NSSCTF-2nd-2023

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Created At : 2023-08-29 13:46

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Crypto

Crypto

EzRSA

hint用不上，e=3直接开方就好

FunnyEncrypt

✧✡✭
✡✮ ✣✴✯ ✤✶✬✬✱ ✬✤ ✱✦✢✥✮✯✧✧, ✴✬✷✯ ✡✧ ✣✴✯ ✶✡✰✴✣. ✡✣ ❂✢✡✮✰✧ ✩✬✸✤✬✢✣, ✤✦✡✣✴, ✦✮✱ ✩✬✮✤✡✱✯✮✩✯. ✡✣ ✰✡✲✯✧ ✳✧ ✰✳✡✱✦✮✩✯ ★✴✯✮ ★✯ ✦✢✯ ✶✬✧✣, ✦✮✱ ✰✡✲✯✧ ✧✳✷✷✬✢✣ ★✴✯✮ ★✯ ✦✢✯ ✦✤✢✦✡✱. ✦✮✱ ✣✴✯ ✸✬✸✯✮✣ ★✯ ✰✡✲✯ ✳✷ ✴✬✷✯, ★✯ ✰✡✲✯ ✳✷ ✬✳✢ ✶✡✲✯✧. ✣✴✯ ★✬✢✶✱ ★✯ ✶✡✲✯ ✡✮ ✡✧ ✱✡✧✡✮✣✯✰✢✦✣✡✮✰ ✡✮✣✬ ✦ ✷✶✦✩✯ ✬✤ ✸✦✶✡✩✯ ✦✮✱ ✴✦✣✢✯✱, ★✴✯✢✯ ★✯ ✮✯✯✱ ✴✬✷✯ ✦✮✱ ✤✡✮✱ ✡✣ ✴✦✢✱✯✢. ✡✮ ✣✴✡✧ ★✬✢✶✱ ✬✤ ✤✯✦✢, ✴✬✷✯ ✣✬ ✤✡✮✱ ❂✯✣✣✯✢, ❂✳✣ ✯✦✧✡✯✢ ✧✦✡✱ ✣✴✦✮ ✱✬✮✯, ✣✴✯ ✸✬✢✯ ✸✯✦✮✡✮✰✤✳✶ ✶✡✤✯ ✬✤ ✤✦✡✣✴ ★✡✶✶ ✸✦✥✯ ✶✡✤✯ ✸✯✦✮✡✮✰✤✳✶.
✧✬✸✯✣✡✸✯✧ ★✯ ✣✴✡✮✥ ✬✤ ✱✢✯✦✸✧ ✦✧ ✤✦✮✣✦✧✡✯✧ - ✡✣'✧ ✯✦✧✵ ✣✬ ✱✬ ★✴✯✮ ✵✬✳ ✴✦✲✯ ✸✬✮✯✵, ✢✯✮✣, ✦✮✱ ★✬✢✥. ❂✳✣ ✵✬✳ ✩✦✮'✣ ✷✢✯✷✦✢✯ ✵✬✳✢✧✯✶✤ ✦✮✱ ✫✳✸✷ ✬✤✤ ✣✴✯ ✩✶✡✤✤: ✵✬✳ ✧✴✬✳✶✱ ✰✢✬★ ✵✬✳✢ ★✡✮✰✧ ✤✡✢✧✣. ✦ ✶✡✣✣✶✯ ❂✡✣ ✣✬★✦✢✱ ✣✴✯ ✱✢✯✦✸. ✧✣✯✷ ❂✵ ✧✣✯✷. ✣✦✥✯ ✦ ✧✣✯✷ ✤✬✢★✦✢✱. ✦✤✣✯✢ ✦✶✶, ✡✣'✧ ✵✬✳✢ ✸✡✧✧✡✬✮.
