击，遍历所有e

参考Franklin-Reiter相关消息攻击_Emmaaaaaaaaaa的博客-CSDN博客

exp:

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# sage
import libnum
from Crypto.Util.number import *

n = 
c1 = 
c2 = 
E = []

for i in range(2^9,2^10):
    if isPrime(i):
        E.append(i)
        
def franklinReiter(n,e,c1,c2):
    PR.<x> = PolynomialRing(Zmod(n))
    g1 = (114*x + 2333)^e - c1
    g2 = (514*x + 4555)^e - c2

    def gcd(g1, g2):
        while g2:
            g1, g2 = g2, g1 % g2
        return g1.monic()
    return -gcd(g1, g2)[0]

# print(E)
for e in E:
    m = franklinReiter(n,e,c1,c2)
    flag = long_to_bytes(int(m))
    if b'SICTF' in flag:
        print(flag)
        break
    else:
        continue

small_e

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import libnum
from Crypto.Util.number import *
import uuid
flag="SICTF{"+str(uuid.uuid4())+"}"
m=libnum.s2n(flag)
p=getPrime(1024)
q=getPrime(1024)
n=p*q
e=3
c=pow(m,e,n)
m1=((m>>60)<<60)
print("n=",n)
print("e=",e)
print("c=",c)
print("((m>>60)<<60)=",m1)
print(flag)
'''
n= 23407088262641313744603678186127228163189328033499381357614318160776774708961658114505773173784501557046914457908828086210961235530240151825359345210845219656000760996670856300710703016947799649686427460688236465568188205550456293373157997725204643414082796492333552579250010906010553831060540937802882205118399938918764313169385349293602085310111289583058965780887097301702677087443291977479125263301000328313103296364864396361278863921717374909215078711198899810620522933994481419395021233240234478331179727351050575360886334237633420906629984625441302945112631166021776379103081857393866576659121443879590011160797
e= 3
c= 1584727211980974717747362694412040878682966138197627512650829607105625096823456063149392973232737929737200028676411430124019573130595696272668927725536797627059576270068695792221537212669276826952363636924278717182163166234322320044764324434683614360641636360301452618063418349310497430566465329766916213742181
((m>>60)<<60)= 11658736990073967239197168945911788935424691658202162501032766529463315401599017877851823976178979438592
'''

小e秒出

exp1:

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from Crypto.Util.number import *
import gmpy2

n= 
e= 3
c= 
k = 0


t = gmpy2.iroot(c,e)
if t[1]:
    print(long_to_bytes(t[0]))
else:
    while 1:
        m = gmpy2.iroot(k*n+c,e)
        if m[1]:
            print(m[0])
            print(long_to_bytes(m[0]))
            break
        else:
            k += 1

预期考点应该是m高位泄露

exp:

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#sage
from Crypto.Util.number import long_to_bytes

n= 
e= 3
c= 
m_high= 11658736990073967239197168945911788935424691658202162501032766529463315401599017877851823976178979438592

R.<x> = PolynomialRing(Zmod(n))
m = m_high + x
f = m^e - c
x = f.small_roots(X = 2^60,beta = 0.4)
if x:
	m = m_high + x[0]
	print(long_to_bytes(int(m)))

easy_math

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from secret import flag
from  Crypto.Util.number import *


m = bytes_to_long(flag)
p = getPrime(512)
q = getPrime(512)
n = p * q
e = 65537
hint1 = getPrime(13)*p+getPrime(256)*q
hint2 = getPrime(13)*p+getPrime(256)*q
c = pow(m,e,n)
print(f'n = {n}')
print(f'hint1 = {hint1}')
print(f'hint2 = {hint2}')
print(f'c = {c}')

'''
n = 68123067052840097285002963401518347625939222208495512245264898037784706226045178539672509359795737570458454279990340789711761542570505016930986418403583534761200927746744298082254959321108829717070206277856970403191060311901559017372393931121345743640657503994132925993800497309703877076541759570410784984067
hint1 = 564294243979930441832363430202216879765636227726919016842676871868826273613344463155168512928428069316237289920953421495330355385445649203238665802121198919543532254290185502622234014832349396422316629991217252686524462096711723580
hint2 = 484307144682854466149980416084532076579378210225500554261260145338511061452958092407101769145891750844383042274498826787696953308289632616886162073232218214504005935332891893378072083589751354946391146889055039887781077066257013110
c = 57751903193610662622957432730720223801836323458721550133101805763463060486486266309568004721657732742899781400754207249733137375171400440423755473421971160000575072519031824740691618617905549725344323721903857290320737224300672847773455169809689188843070599176261204013341324705808617411345132933937680951713
'''

$hint_1 = a_1p+b_1q$，$hint_2 = a_2p+b_2q$

则$a_2hint_1 = a_1a_2p+b_1a_2q$，$a_1hint_2 = a_1a_2p + b_2a_1q$

两式相减得$a_2hint_1-a_1hint_2 = (b_1a_2-b_2a_1)q$

于是$q = gcd(n,a_2hint1-a_1hint_2)$，可以做检验

$a_1,a_2$不大，可以爆破

exp:

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from Crypto.Util.number import *
from itertools import product
from tqdm import *
import gmpy2


n = 
hint1 = 
hint2 = 
c = 

hints = [hint1,hint2]
A = []
for i in range(2**12,2**13):
    if isPrime(i):
        A.append(i)

for a1, a2 in tqdm(list(product(A, repeat=2))):
    kq = gmpy2.gcd(a1 * hints[0] - a2 * hints[1], n)
    if 1 < kq < n:
        print('ffind!')
        print("kq=",kq)
        print("a1=",a1)
        print("a2=",a2)
        break

q = kq
p = n // kq
d = gmpy2.invert(65537,(p-1)*(q-1))
m = pow(c, d, n)
flag = long_to_bytes(m)
print(flag)

参考：Challenges_2023_Public/crypto/apbq-rsa-i/solve/solv.py at main · DownUnderCTF/Challenges_2023_Public (github.com)

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