2025H&NCTF
2025H&NCTFCrypto
lcgp
题目：
from Crypto.Util.number import *
import gmpy2
import random
n = getPrime(1024)
flag = b'H&NCTF{' + str(uuid.uuid4()).encode() + b'}'
flag=bytes_to_long(flag)
e = 2024
c=pow(e, flag, n)
class LCG:
def __init__(self, seed, a, b, m):
self.seed = seed
self.a = a
self.b = b
self.m = m
def generate(self):
self.seed = (self.a * self.seed + self.b) % self.m
return self.seed
lcg = LCG(c, getPrime(256), getPrime(256), getPrime(2048))
random =
print(random)
print("n=",n)
#
#n=604805773885048132038788501528078428693141138274580426531445179173412328238102786863592612653315029009606622583856638282837864213048342883583286440071990592001905867027978355755042060684149344414810835371740304319571184567860694439564098306766474576403800046937218588251809179787769286393579687694925268985445059思路：
LCG | DexterJie'Blog，利用 Gröbner基 解 LCG 问题，可以还原出 c，然后再解一个 DLP 问题
exp：
from Crypto.Util.number import *
import gmpy2
output =
N = 604805773885048132038788501528078428693141138274580426531445179173412328238102786863592612653315029009606622583856638282837864213048342883583286440071990592001905867027978355755042060684149344414810835371740304319571184567860694439564098306766474576403800046937218588251809179787769286393579687694925268985445059
e = 2024
R. = PolynomialRing(ZZ)
f1 = output*a + b - output
f2 = output*a + b - output
f3 = output*a + b - output
f4 = output*a + b - output
F =
# 使用F构建一个理想的Ideal。
ideal = Ideal(F)
# 计算Ideal的Gröbner基I
I = ideal.groebner_basis()
a = ZZ(-I.univariate_polynomial()(0))
b = ZZ(-I.univariate_polynomial()(0))
n = ZZ(I)
print(a)
print(b)
print(n)
enc = (output - b) * inverse(a,n) % n
m = discrete_log(enc ,mod(e, N))
print(long_to_bytes(m))flag：H&NCTF{7ecf4c8c-e6a5-45c7-b7de-2fecc31d8511}
哈基coke
题目：
import matplotlib.pyplot as plt
import cv2
import numpy as np
from PIL import Image
def arnold_encode(image, shuffle_times, a, b):
""" Arnold shuffle for rgb image
Args:
image: input original rgb image
shuffle_times: how many times to shuffle
Returns:
Arnold encode image
"""
arnold_image = np.zeros(shape=image.shape)
h, w = image.shape, image.shape
N = h
for time in range(shuffle_times):
for ori_x in range(h):
for ori_y in range(w):
new_x = (1*ori_x + b*ori_y)% N
new_y = (a*ori_x + (a*b+1)*ori_y) % N
arnold_image = image
image = np.copy(arnold_image)
cv2.imwrite('en_flag.png', arnold_image, )
return arnold_image
img = cv2.imread('coke.png')
arnold_encode(img,6,9,1)思路：
一个猫脸变换的加密，直接解密即可
import matplotlib.pyplot as plt
import cv2
import numpy as np
from PIL import Image
def arnold_decode(image, shuffle_times, a, b):
""" decode for rgb image that encoded by Arnold
Args:
image: rgb image encoded by Arnold
shuffle_times: how many times to shuffle
Returns:
decode image
"""
# 1:创建新图像
decode_image = np.zeros(shape=image.shape, dtype=np.uint8)
# 2：计算N
h, w = image.shape, image.shape
N = h# 或N=w
# 3：遍历像素坐标变换
for time in range(shuffle_times):
for ori_x in range(h):
for ori_y in range(w):
# 按照公式坐标变换
new_x = ((a * b + 1) * ori_x + (-b) * ori_y) % N
new_y = ((-a) * ori_x + ori_y) % N
decode_image = image
image = np.copy(decode_image)
