标签：self range 2024 print flag seed DragonKnightCTF import

misc

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用stegsolve发现二维码，扫码关注中学生CTF 发送 Dragon Knight 2024 即可获得flag

神秘文字

一开始想着埃及文，找了半天发现很多都没有对应的字符，都开始摆了。

但是，当我发给gpt后，发现是js混淆类的

直接运行就拿到了压缩包密码了，然后解压就是flag

crypto

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分析之后是每次进行了两次普通的LCG，但其实没有影响

Xn+1=a²*Xn+a*b+b mod m

还是把每个部分当成一个整体计算就可以了

参考资料ctf之lcg算法_ctf lcg-CSDN博客

from Crypto.Util.number import *

"""m = b'flag{********}'
a =  getPrime(247)
b =  getPrime(247)
n =  getPrime(247)

seed = bytes_to_long(m)

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
        self.seed = (self.a * self.seed + self.b) % self.m
        return self.seed

seed = bytes_to_long(m)

output = LCG(seed,a,b,n)

for i in range(getPrime(16)):
    output.generate()

print(output.generate())
print(output.generate())
print(output.generate())
print(output.generate())
print(output.generate())"""

'''
5944442525761903973219225838876172353829065175803203250803344015146870499
141002272698398325287408425994092371191022957387708398440724215884974524650
42216026849704835847606250691811468183437263898865832489347515649912153042
67696624031762373831757634064133996220332196053248058707361437259689848885
19724224939085795542564952999993739673429585489399516522926780014664745253
'''
from gmpy2 import *

s = [5944442525761903973219225838876172353829065175803203250803344015146870499,141002272698398325287408425994092371191022957387708398440724215884974524650,42216026849704835847606250691811468183437263898865832489347515649912153042,67696624031762373831757634064133996220332196053248058707361437259689848885,19724224939085795542564952999993739673429585489399516522926780014664745253]
t = []
l=len(s)
for i in range(1,l):
    t.append(s[i]-s[i-1])

all_n = []
for i in range(1,l-3):
    all_n.append(gcd((t[i+1]*t[i-1]-t[i]*t[i]), (t[i+2]*t[i]-t[i+1]*t[i+1])))

for n in all_n:
    n=abs(n)
    if n==1:
        continue
    
    a=(s[2]-s[1])*invert((s[1]-s[0]),n)%n

    ani=invert(a,n)
    
    b=(s[1]-a*s[0])%n

    seed = (ani*(s[0]-b))%n
    for i in range(2**16):
        if b'flag' in long_to_bytes(seed):
            print(long_to_bytes(seed))
        seed = (ani*(seed-b))%n

#b'flag{Hello_CTF}'

MatrixRSA_Revenge

之前H&NCTF中有一题一样的，具体步骤跟求RSA相同

上次的题H&NCTF 密码出题记录-CSDN博客

只有求phin时不同，下面是参考论文

pangia.pdf (gcsu.edu)

关于phin的问题，之前H&NCTF中糖醋小鸡块师傅有解答一部分

具体是对于求phin其实并没有那么严格

因此，该题可以用上面那题的代码也可以实现

EXP:

#sagemath
from Crypto.Util.number import *
import os
from gmpy2 import *

"""flag = b"DRKCTF{??????????????}" + os.urandom(212)

p = getPrime(120)
q = getPrime(120)
print(f"p = {p}")
print(f"q = {q}")

part = [bytes_to_long(flag[16*i:16*(i+1)]) for i in range(16)]