✥✯✯✷ ✤✦✡✣✴ ✦✮✱ ✴✬✷✯ ✤✬✢ ✣✴✯ ✤✳✣✳✢✯. ✸✦✥✯ ✵✬✳✢ ✸✬✧✣ ✧✡✮✩✯✢✯ ✱✢✯✦✸✧, ✦✮✱ ★✴✯✮ ✣✴✯ ✬✷✷✬✢✣✳✮✡✣✡✯✧ ✩✬✸✯, ✣✴✯✵ ★✡✶✶ ✤✡✰✴✣ ✤✬✢ ✣✴✯✸. ✡✣ ✸✦✵ ✣✦✥✯ ✦ ✧✯✦✧✬✮ ✬✢ ✸✬✢✯, ❂✳✣ ✣✴✯ ✯✮✱✡✮✰ ★✡✶✶ ✮✬✣ ✩✴✦✮✰✯. ✦✸❂✡✣✡✬✮, ❂✯✧✣, ❂✯✩✬✸✯ ✦ ✢✯✦✶✡✣✵. ✦✮ ✳✮✩✯✢✣✦✡✮ ✤✳✣✳✢✯, ✬✮✶✵ ✬✮✯ ✧✣✯✷ ✦✣ ✦ ✣✡✸✯, ✣✴✯ ✴✬✷✯ ✩✦✮ ✢✯✦✶✡✪✯ ✣✴✯ ✱✢✯✦✸ ✬✤ ✣✴✯ ✴✡✰✴✯✧✣. ★✯ ✸✳✧✣ ✣✢✯✦✧✳✢✯ ✣✴✯ ✱✢✯✦✸, ✣✬ ✷✢✬✣✯✩✣ ✡✣ ✦ ✧✯✦✧✬✮, ✶✯✣ ✡✣ ✡✮ ✣✴✯ ✴✯✦✢✣ ❋✳✡✯✣✶✵ ✰✯✢✸✡✮✦✶.
✬✮✶✵ ★✴✯✮ ✵✬✳ ✳✮✱✯✢✧✣✦✮✱ ✣✴✯ ✣✢✳✯ ✸✯✦✮✡✮✰ ✬✤ ✶✡✤✯ ✩✦✮ ✵✬✳ ✶✡✲✯ ✣✢✳✶✵. ❂✡✣✣✯✢✧★✯✯✣ ✦✧ ✶✡✤✯ ✡✧, ✡✣'✧ ✧✣✡✶✶ ★✬✮✱✯✢✤✳✶, ✦✮✱ ✡✣'✧ ✤✦✧✩✡✮✦✣✡✮✰ ✯✲✯✮ ✡✮ ✣✢✦✰✯✱✵. ✡✤ ✵✬✳'✢✯ ✫✳✧✣ ✦✶✡✲✯, ✣✢✵ ✴✦✢✱✯✢ ✦✮✱ ✣✢✵ ✣✬ ✶✡✲✯ ★✬✮✱✯✢✤✳✶✶✵.
✡ ❂✯✶✡✯✲✯ ✣✴✯✢✯ ✡✧ ✦ ✷✯✢✧✬✮ ★✴✬ ❂✢✡✮✰✧ ✧✳✮✧✴✡✮✯ ✡✮✣✬ ✵✬✳✢ ✶✡✤✯. ✣✴✦✣ ✷✯✢✧✬✮ ✸✦✵ ✴✦✲✯ ✯✮✬✳✰✴ ✣✬ ✧✷✢✯✦✱ ✦✢✬✳✮✱. ❂✳✣ ✡✤ ✵✬✳ ✢✯✦✶✶✵ ✴✦✲✯ ✣✬ ★✦✡✣ ✤✬✢ ✧✬✸✯✬✮✯ ✣✬ ❂✢✡✮✰ ✵✬✳ ✣✴✯ ✧✳✮ ✦✮✱ ✰✡✲✯ ✵✬✳ ✦ ✰✬✬✱ ✤✯✯✶✡✮✰, ✣✴✯✮ ✵✬✳ ✸✦✵ ✴✦✲✯ ✣✬ ★✦✡✣ ✦ ✶✬✮✰ ✣✡✸✯.
✡✮ ✦ ★✬✢✱,✡ ✴✬✷✯ ✵✬✳ ★✡✶✶ ✶✡✥✯ ✩✢✵✷✣✬✰✢✦✷✴✵.✣✴✡✧ ✡✧ ✵✬✳✢ ✤✶✦✰:✮✧✧✩✣✤{✩✢✵✷✣✬_✡✧_✧✬_✡✮✣✯✢✯✧✣✡✮✰_★✴✵_✱✬✮'✣_✵✬✳_✫✬✡✮_✳✧}