cv2.imwrite('flag.png', decode_image, )
return decode_image
image = cv2.imread('en_flag.png')
arnold_decode(image, 6, 9, 1)
# H&NCTF{haji_coke_you_win}flag：H&NCTF{haji_coke_you_win}
数据处理
题目：
from Crypto.Util.number import bytes_to_long
import random
flag = b"H&NCTF{}"
btl = str(bytes_to_long(flag))
lowercase = '0123456789'
uppercase = '7***4****5'
table = ''.maketrans(lowercase, uppercase)
new_flag = btl.translate(table)
n = 2 ** 512
m = random.randint(2, n - 1) | 1
c = pow(m, int(new_flag), n)
print('m = ' + str(m))
print('c = ' + str(c))
# m = 5084057673176634704877325918195984684237263100965172410645544705367004138917087081637515846739933954602106965103289595670550636402101057955537123475521383
# c = 2989443482952171039348896269189568991072039347099986172010150242445491605115276953489889364577445582220903996856271544149424805812495293211539024953331399思路：
先根据离散对数求出 new_flag ，然后再爆破 uppercase，再去逆推爆破出 table ，最后还原 flag
from Crypto.Util.number import *
import gmpy2
import itertools
m = 5084057673176634704877325918195984684237263100965172410645544705367004138917087081637515846739933954602106965103289595670550636402101057955537123475521383
c = 2989443482952171039348896269189568991072039347099986172010150242445491605115276953489889364577445582220903996856271544149424805812495293211539024953331399
n = 2 ** 512
new_flag = discrete_log(c, mod(m, n))
num = '0123456789'
lowercase = '0123456789'
# uppercase = '2***7****0'
# 生成所有可能的替换表
for i in itertools.product(num, repeat=3):
for j in itertools.product(num, repeat=4):
uppercase = '7' + ''.join(i) + '4' + ''.join(j) + '5'
table = ''.maketrans(uppercase, lowercase)
m = str(new_flag).translate(table)
flag = long_to_bytes(int(m))
if b'H&NCTF{' in flag and b'}' in flag:
print(f'uppercase = {uppercase}, flag = {flag}')
flag：H&NCTF{cut_cut_rrioajtfijrwegeriogjiireigji}
three vertical lines
题目：
from Crypto.Util.number import *
from secret import flag
from rsa.prime import getprime
while(1):
p=getprime(256)
q=getprime(256)
if isPrime(3*p**5+4*q**5):
print(3*p**5+4*q**5)
break
e = 65537
print(pow(bytes_to_long(flag), e, p * q))
#72063558451087451183203801132459543552092564094711815404066471440396765744526854383117910805713050240067432476705168314622044706081669935956972031037827580519320550326077291392722314265758802332280697884744792689996718961355845963752788234205565249205191648439412084543163083032775054018324646541875754706761793307667356964825613429368358849530455220484128264690354330356861777561511117
#2864901454060087890623075705953001126417241189889895476561381971868301515757296100356013797346138819690091860054965586977737630238293536281745826901578223思路：
审计代码得：
\
那么：
\
两边同时开 5 次方，有：
\
先求出 \(t\) ，因为 \(t^5 \equiv -\frac{4}{3} \quad (\mathrm{mod~} k)\)。也就是说，我们需要在模\(k\) 下计算 \(-\frac{4}{3}\) 的逆元，然后找到一个 \(t\) 使得它的五次方等于这个值。
根据模运算的定义，这等价于存在整数 \(s\) 使得
\
为了消除分数，两边同时乘以 3：
\
然后求出 \(t ^ 5\)，我们要先找到 3 在模 \(k\) 下的逆元，假设位 \(x = \text { gmpy2.invert}(3, k)\)：则有
\
然后就能求出 \(t^5\\mathrm{mod~} k\)的值
\
然后就是模 \(k\) 下寻找五次方根，利用 sage 就行
R = Zmod(n)
roots = []
try:
roots = R(c).nth_root(5, all=True)
print("Roots:", roots)
except ValueError:
print("No fifth roots found.")
