fizzbuzz100
题目代码：
#!/usr/local/bin/python
from Crypto.Util.number import *
from os import urandom
flag = open("flag.txt", "rb").read()
flag = bytes_to_long(urandom(16) + flag + urandom(16))
p = getPrime(512)
q = getPrime(512)
n = p * q
e = 0x10001
d = pow(e, -1, (p-1)*(q-1))
assert flag < n
ct = pow(flag, e, n)
print(f"{n = }")
print(f"{e = }")
print(f"{ct = }")
while True:
ct = int(input("> "))
pt = pow(ct, d, n)
out = ""
if pt == flag:
exit(-1)
if pt % 3 == 0:
out += "Fizz"
if pt % 5 == 0:
out += "Buzz"
if not out:
out = pt
print(out)
典型的选择密文攻击,由于对解密过后的pt有限制，直接写脚本一直传输各种我们所构造的ct，直到回显数字过来，然后手动解密一下即可
from Crypto.Util.number import *
from pwn import *
# a=remote("be.ax",31100)
# l=a.recv().split(b'=')
# print(l)
# while True:
# X = getPrime(10)
# c = int(l[3][:-2])
# n = int(l[1][:-2])
# Y = (c * (X ** 65537)) % n
# a.sendline(str(Y).encode())
# print(X)
# print(Y)
# print(a.recvline())
n=109235908866342288271346584662343329621605469702598068759472999551170947232996590557964241534101853321789427925631360625471407190264507981047481840604577029503316535875248549400561470650694604159439358769594819721166974367763712018514398742922521556119264931745691626361879268890616276343169833999231861189833
c=54154300348903111005066683187322110240325256042602814495907789220256344603932965149221705493685749178531643137134769583336652306519327753298004337131622195496840709892914894693840688844013339917136532406319553405215428628337555863105880457193242611315359206147315741465280238895273742546724024586290588394695
w=907
y=26641182241400186140764833732531788410188640530460941989584702007887972922089308897736657230632495490138397006968382235153358170724154868714588801750507397028728784901215019435825808542306682388514239090900283743849756631160550184102438364341973285800800919002105937122573558829974207210810107339187277577768
ct=1617872921486161438959529626097492037660629295223293116750255222467849722125817349217082369174984216697864465452996697846569528891027592080706141175012700498085383500035413467709314885420625836615662777908956038879208002465859460678671983566645413919704836898065353721740490145081034
flag=ct*inverse(w,n)%n
print(long_to_bytes(flag))
#corctf{h4ng_0n_th15_1s_3v3n_34s13r_th4n_4n_LSB_0r4cl3...4nyw4y_1snt_f1zzbuzz_s0_fun}
eyes
题目代码：
from Crypto.Util.number import bytes_to_long, getPrime
# my NEW and IMPROVED secret sharing scheme!! (now with multivariate quadratics)
with open('flag.txt', 'rb') as f:
flag = f.read()
s = bytes_to_long(flag)
p = getPrime(len(bin(s)))
print(p)
F = GF(p)
N = 1024
conv = lambda n: matrix(F, N, 1, [int(i) for i in list(bin(n)[2:][::-1].ljust(N, '0'))])
A = random_matrix(F, N, N)
for i in range(0, N):
for j in range(0, i):
A[i, j] = 0
B = random_matrix(F, N, 1)
C = matrix(F, [F(s)])
fn = lambda x: (x.T * A * x + B.T * x + C)[0][0]
L = []
for i in range(7):
L.append(fn(conv(i + 1)))
print(L)
这个题目看起来很复杂，其实实质很简单，通过观察conv矩阵可以发现可以将每一次计算出的L用A,B,C里面的元素表达出来，具体表达如下：
通过等式变形可以得到：c=(L[6]-(L[2]+L[4]+L[5]-2*(L[0]+L[1]+L[3]))-(L[0]+L[1]+L[3]))%p
我们便可以轻松求出flag了
from Crypto.Util.number import *
