[签到] Vigenere

维吉尼亚，没有 Key，找个在线网站就好了：

Screenshot-202402191617

签到，确信！

p 进制分解：

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n = 8361361624563191168612863710516449028280757632934603412143152925186847721821552879338608951120157631182699762833743097837368740526055736516080136520584848113137087581886426335191207688807063024096128001406698217998816782335655663803544853496060418931569545571397849643826584234431049002394772877263603049736723071392989824939202362631409164434715938662038795641314189628730614978217987868150651491343161526447894569241770090377633602058561239329450046036247193745885174295365633411482121644408648089046016960479100220850953009927778950304754339013541019536413880264074456433907671670049288317945540495496615531150916647050158936010095037412334662561046016163777575736952349827380039938526168715655649566952708788485104126900723003264019513888897942175890007711026288941687256962012799264387545892832762304320287592575602683673845399984039272350929803217492617502601005613778976109701842829008365226259492848134417818535629827769342262020775115695472218876430557026471282526042545195944063078523279341459199475911203966762751381334277716236740637021416311325243028569997303341317394525345879188523948991698489667794912052436245063998637376874151553809424581376068719814532246179297851206862505952437301253313660876231136285877214949094995458997630235764635059528016149006613720287102941868517244509854875672887445099733909912598895743707420454623997740143407206090319567531144126090072331
e = 65537
c = 990174418341944658163682355081485155265287928299806085314916265580657672513493698560580484907432207730887132062242640756706695937403268682912083148568866147011247510439837340945334451110125182595397920602074775022416454918954623612449584637584716343806255917090525904201284852578834232447821716829253065610989317909188784426328951520866152936279891872183954439348449359491526360671152193735260099077198986264364568046834399064514350538329990985131052947670063605611113730246128926850242471820709957158609175376867993700411738314237400038584470826914946434498322430741797570259936266226325667814521838420733061335969071245580657187544161772619889518845348639672820212709030227999963744593715194928502606910452777687735614033404646237092067644786266390652682476817862879933305687452549301456541574678459748029511685529779653056108795644495442515066731075232130730326258404497646551885443146629498236191794065050199535063169471112533284663197357635908054343683637354352034115772227442563180462771041527246803861110504563589660801224223152060573760388045791699221007556911597792387829416892037414283131499832672222157450742460666013331962249415807439258417736128976044272555922344342725850924271905056434303543500959556998454661274520986141613977331669376614647269667276594163516040422089616099849315644424644920145900066426839607058422686565517159251903275091124418838917480242517812783383
k = 7

from Crypto.Util.number import *

R = Zmod(n)["x"]
while True:
    Q = R.quo(R.random_element(k))
    pp = gcd(ZZ(list(Q.random_element() ^ n)[1]), n)
    if pp != 1:
        qq = sum([pp**i for i in range(k)])
        rr = n // (pp * qq)
        assert n == pp * qq * rr
        break
phi = (pp - 1) * (qq - 1) * (rr - 1)
d = pow(e, -1, phi)
m = pow(c, d, n)
print(long_to_bytes(m))

# SICTF{d9428fc7-fa3a-4096-8ec9-191c0a4562ff}

gggcccddd

相关消息攻击，但是 e 很大，所以用 HGCD 加个速：

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import sys

def HGCD(a, b):
    if 2 * b.degree() <= a.degree() or a.degree() == 1:
        return 1, 0, 0, 1
    m = a.degree() // 2
    a_top, a_bot = a.quo_rem(x^m)
    b_top, b_bot = b.quo_rem(x^m)
    R00, R01, R10, R11 = HGCD(a_top, b_top)
    c = R00 * a + R01 * b
    d = R10 * a + R11 * b
    q, e = c.quo_rem(d)
    d_top, d_bot = d.quo_rem(x^(m // 2))
    e_top, e_bot = e.quo_rem(x^(m // 2))
    S00, S01, S10, S11 = HGCD(d_top, e_top)
    RET00 = S01 * R00 + (S00 - q * S01) * R10
    RET01 = S01 * R01 + (S00 - q * S01) * R11
    RET10 = S11 * R00 + (S10 - q * S11) * R10
    RET11 = S11 * R01 + (S10 - q * S11) * R11

    return RET00, RET01, RET10, RET11
    
def GCD(a, b):
    print(a.degree(), b.degree())
    q, r = a.quo_rem(b)
    if r == 0:
        return b
    R00, R01, R10, R11 = HGCD(a, b)
    c = R00 * a + R01 * b
    d = R10 * a + R11 * b
    if d == 0:
        return c.monic()
    q, r = c.quo_rem(d)
    if r == 0:
        return d

    return GCD(d, r)

sys.setrecursionlimit(500000)
n = 71451784354488078832557440841067139887532820867160946146462765529262021756492415597759437645000198746438846066445835108438656317936511838198860210224738728502558420706947533544863428802654736970469313030584334133519644746498781461927762736769115933249195917207059297145965502955615599481575507738939188415191
c1 = 60237305053182363686066000860755970543119549460585763366760183023969060529797821398451174145816154329258405143693872729068255155086734217883658806494371105889752598709446068159151166250635558774937924668506271624373871952982906459509904548833567117402267826477728367928385137857800256270428537882088110496684
c2 = 20563562448902136824882636468952895180253983449339226954738399163341332272571882209784996486250189912121870946577915881638415484043534161071782387358993712918678787398065688999810734189213904693514519594955522460151769479515323049821940285408228055771349670919587560952548876796252634104926367078177733076253
e = 65537
a = 233
b = 9527

R.<x> = PolynomialRing(Zmod(n))
f = x^e - c1
g = (a*x+b)^e - c2
res = GCD(f,g)

print(int(-res.monic().coefficients()[0]))

# SICTF{45115fb2-84d6-4369-88c2-c8c3d72b4c55}

SuperbRSA

e 不互素情况下的共模攻击：

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from Crypto.Util.number import long_to_bytes
from gmpy2 import gcd, gcdext, iroot

g = gcd(e1, e2)
l, x, y = gcdext(e1//g, e2//g)
m = pow(c1, x, n) * pow(c2, y, n) % n
m = iroot(m, g)[0]

print(long_to_bytes(m).decode())

# SICTF{S0_Great_RSA_Have_Y0u_Learned?}

easyLattice

格密码基础题，只不过 Flag 稍微有些大，问题不大：

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rate = 2**256
Ge = Matrix(ZZ, [[1, rate*h], [0, rate*p]])
print(Ge.LLL())
f, g = Ge.LLL()[0]
f, g = abs(f), abs(g)
print(libnum.n2s(int(f)))

# SICTF{e3fea01c-18f3-4638-9544-9201393940a9}

This article is authored by Luoingly and is licensed under CC BY-NC 4.0

Permalink: https://luoingly.top/post/sictf-2023-round3-crypto-writeup/