exit()
求出 t 后，有：
\
然后造格，并应用 LLL 算法进行格基规约，找到满足条件的 p，q
\[\left(\begin{array}{ll}1 & p\end{array}\right)\left(\begin{array}{ll}1 & t \\0 & k\end{array}\right)=\left(\begin{array}{ll}q & p\end{array}\right)\]
exp：
from Crypto.Util.number import *
n = 72063558451087451183203801132459543552092564094711815404066471440396765744526854383117910805713050240067432476705168314622044706081669935956972031037827580519320550326077291392722314265758802332280697884744792689996718961355845963752788234205565249205191648439412084543163083032775054018324646541875754706761793307667356964825613429368358849530455220484128264690354330356861777561511117
ciphertext = 2864901454060087890623075705953001126417241189889895476561381971868301515757296100356013797346138819690091860054965586977737630238293536281745826901578223
e = 65537
# 计算 -4/3 mod n
c = (-4) * inverse_mod(3, n) % n
# 在模n下寻找五次方根
R = Zmod(n)
roots = []
try:
roots = R(c).nth_root(5, all=True)
except ValueError:
print("No fifth roots found.")
exit()
found = False
for t in roots:
t = int(t)
# 构造格
M = matrix(ZZ, [, ])
L = M.LLL()
# 检查每一行向量
for row in L:
q, p = row
# 取绝对值处理可能的负值
q_abs = abs(q)
p_abs = abs(p)
if q_abs == 0 or p_abs == 0:
continue
# 检查是否为素数且满足方程
if is_prime(q_abs) and is_prime(p_abs) and 3*p_abs**5 + 4*q_abs**5 == n:
print(f"Found p = {p_abs}, q = {q_abs}")
found = True
break
if found:
break
if not found:
print("Failed to find p and q.")
exit()
N = p_abs * q_abs
phi = (p_abs - 1) * (q_abs - 1)
d = inverse_mod(e, phi)
plaintext = pow(ciphertext, d, N)
print("flag:", long_to_bytes(plaintext))flag：H&NCTF{You_learned_the_code_well}
为什么出题人的rsa总是ez
题目：
#part 1
def pad(flag, bits=1024):
pad = os.urandom(bits//8 - len(flag))
return int.from_bytes(flag + pad, "big")
p = random_prime(2**1024)
q = random_prime(2**1024)
a = randint(0, 2**1024)
b = randint(0, 2**1024)
n = p * q
e = 0x10001
flag = b''
m = pad(flag)
assert m < n
c = pow(m, e, n)
print(f"c={c}")
print(f"n={n}")
print(f"h1={p + b * q}")
print(f"h2={a * p + q}")
# c=13148687178480196374316468746303529314940770955906554155276099558796308164996908275540972246587924459788286109602343699872884525600948529446071271042497049233796074202353913271513295267105242313572798635502497823862563815696165512523074252855130556615141836416629657088666030382516860597286299687178449351241568084947058615139183249169425517358363928345728230233160550711153414555500038906881581637368920188681358625561539325485686180307359210958952213244628802673969397681634295345372096628997329630862000359069425551673474533426265702926675667531063902318865506356674927615264099404032793467912541801255735763704043
# n=13718277507497477508850292481640653320398820265455820215511251843542886373380880887850571647060788265498378060163112689840208264538965960596605641194331300743676780910818492860412739541418029075802834265712602393103809065720527365081016381358333378953245379751008531500896923727040455566953960991908174586311899809864209624888469263612475732913062035036254077225370843701146080145441104733074178115602425412116325647598625157922655504918118208783230138448694045386019901732846478340735331718476554208157393418221315041837392020742062275999319586357229583509788489495876723122993592623230858393165458733055504467513549
# h1=6992022576367328281523272055384380182550712894467837916200781058620282657859189270338635886912232754034211897894637971546032107000253692739473463119025570291091085702056938901846349325941043398928197991115231668917435951127329817379935880511925882734157491821315858319170121031835598580384038723788681860763814776365440362143661999054338470989558459179388468943933975861549233231199667742564080001256192881732567616103760815633265325456143601649393547666835326272408622540044065067528568675569233240785553062685974593620235466519632833169291153478793523397788719000334929715524989845012633742964209311952378479134661
# h2=16731800146050995761642066586565348732313856101572403535951688869814016691871958158137790504490910445304384109605408840493227057830017039824412834989258703833576252634055087138315434304691218949240382395879124201923060510497916818961571111218224960267593032380037212325935576750663442553781924370849537501656957488833521657563900462052017695599020610911371304659875887924695896434699048696392210066253577839887826292569913713802634067508141124685789817330268562127695548527522031774601654778934513355315628270319037043809972087930951609429846675450469414212384044849089372435124609387061864545559812994515828333828939
#part 2