M = Matrix(Zmod(n),[
    [part[4*i+j] for j in range(4)] for i in range(4)
])

e = 65537
C = M ** e

print(f"C = {list(C)}")"""


p = 724011645798721468405549293573288113   
q = 712853480230590736297703668944546433
c = [(354904294318305224658454053059339790915904962123902870614765704810196137, 307912599668649689143528844269686459695648563337814923172488152872006235, 143644686443811064172873392982322248654471792394264352463341325181752577, 22995887787365556743279529792687264972121816670422146768160153217903088), (111349308911096779758451570594323566987628804920420784718347230085486245, 370237591900013263581099395076767089468466012835217658851568690263421449, 305451886364184428434479088589515273362629589399867618474106045683764897, 454103583344277343974714791669312753685583930212748198341578178464249150), (168497521640129742759262423369385500102664740971338844248603058993335309, 228941893018899960301839898935872289351667488000643221589230804176281482, 340080333594340128998141220817079770261711483018587969623825086357581002, 122922413789905368029659916865893297311475500951645918611759627764993518), (10332477229415731242316540110058798318207430965033579181240340751539101, 238172263316130590821973648893483382211906906298557131324791759298887701, 487586702165464601760230182019344052665496627253691596871783460314632260, 12238020921585443139790088280608644406695242899000592355653073240122626)]
e = 65537
n=p*q
C = Matrix(Zmod(n),c)
order_p = (p**4-p**3)*(p**4-p**2)*(p**4-p)*(p**4-1)
order_q = (q**4-q**3)*(q**4-q**2)*(q**4-q)*(q**4-1)
order = order_p * order_q

d = gmpy2.invert(e,order)
M = C ** d

flag = b""
for i in range(4):
    for j in range(4):
        m = int(M[i,j])
        flag += long_to_bytes(m)
        
print(flag)

EzDES

不是哥们，我都没想到能出来

太抽象了把

from Crypto.Cipher import DES
from Crypto.Util.Padding import pad
from Crypto.Util.number import *
"""from secret import flag,key

key = bytes.fromhex(key)
des = DES.new(key, DES.MODE_ECB)

enc = des.encrypt(pad(flag,64))

print(enc)
"""

a=b't\xe4f\x19\xc6\xef\xaaL\xc3R}\x08;K\xc9\x88\xa6|\nF\xc3\x12h\xcd\xd3x\xc3(\x91\x08\x841\xca\x8b\xc1\x94\xb5\x9f[\xcd\xc6\x9f\xf9\xf6\xca\xf5\x1a\xda\x16\xcf\x89\x154\xa1\xfe\xc5\x16\xcf\x89\x154\xa1\xfe\xc5'
i=1<<56
while 1:
    key=long_to_bytes(i)
    des = DES.new(key, DES.MODE_ECB)
    dec = des.decrypt(a)
    if b'DRKCTF' in dec:
        print(dec)
        break
    i=i+1
    if i>=(1<<64):
        print("no")
        break
#b'DRKCTF{We4k_K3y_1s_V3ry_D4nger0us_In_DES}\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17\x17'

MyEncrypto

首先利用z3解方程求出r，然后就是简单的RSA求出P

from Crypto.Util.number import *
import random

"""def getMyPrime():              
    while True:              
        r = random.getrandbits(64)              
        _p = r**6 -3*r**5 - r**4 + r**2 - r - 6
        _q = r**7 + 2*r**6 + r**5 + 4*r**4 + 7*r**2 + r + 4653
        if isPrime(_p) and isPrime(_q):              
            return _p, _q
                       
def enc(m, n):                        
    return pow(m, 65537, n)              

def LCG(s,a,b,n):
    return (a*s + b) % n

flag=b'DRKCTF{******}'
seed = bytes_to_long(flag)
P = getPrime(512)
a = random.randrange(0,P)
b = random.randrange(0,P)

def Roll():
    global seed
    seed = LCG(seed,a,b,P)
    return seed % 2**16

p, q = getMyPrime()
n = p * q              
enc_P = enc(P, n)
print(f"n = {n}")              
print(f"enc_P = {enc_P}")

out = []
for _ in range(40):
    out.append(Roll())

print(f"a = {a}")
print(f"b = {b}")
print(f"out = {out}")"""