按箭头种类的顺序替换为a-z，然后词频分析即可

LatticeLCG

from Crypto.Util.number import *

flag = b'NSSCTF{******************************}'

a = getPrime(512)
seed = getPrime(512)
b = bytes_to_long(flag)
n = getPrime(1024)

e1 = 2333
e2 = 23333
c1 = pow(a,e1,n)
c2 = pow(a,e2,n)

output = []
for i in range(10):
    seed = (a*seed+b)%n
    output.append(seed)

print("c1 = ",c1)
print("c2 = ",c2)
print("output1 = ",output[0])
print("output2 = ",output[1])


e = [getPrime(128) for _ in range(20)]
out = []
m = getPrime(64)

for i in e:
    out.append(pow(m,i,n))

print("e=",e)
print("out=",out)

最后的for循环有一组关系：$out \equiv m^e \pmod n$

知道out、e，格基规约求出m和n。

es = []
cs = []

L = matrix(es).T.augment(matrix.identity(len(es)))
L[:, 0] *= 2 ^ 2048
L = L.LLL()
print(L[0][1:])
print(L[1][1:])
xx = product([ZZ(y) ^ x for x, y in zip(L[0][1:], cs)])
yy = product([ZZ(y) ^ x for x, y in zip(L[1][1:], cs)])
n = gcd(xx.numer() - xx.denom(), yy.numer() - yy.denom())
print(n)

g, x, y = xgcd(es[0], es[1])
m = ZZ(pow(cs[0], x, n) * pow(cs[1], y, n)) % n
print(m)

接着共模求a，最后就是LCG里的未知b的类型，没啥好说的。

Math

from secret import flag
from Crypto.Util.number import *
import gmpy2

length = len(flag)
flag1 = flag[:length//2]
flag2 = flag[length//2:]
e = 65537

m1 = bytes_to_long(flag1)
p = getPrime(512)
q = getPrime(512)
n = p*q
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)

p1 = gmpy2.invert(p,q)
q1 = gmpy2.invert(q,p)
c = pow(m1,e,n)

print("p1=",p1)
print("q1=",q1)
print("c=",c)
print("phi=",phi)

"""
p1= 3020925936342826638134751865559091272992166887636010673949262570355319420768006254977586056820075450411872960532347149926398408063119965574618417289548987
q1= 4671408431692232396906683283409818749720996872112784059065890300436550189441120696235427299344866325968178729053396743472242000658751114391777274910146291
c= 25112054943247897935419483097872905208058812866572413543619256987820739973912338143408907736140292730221716259826494247791605665059462509978370784276523708331832947651238752021415405546380682507724076832547566130498713598421615793975775973104012856974241202142929158494480919115138145558312814378701754511483
phi= 57503658815924732796927268512359220093654065782651166474086873213897562591669139461637657743218269483127368502067086834142943722633173824328770582751298229218384634668803018140064093913557812104300156596305487698041934061627496715082394633864043543838906900101637618600513874001567624343801197495058260716932
"""

m2 = bytes_to_long(flag2)
p = getPrime(1024)
q = getPrime(1024)
n = p * q
c = pow(m2, e, n)
hint = pow(2023 * p + 114514, q, n)
print("n=",n)
print("c=",c)
print("hint=",hint)

"""
n= 12775720506835890504634034278254395430943267336816473660983646973423280986156683988190224391394224069040565587173690009193979401332176772774003070053150665425296356891182224095151626957780349726980433545162004592720236315207871365869074491602494662741551613634958123374477023452496165047922053316939727488269523121920612595228860205356006298829652664878874947173274376497334009997867175453728857230796230189708744624237537460795795419731996104364946593492505600336294206922224497794285687308908233911851722675754289376914626682400586422368439122244417279745706732355332295177737063024381192630487607768783465981451061
c= 11915755246503584850391275332434803210208427722294114071001100308626307947436200730224125480063437044802693983505018296915205479746420176594816835977233647903359581826758195341201097246092133133080060014734506394659931221663322724002898147351352947871411658624516142945817233952310735792476179959957816923241946083918670905682025431311942375276709386415064702578261223172000098847340935816693603778431506315238612938066215726795441606532661443096921685386088202968978123769780506210313106183173960388498229061590976260661410212374609180449458118176113016257713595435899800372393071369403114116302366178240855961673903
hint= 3780943720055765163478806027243965253559007912583544143299490993337790800685861348603846579733509246734554644847248999634328337059584874553568080801619380770056010428956589779410205977076728450941189508972291059502282197067064652703679207594494311426932070873126291964667101759741689303119878339091991064473009603015444698156763131697516348762529243379294719509271792197450290763350043267150173332933064667716343268081089911389405010661267902446894363575630871542572200564687271311946580866369204751787686029541644463829030926902617740142434884740791338666415524172057644794094577876577760376741447161098006698524808
"""