from Crypto.Util.number import *
from gmpy2 import *
a = random_prime()
b = random_prime()
g = random_prime()
h = 2*g*a*b+a+b
while not is_prime(h):
a = random_prime()
b = random_prime()
g = random_prime()
h = 2*g*a*b+a+b
N = 2*h*g+1
e from part1's flag
flag=b''
c=pow(bytes_to_long(flag),e,N)
print(N)
print(g)
print(c)
#N=10244621233521168199001177069337072125430662416754674144307553476569744623474797179990380824494968546110022341144527766891662229403969035901337876527595841503498459533492730326942662450786522178313517616168650624224723066308178042783540825899502172432884573844850572330970359712379107318586435848029783774998269247992706770665069866338710349292941829996807892349030660021792813986069535854445874069535737849684959397062724387110903918355074327499675776518032266136930264621047345474782910332154803497103199598761422179303240476950271702406633802957400888398042773978322395227920699611001956973796492459398737390290487
#g=2296316201623391483093360819129167852633963112610999269673854449302228853625418585609211427788830598219647604923279054340009043347798635222302374950707
#c=7522161394702437062976246147354737122573350166270857493289161875402286558096915490526439656281083416286224205494418845652940140144292045338308479237214749282932144020368779474518032067934302376430305635297260147830918089492765917640581392606559936829974748692299762475615766076425088306609448483657623795178727831373194757182797030376302086360751637238867384469269953187938304369668436238848537646544257504724753333177938997524154486602644412199535102323238852958634746165559537630341890450666170836721803871120344373143081664567068672230842855208267929484000179260292518351155693154372172449820053764896414799137097思路：
第一部分：
就是强网杯 2024-apbq 这题 2024-强网杯-wp-crypto | 糖醋小鸡块的blog
已知：
\
exp：
from Crypto.Util.number import *
c = 13148687178480196374316468746303529314940770955906554155276099558796308164996908275540972246587924459788286109602343699872884525600948529446071271042497049233796074202353913271513295267105242313572798635502497823862563815696165512523074252855130556615141836416629657088666030382516860597286299687178449351241568084947058615139183249169425517358363928345728230233160550711153414555500038906881581637368920188681358625561539325485686180307359210958952213244628802673969397681634295345372096628997329630862000359069425551673474533426265702926675667531063902318865506356674927615264099404032793467912541801255735763704043
n = 13718277507497477508850292481640653320398820265455820215511251843542886373380880887850571647060788265498378060163112689840208264538965960596605641194331300743676780910818492860412739541418029075802834265712602393103809065720527365081016381358333378953245379751008531500896923727040455566953960991908174586311899809864209624888469263612475732913062035036254077225370843701146080145441104733074178115602425412116325647598625157922655504918118208783230138448694045386019901732846478340735331718476554208157393418221315041837392020742062275999319586357229583509788489495876723122993592623230858393165458733055504467513549
h2 = 6992022576367328281523272055384380182550712894467837916200781058620282657859189270338635886912232754034211897894637971546032107000253692739473463119025570291091085702056938901846349325941043398928197991115231668917435951127329817379935880511925882734157491821315858319170121031835598580384038723788681860763814776365440362143661999054338470989558459179388468943933975861549233231199667742564080001256192881732567616103760815633265325456143601649393547666835326272408622540044065067528568675569233240785553062685974593620235466519632833169291153478793523397788719000334929715524989845012633742964209311952378479134661