p=1873089703968291141600166892623234932796169766648225659075834963115683566265697596115468506218441065194050127470898727249982614285036691594726454694776985338487833409983284911305295748861807972501521427415609
L=[676465814304447223312460173335785175339355609820794166139539526721603814168727462048669021831468838980965201045011875121145342768742089543742283566458551844396184709048082643767027680757582782665648386615861, 1472349801957960100239689272370938102886275962984822725248081998254467608384820156734807260120564701715826694945455282899948399224421878450502219353392390325275413701941852603483746312758400819570786735148132, 202899433056324646894243296394578497549806047448163960638380135868871336000334692955799247243847240605199996942959637958157086977051654225700427599193002536157848015527462060033852150223217790081847181896018, 1065982806799890615990995824412253076607488063240855100580513221962298598002468338823225586171107539104635808108356492123167315175110515086192932230998426512947581115358738651206273178867911944034690138825583, 1676559204037482856674710667663849447914859348633288513196735253002541076530170853584406282605482862202276451646974549657672382936948091649764874334064431407644457518190694888175499630744741620199798070517691, 13296702617103868305327606065418801283865859601297413732594674163308176836719888973529318346255955107009306239107173490429718438658382402463122134690438425351000654335078321056270428073071958155536800755626, 1049859675181292817835885218912868452922769382959555558223657616187915018968273717037070599055754118224873924325840103339766227919051395742409319557746066672267640510787473574362058147262440814677327567134194]
F = GF(p)
N = 1024
conv = lambda n: matrix(F, N, 1, [int(i) for i in list(bin(n)[2:][::-1].ljust(N, '0'))])
c=(L[6]-(L[2]+L[4]+L[5]-2*(L[0]+L[1]+L[3]))-(L[0]+L[1]+L[3]))%p
print(long_to_bytes(c))
#corctf{mind your ones and zeroes because zero squared is zero and one squared is one}
cbc
题目代码：
import random
def random_alphastring(size):
return "".join(random.choices(alphabet, k=size))
def add_key(key, block):
ct_idxs = [(k_a + pt_a) % len(alphabet) for k_a, pt_a in zip([alphabet.index(k) for k in key], [alphabet.index(pt) for pt in block])]
return "".join([alphabet[idx] for idx in ct_idxs])
def cbc(key, plaintext):
klen = len(key)
plaintext = pad(klen, plaintext)
iv = random_alphastring(klen)
blocks = [plaintext[i:i+klen] for i in range(0, len(plaintext), klen)]
prev_block = iv
ciphertext = ""
for block in blocks:
block = add_key(prev_block, block)
prev_block = add_key(key, block)
ciphertext += prev_block
return iv, ciphertext
def pad(block_size, plaintext):
plaintext += "X" * (-len(plaintext) % block_size)
return plaintext
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
bs = 16
message = open("message.txt").read().upper()
message = "".join([char for char in message if char in alphabet])
flag = open("flag.txt").read()
flag = flag.lstrip("corctf{").rstrip("}")
message += flag
assert all([char in alphabet for char in message])
