我和小蓝鲨的秘密

task:

from PIL import Image
from Crypto.Util.number import bytes_to_long, long_to_bytes
import numpy as np

n = 29869349657224745144762606999
e = 65537

original_image_path = "flag.jpg"
img = Image.open(original_image_path)
img = img.convert("RGB")

img_array = np.array(img)
h, w, _ = img_array.shape

encrypted_array = np.zeros((h, w, 3), dtype=object)
for i in range(h):
    for j in range(w):
        r, g, b = int(img_array[i, j, 0]), int(img_array[i, j, 1]), int(img_array[i, j, 2])

        encrypted_array[i, j, 0] = pow(r, e, n)
        encrypted_array[i, j, 1] = pow(g, e, n)
        encrypted_array[i, j, 2] = pow(b, e, n)

np.save("encrypted_image.npy", encrypted_array)
print("图片已加密并保存为 encrypted_image.npy")

分析代码：

这个主要是对图片的每个像素的红色、绿色和蓝色进行了rsa加密，然后保存为一个.npy文件，思路很清晰，直接逆反回去即可。

exp:

from Crypto.Util.number import inverse
import numpy as np
from PIL import Image


n = 29869349657224745144762606999
e = 65537

p = 160216064374859
q=n//p


phi_n = (p - 1) * (q - 1)
d = inverse(e, phi_n)

# 加载加密的图像数组
encrypted_array = np.load("encrypted_image.npy", allow_pickle=True)
h, w, _ = encrypted_array.shape

# 创建一个空的数组来存储解密后的图像数据
decrypted_array = np.zeros((h, w, 3), dtype=np.uint8)

# 解密每个像素
for i in range(h):
    for j in range(w):
        r_enc = encrypted_array[i, j, 0]
        g_enc = encrypted_array[i, j, 1]
        b_enc = encrypted_array[i, j, 2]

        # 使用 d 解密每个通道
        r = pow(r_enc, d, n)
        g = pow(g_enc, d, n)
        b = pow(b_enc, d, n)

        # 将解密后的值存入图像数组
        decrypted_array[i, j] = (r, g, b)

# 将解密后的图像保存为图片文件
decrypted_image = Image.fromarray(decrypted_array, "RGB")
decrypted_image.save("decrypted_flag.jpg")
print("图片已解密并保存为 decrypted_flag.jpg")

最后得到一个图片，就是flag。

ChaCha20-Poly1305

打开题目有三个附件

task:

from Crypto.Cipher import ChaCha20_Poly1305
import os

key = os.urandom(32)
nonce = os.urandom(12)

with open('flag.txt', 'rb') as f:
    plaintext = f.read()

cipher = ChaCha20_Poly1305.new(key=key, nonce=nonce)

ct, tag = cipher.encrypt_and_digest(plaintext)

print(f"Encrypted Flag: {ct.hex()}")
print(f"Tag: {tag.hex()}")
print(f"Nonce: {nonce.hex()}")

with open('key.txt', 'w') as key_file:
    key_file.write(key.hex())

key?.txt:

3=t#sMX3?9GHSPdi4i^gk!3*(cH8S8XT2y&?Tv4!?AGG=R]ZDy/PVVa+DqiXAH*}DS&Nn*a+@<H,=!L

output.txt:

Encrypted Flag: 20408b9fc498063ad53a4abb53633a6a15df0ddaf173012d620fa33001794dbb8c038920273464e13170e26d08923aeb Tag: 70ffcc508bf4519e7616f602123c307b Nonce: d8ebeedec812a6d71240cc50

分析题目：

第一次看这种加密，但细看和AES加密很像，那么问题就很好解决了。

首先先把key求出来，试了一下是base92解码，然后得到正常的key，就可以还原flag了。

exp:

from Crypto.Cipher import ChaCha20_Poly1305
import binascii


key_hex="173974535637a5ef30a116b03d00bd2fe751951ca3eaa62daec2b8f5ca5b6135"
ciphertext_hex = "20408b9fc498063ad53a4abb53633a6a15df0ddaf173012d620fa33001794dbb8c038920273464e13170e26d08923aeb"
tag_hex = "70ffcc508bf4519e7616f602123c307b"
nonce_hex = "d8ebeedec812a6d71240cc50"

key = binascii.unhexlify(key_hex)
ciphertext = binascii.unhexlify(ciphertext_hex)
tag = binascii.unhexlify(tag_hex)
nonce = binascii.unhexlify(nonce_hex)

cipher = ChaCha20_Poly1305.new(key=key, nonce=nonce)
plaintext = cipher.decrypt_and_verify(ciphertext, tag)
print(f"Decrypted Flag: {plaintext.decode('utf-8')}")

蓝鲨的费马

task:

import libnum
import gmpy2
from Crypto.Util.number import *

flag=b'ISCTF{********}'
m=bytes_to_long(flag)

p=libnum.generate_prime(1024)
q=libnum.generate_prime(1024)
n=p*q
e=0x10001
c=pow(m,e,n)
d=inverse(e,(p-1)*(q-1))
leak = (d+(pow(p,q,n)+pow(q,p,n)))%n

print("c=", c)
print("n=", n)
print("leak=", leak)