h1 = 16731800146050995761642066586565348732313856101572403535951688869814016691871958158137790504490910445304384109605408840493227057830017039824412834989258703833576252634055087138315434304691218949240382395879124201923060510497916818961571111218224960267593032380037212325935576750663442553781924370849537501656957488833521657563900462052017695599020610911371304659875887924695896434699048696392210066253577839887826292569913713802634067508141124685789817330268562127695548527522031774601654778934513355315628270319037043809972087930951609429846675450469414212384044849089372435124609387061864545559812994515828333828939
e = 0x10001
brute = 2
for i in range(2^brute):
for j in range(2^brute):
L = Matrix(ZZ, [
,
,
,
])
L[:,-1:] *= n
res = L.LLL()
p = 2^brute*abs(res)+i
if(n % p == 0):
print(p)
q = n//p
phi = (p-1)*(q-1)
d = inverse_mod(e, phi)
print(long_to_bytes(pow(c, d, n)))
print(pow(c, d, n))flag{e_is_xevaf-cityf-fisof-ketaf-metaf-disef-nuvaf-cysuf-dosuf-getuf-cysuf-dasix,bubbleBabble}
古典密码 bubbleBabble 解一下得到 e
xevaf-cityf-fisof-ketaf-metaf-disef-nuvaf-cysuf-dosuf-getuf-cysuf-dasix
第二部分
就是 Common Prime RSA 笔记 | 独奏の小屋，这里我不知道为啥不能直接在这个脚本中直接去计算 rsa 一直解不出来，想了半天不知道为啥，害，最后要单独把 p，q 取出来才能解密，参考 强网杯 2024-EasyRSA2024-强网杯-wp-crypto | 糖醋小鸡块的blog
from sage.groups.generic import bsgs
from Crypto.Util.number import *
from sage.all import ZZ, Zmod, ceil
import gmpy2
N=10244621233521168199001177069337072125430662416754674144307553476569744623474797179990380824494968546110022341144527766891662229403969035901337876527595841503498459533492730326942662450786522178313517616168650624224723066308178042783540825899502172432884573844850572330970359712379107318586435848029783774998269247992706770665069866338710349292941829996807892349030660021792813986069535854445874069535737849684959397062724387110903918355074327499675776518032266136930264621047345474782910332154803497103199598761422179303240476950271702406633802957400888398042773978322395227920699611001956973796492459398737390290487
g=2296316201623391483093360819129167852633963112610999269673854449302228853625418585609211427788830598219647604923279054340009043347798635222302374950707
c=7522161394702437062976246147354737122573350166270857493289161875402286558096915490526439656281083416286224205494418845652940140144292045338308479237214749282932144020368779474518032067934302376430305635297260147830918089492765917640581392606559936829974748692299762475615766076425088306609448483657623795178727831373194757182797030376302086360751637238867384469269953187938304369668436238848537646544257504724753333177938997524154486602644412199535102323238852958634746165559537630341890450666170836721803871120344373143081664567068672230842855208267929484000179260292518351155693154372172449820053764896414799137097
e = 81733668723981020451323
print(N.bit_length())
print(g.bit_length())
nbits = N.bit_length()
gamma = 500/(1024*2)
cbits = ceil(nbits * (0.5 - 2 * gamma))
M = (N - 1) // (2 * g)
u = M // (2 * g)
v = M - 2 * g * u
GF = Zmod(N)
x = GF(2)
y = x ** (2 * g)
c = bsgs(y, y ** u, (2**(cbits - 1), 2**(cbits + 1)))
ab = u - c
apb = v + 2 * g * c
# 定义多项式并求解根
P.<x> = ZZ[]
f = x^2 - apb * x + ab
a_roots = f.roots()
if a_roots:
a, b = a_roots, a_roots
p = 2 * g * a + 1
q = 2 * g * b + 1
print(p)
print(q)
assert p * q == N
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))from Crypto.Util.number import long_to_bytes, isPrime
import gmpy2
p=114223230692329221233593176873809300402703143063931397822580003361438712054166147412606967205124955861521969519047697209787311806459638088623102779532442176190528211946769436968799761724667716082655997119572158947882963880778970164432326393480858337150525471346874196183962785321409039116916910681846888186947
q=89689471847596377741222144429998037990667180177357564376140367776414308387012923411995322262858328353318533567758158657166891766790825298883408932478365300323143798943782413227199460096726803837011408747331311614322911225425264615177532865042145570988661750629687056545931523546752474174278821175282246381821