flag分两部分。

part1

$phi = (p-1)\cdot (q-1)$
$p1\cdot p \equiv 1 \pmod q$
$q1\cdot q \equiv 1 \pmod p$
$c \equiv m1^e \pmod n$

已知phi、p1、q1、c

2021年中国能源网络安全大赛 | Lazzaro (lazzzaro.github.io)

"""
alpha = p' * q' - l
beta = l^2 * [(e * d - 1) / s] + q' * l + p' * l - p' * q' - alpha - l^2
i.e.:
beta = l^2 * {[(e * d - 1) / s] - 1} + l * (q' + p') - alpha - p' * q'
if l,s are correct:
    alpha = k * t
    beta = k * (p' - l) + t * (q' - l)
i.e:
"""
import gmpy2
from itertools import product

def alpha_from_pprime_qprime_l(pprime, qprime, l):
    return pprime * qprime - l

def beta_from_pprime_qprime_e_d_l_s_alpha(pprime, qprime, e, d, l, s, alpha):
    temp1 = e * d - 1
    assert temp1 % s == 0
    temp2 = ((temp1 // s) - 1) * l * l
    temp3 = temp2 + l * (pprime + qprime)
    return temp3 - alpha - (pprime * qprime)

def k_t_from_pprime_qprime_l_alpha_beta(pprime, qprime, l, alpha, beta):
    a = pprime - l
    b = -beta
    c = alpha * (qprime - l)
    disc = b * b - 4 * a * c
    assert gmpy2.is_square(disc)
    temp = -b + gmpy2.isqrt(disc)
    assert temp % (2 * a) == 0
    k = temp // (2 * a)
    assert alpha % k == 0
    return k, alpha // k

def brute_k_t_l(pprime, qprime, e, d):
    # l, s = 2, 2

    ss = [s for s in range(e - 100000, e + 1000000) if s != 0 and (e * d - 1) % s == 0]

    for l, s in product(range(1, 5000), ss):
        # print(f'l = {l}, s = {s}')
        try:
            alpha = alpha_from_pprime_qprime_l(pprime, qprime, l)
            beta = beta_from_pprime_qprime_e_d_l_s_alpha(pprime, qprime, e, d, l, s, alpha)
            k, t = k_t_from_pprime_qprime_l_alpha_beta(pprime, qprime, l, alpha, beta)
            return k, t, l

        except AssertionError:
            continue
p1= ...
q1= ...
c= ...
phi= ...

if __name__ == "__main__":
    e = 65537
    fn = phi
    d = gmpy2.invert(e, fn)
    pprime = q1
    qprime = p1

    k, t, l = brute_k_t_l(pprime, qprime, e, d)

    lp, lq = qprime + k, pprime + t
    assert lp % l == 0, lq % l == 0
    p, q = lp // l, lq // l

    assert gmpy2.invert(p, q) == pprime, gmpy2.invert(q, p) == qprime
    assert gmpy2.is_prime(p), gmpy2.is_prime(q)
    N = int(p * q)

    d = int(invert(e,(p-1)*(q-1)))
    m1 = pow(c,d,N)
    print(long_to_bytes(m1))

part2

$hint = (2023p + 114514)^q \pmod n$
$c \equiv m2^e \pmod n$

化简hint得：

$hint = 114514^q \pmod p$

费马小定理逆推：

$114514^n = 114514^{pq} = 114514^{q^p} = 114514^q \pmod p$

故$p = gcd(114514^n\pmod p, n)$

p = gcd(pow(114514,n,n) - hint, n)

然后就是基础RSA。

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