n = 17959692613208124553115435318871530105762927141420294800783695207170608966804977782615874404539156257549097962410144332053383210075663138848832474791712256427111304125146378883542387121684653496644116081809328796925343393644118376497507
enc_P = 17215745298239635988196009014709535403293865406390546681749129213899045156482782458937447412919331336842808052179915132663427715069134196783415529688715962754860563850858056507148936427379551986735103284388996678146580229028006898491552
a = 2759277675743644814124420138047586760905070650864591936190199977578763421196999718749092450720072564786874114432179104175692800471519816376692104515142375
b = 8111240578821759579875175166986910195923820191652867334412871591814076020421468033017946066268237980082938735686222173713853299600396887041341974719819186
out = [39566, 15295, 19818, 55685, 49100, 6517, 2675, 9567, 37243, 40312, 42906, 35874, 44178, 1256, 40298, 29149, 35721, 19886, 63020, 50116, 6844, 39897, 16134, 50218, 44609, 46188, 52712, 49903, 20933, 5441, 19411, 8330, 6904, 39350, 60853, 43446, 35910, 43728, 61533, 13757]

"""from z3 import *
r=Int('r')
s=Solver()
s.add(r<2**64,(r**6 -3*r**5 - r**4 + r**2 - r - 6)*(r**7 + 2*r**6 + r**5 + 4*r**4 + 7*r**2 + r + 4653)==n)
s.check()
m=s.model()
print(m[m.decls()[0]])"""

from gmpy2 import *
r=1248775963213848425
p = r**6 -3*r**5 - r**4 + r**2 - r - 6
q = r**7 + 2*r**6 + r**5 + 4*r**4 + 7*r**2 + r + 4653
e=65537
P=pow(enc_P,invert(e,(p-1)*(q-1)),n)
#10679387699123200522776360035184725927822172255453595568464894884736102462568579313264894449779104030120028056158023524486966766295648236135714849745610937

格密码我是一点没看懂，慢慢来学，但是找了类似的题

格两例12345-CSDN博客

把这个例子EXP中的100换成2**16即可

#sagemath
from sage.all import *
from Crypto.Util.number import bytes_to_long, long_to_bytes


a = 2759277675743644814124420138047586760905070650864591936190199977578763421196999718749092450720072564786874114432179104175692800471519816376692104515142375
b = 8111240578821759579875175166986910195923820191652867334412871591814076020421468033017946066268237980082938735686222173713853299600396887041341974719819186
m = 10679387699123200522776360035184725927822172255453595568464894884736102462568579313264894449779104030120028056158023524486966766295648236135714849745610937
l = [0,39566, 15295, 19818, 55685, 49100, 6517, 2675, 9567, 37243, 40312, 42906, 35874, 44178, 1256, 40298, 29149, 35721, 19886, 63020, 50116, 6844, 39897, 16134, 50218, 44609, 46188, 52712, 49903, 20933, 5441, 19411, 8330, 6904, 39350, 60853, 43446, 35910, 43728, 61533, 13757]
A = [1]
B = [0]
for i in range(1, len(l)-1):
    A.append(a*A[i-1] % m)
    B.append((a*B[i-1]+(a*l[i]+b-l[i+1])*inverse_mod(2**16,m)) % m)
A = A[1:]
B = B[1:]
dim = len(l)
M = matrix(ZZ, dim, dim)
for i in range(dim-2):
    M[i, i] = m
    M[-2, i] = A[i]
    M[-1, i] = B[i]
    M[i, -2] = M[i, -1] = 0
M[-2,-2] =  1
M[-1,-1] = m//(2**16)
 
ll = M.LLL()[0]
l1 = ll[-2]
h1 = l[1]
s1 = l1*2**16+h1
#for s1=a*seed+b%m
seed = ((s1-b)*inverse_mod(a,m))%m
print(seed)
print(long_to_bytes(seed))
#2448362126058501250353008578968898134742331314317704567359511706590811819124428162365725464780212201207933
#b'DRKCTF{a57b63a6-ecf5-46d3-a501-2d359a4fd168}'

OSINT

羡慕群友每一天

谷歌识图，每个点进去看看，最后发现有个视频很像

这里可以发现低下有一个锯齿状，旁边还有树

最后查找摩天轮的英文有点坑了

DRKCTF{美国-佛罗里达州-迈阿密市-Skyviews}

标签：self,range,2024,print,flag,seed,DragonKnightCTF,import
From： https://www.cnblogs.com/naby/p/18214164

DragonKnightCTF 2024