c=7522161394702437062976246147354737122573350166270857493289161875402286558096915490526439656281083416286224205494418845652940140144292045338308479237214749282932144020368779474518032067934302376430305635297260147830918089492765917640581392606559936829974748692299762475615766076425088306609448483657623795178727831373194757182797030376302086360751637238867384469269953187938304369668436238848537646544257504724753333177938997524154486602644412199535102323238852958634746165559537630341890450666170836721803871120344373143081664567068672230842855208267929484000179260292518351155693154372172449820053764896414799137097
n=p*q
e=81733668723981020451323
phi=(p-1)*(q-1)
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))flag：flag{I wish you success in your cryptography career}
ez-factor
题目：
from Crypto.Util.number import *
import uuid
rbits = 248
Nbits = 1024
p = getPrime(Nbits // 2)
q = getPrime(Nbits // 2)
N = p * q
r = getPrime(rbits)
hint = getPrime(Nbits // 2) * p + r
R = 2^rbits
e=0x10001
n=p*q
phi=(p-1)*(q-1)
flag = b'H&NCTF{' + str(uuid.uuid4()).encode() + b'}'
m=bytes_to_long(flag)
c=pow(m,e,n)
print("N=",N)
print("hint=",hint)
print(c)
N= 155296910351777777627285876776027672037304214686081903889658107735147953235249881743173605221986234177656859035013052546413190754332500394269777193023877978003355429490308124928931570682439681040003000706677272854316717486111569389104048561440718904998206734429111757045421158512642953817797000794436498517023
hint= 128897771799394706729823046048701824275008016021807110909858536932196768365642942957519868584739269771824527061163774807292614556912712491005558619713483097387272219068456556103195796986984219731534200739471016634325466080225824620962675943991114643524066815621081841013085256358885072412548162291376467189508
c=32491252910483344435013657252642812908631157928805388324401451221153787566144288668394161348411375877874802225033713208225889209706188963141818204000519335320453645771183991984871397145401449116355563131852618397832704991151874545202796217273448326885185155844071725702118012339804747838515195046843936285308思路：
已知：
\
其中 k 是 512 位，r 只有 248 位，r 远远⼩于 hint，那么有
\
r是⼀个小根，使⽤ CopperSmith 慢慢调参数即可，这里一开始 epsilon 开到 0.01 很慢，然后慢慢加 0.015 会快一点，这里也可以使用 flatter 加速⼀下（会快好多，这里同时把 epsilon 设置成 0.015 只需要 3 秒），用大佬的脚步即可 （SEETF 2023 WriteUps | 廢文集中區）
exp：
from Crypto.Util.number import *
import gmpy2
N = 155296910351777777627285876776027672037304214686081903889658107735147953235249881743173605221986234177656859035013052546413190754332500394269777193023877978003355429490308124928931570682439681040003000706677272854316717486111569389104048561440718904998206734429111757045421158512642953817797000794436498517023
hint = 128897771799394706729823046048701824275008016021807110909858536932196768365642942957519868584739269771824527061163774807292614556912712491005558619713483097387272219068456556103195796986984219731534200739471016634325466080225824620962675943991114643524066815621081841013085256358885072412548162291376467189508
c = 32491252910483344435013657252642812908631157928805388324401451221153787566144288668394161348411375877874802225033713208225889209706188963141818204000519335320453645771183991984871397145401449116355563131852618397832704991151874545202796217273448326885185155844071725702118012339804747838515195046843936285308
e = 0x10001
R = 2^248 # r 的上界
P.<x> = PolynomialRing(Zmod(N))
f = hint - x
f = f.monic()
r = f.small_roots(X=R, beta=0.495, epsilon=0.015)
print(r)
p = gmpy2.gcd(int(hint - r), int(N))
q = int(N)// p
phi = (p - 1) * (q - 1)
d = inverse_mod(e, int(phi))
m = pow(c, d, N)
print(long_to_bytes(m))
from sage.all import *from Crypto.Util.number import *from subprocess import check_outputfrom re import findallimport gmpy2def flatter(M): # compile https://github.com/keeganryan/flatter and put it in $PATH z = "[[" + "]\n[".join(" ".join(map(str, row)) for row in M) + "]]" ret = check_output(["flatter"], input=z.encode()) return matrix(M.nrows(), M.ncols(), map(int, findall(b"-?\\d+", ret)))def small_roots(self, X=None, beta=1.0, epsilon=None, **kwds): from sage.misc.verbose import verbose from sage.matrix.constructor import Matrix from sage.rings.real_mpfr import RR N = self.parent().characteristic() if not self.is_monic(): raise ArithmeticError("Polynomial must be monic.") beta = RR(beta) if beta1.0: raise ValueError("0.0 < beta
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