key = random_alphastring(bs)
iv, ct = cbc(key, pad(bs, message))
print(f"{iv = }")
print(f"{ct = }")
很有趣的一个题目，我们可以先用ct-iv这样可以获得pt+key，由于key不知道，但是我们可以把pt+key当作维吉尼亚加密过后的密文，由于文本长度很长，直接使用在线工具对其分析就可以还原出flag：https://www.guballa.de/vigenere-solver
iv = 'RLNZXWHLULXRLTNP'
ct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
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"
bs = 16
import random
def random_alphastring(size):
return "".join(random.choices(alphabet, k=size))
def add_key(key, block):
ct_idxs = [(k_a + pt_a) % len(alphabet) for k_a, pt_a in
zip([alphabet.index(k) for k in key], [alphabet.index(pt) for pt in block])]
return "".join([alphabet[idx] for idx in ct_idxs])
def sub_key(key, block):
ct_idxs = [(pt_a - k_a) % len(alphabet) for k_a, pt_a in
zip([alphabet.index(k) for k in key], [alphabet.index(pt) for pt in block])]
return "".join([alphabet[idx] for idx in ct_idxs])
def cbc(key, plaintext):
klen = len(key)
plaintext = pad(klen, plaintext)
iv = random_alphastring(klen)
blocks = [plaintext[i:i + klen] for i in range(0, len(plaintext), klen)]
prev_block = iv
ciphertext = ""
for block in blocks:
block = add_key(prev_block, block)
prev_block = add_key(key, block)
ciphertext += prev_block
return iv, ciphertext
def decode(key, ciphertext):
klen = len(key)
blocks = [ciphertext[i:i + klen] for i in range(0, len(ciphertext), klen)]
plaintext = ""
for block in blocks:
block = sub_key(key, block)
plaintext += block
return plaintext
def pad(block_size, plaintext):
plaintext += "X" * (-len(plaintext) % block_size)
return plaintext
blocks = [ct[i:i + bs] for i in range(0, len(ct), bs)]
ct_noiv = ""
prev_block = iv
for block in blocks:
sblock = sub_key(prev_block, block)
ct_noiv += sblock
prev_block = block
print(ct_noiv)
#IDJUSTLIKETOINTERJECTFORAMOMENTWHATYOUREREFERINGTOASLINUXISINFACTGNULINUXORASIVERECENTLYTAKENTOCALLINGITGNUPLUSLINUXLINUXISNOTANOPERATINGSYSTEMUNTOITSELFBUTRATHERANOTHERFREECOMPONENTOFAFULLYFUNCTIONINGGNUSYSTEMMADEUSEFULBYTHEGNUCORELIBSSHELLUTILITIESANDVITALSYSTEMCOMPONENTSCOMPRISINGAFULLOSASDEFINEDBYPOSIXMANYCOMPUTERUSERSRUNAMODIFIEDVERSIONOFTHEGNUSYSTEMEVERYDAYWITHOUTREALIZINGITTHROUGHAPECULIARTURNOFEVENTSTHEVERSIONOFGNUWHICHISWIDELYUSEDTODAYISOFTENCALLEDLINUXANDMANYOFITSUSERSARENOTAWARETHATITISBASICALLYTHEGNUSYSTEMDEVELOPEDBYTHEGNUPROJECTTHEREREALLYISALINUXANDTHESEPEOPLEAREUSINGITBUTITISJUSTAPARTOFTHESYSTEMTHEYUSELINUXISTHEKERNELTHEPROGRAMINTHESYSTEMTHATALLOCATESTHEMACHINESRESOURCESTOTHEOTHERPROGRAMSTHATYOURUNTHEKERNELISANESSENTIALPARTOFANOPERATINGSYSTEMBUTUSELESSBYITSELFITCANONLYFUNCTIONINTHECONTEXTOFACOMPLETEOPERATINGSYSTEMLINUXISNORMALLYUSEDINCOMBINATIONWITHTHEGNUOPERATINGSYSTEMTHEWHOLESYSTEMISBASICALLYGNUWITHLINUXADDEDORGNULINUXALLTHESOCALLEDLINUXDISTRIBUTIONSAREREALLYDISTRIBUTIONSOFGNULINUXANYWAYHERECOMESTHEFLAGITSEVERYTHINGAFTERTHISATLEASTITSNOTAGENERICROTTHIRTEENCHALLENGEIGUESS
#corctf{ATLEASTITSNOTAGENERICROTTHIRTEENCHALLENGEIGUESS}
后面的题目会逐渐复现的，大家可以期待一波！
标签：pt,plaintext,flag,key,corctf2023,block,ct From： https://www.cnblogs.com/Sone070805/p/17594145.html