"""
c= 8989289659072309605793417141528767265266446236550650613514493589798432446586991233583435051268377555448062724563967695425657559568596372723980081067589103919296476501677424322525079257328042851349095575718347302884996529329066703597604694781627113384086536158793653551546025090807063130353950841148535682974762381044510423210397947080397718080033363000599995100765708244828566873128882878164321817156170983773105693537799111546309755235573342169431295776881832991533489235535981382958295960435126843833532716436804949502318851112378495533302256759494573250596802016112398817816155228378089079806308296705261876583997
n= 13424018200035368603483071894166480724482952594135293395398366121467209427078817227870501294732149372214083432516059795712917132804111155585926502759533393295089100965059106772393520277313184519450478832376508528256865861027444446718552169503579478134286009893965458507369983396982525906466073384013443851551139147777507283791250268462136554061959016630318688169168797939873600493494258467352326974238472394214986505312411729432927489878418792288365594455065912126527908319239444514857325441614280498882524432151918146061570116187524918358453036228204087993064505391742062288050068745930452767100091519798860487150247
leak= 9192002086528025412361053058922669469031188193149143635074798633855112230489479254740324032262690315813650428270911079121913869290893574897752990491429582640499542165616254566396564016734157323265631446079744216458719690853526969359930225042993006404843355356540487296896949431969541367144841985153231095140361069256753593550199420993461786814074270171257117410848796614931926182811404655619662690700351986753661502438299236428991412206196135090756862851230228396476709412020941670878645924203989895008014836619321109848938770269989596541278600166088022166386213646074764712810133558692545401032391239330088256431881
"""

分析代码：

leak = (d+(pow(p,q,n)+pow(q,p,n)))%n

可化简为：

leak=d+p+q

-->d = leak -（p+q）

又因为m=c^d % n

把d代入化简

-->m = c^(leak -(p+q)) mod n

得到式子：c^(leak -(p+q))≡ c^leak * c^-(p+q) mod n ①

又可以又欧拉定理推得：

phi=(p-1)*(q-1)

c^phi≡ 1 mod n

-->1 ≡ c^(p*q + 1 - (p+q)) mod n

-->1 ≡ c^(n+1) * c^-(p+q) mod n

得到式子：c^-(n+1) ≡ c^-(p+q) mod n ②

把①和②式子合并得：

m = c^leak * c^-(n+1) mod n

exp:

from Crypto.Util.number import *
c= 8989289659072309605793417141528767265266446236550650613514493589798432446586991233583435051268377555448062724563967695425657559568596372723980081067589103919296476501677424322525079257328042851349095575718347302884996529329066703597604694781627113384086536158793653551546025090807063130353950841148535682974762381044510423210397947080397718080033363000599995100765708244828566873128882878164321817156170983773105693537799111546309755235573342169431295776881832991533489235535981382958295960435126843833532716436804949502318851112378495533302256759494573250596802016112398817816155228378089079806308296705261876583997
n= 13424018200035368603483071894166480724482952594135293395398366121467209427078817227870501294732149372214083432516059795712917132804111155585926502759533393295089100965059106772393520277313184519450478832376508528256865861027444446718552169503579478134286009893965458507369983396982525906466073384013443851551139147777507283791250268462136554061959016630318688169168797939873600493494258467352326974238472394214986505312411729432927489878418792288365594455065912126527908319239444514857325441614280498882524432151918146061570116187524918358453036228204087993064505391742062288050068745930452767100091519798860487150247
leak= 9192002086528025412361053058922669469031188193149143635074798633855112230489479254740324032262690315813650428270911079121913869290893574897752990491429582640499542165616254566396564016734157323265631446079744216458719690853526969359930225042993006404843355356540487296896949431969541367144841985153231095140361069256753593550199420993461786814074270171257117410848796614931926182811404655619662690700351986753661502438299236428991412206196135090756862851230228396476709412020941670878645924203989895008014836619321109848938770269989596541278600166088022166386213646074764712810133558692545401032391239330088256431881

m=pow(c,leak-n-1,n)
print(long_to_bytes(m))

小蓝鲨的数学题

task:

Base和Ciphertext
m = 5321153468370294351697008906248782883193902636120413346203705810525086437271585682015110123362488732193020749380395419994982400888011862076022065339666193
c = 7383779796712259466884236308066760158536557371789388054326630574611014773044467468610300619865230550443643660647968413988480055366698747395046400909922513

这题给的提示是模数是2**512

题目给了底数，密文。典型的离散对数log

discrete_log_rho(a,base,ord,operation)：

求解以base为底，a的对数；ord为base的阶，可以缺省，operation可以是+与*，默认为*；bounds是一个区间(ld,ud)，需要保证所计算的对数在此区间内。

这里用python中sympy库自带的discrete_log可求得flag。

exp:

m = 5321153468370294351697008906248782883193902636120413346203705810525086437271585682015110123362488732193020749380395419994982400888011862076022065339666193
c = 7383779796712259466884236308066760158536557371789388054326630574611014773044467468610300619865230550443643660647968413988480055366698747395046400909922513

flag = discrete_log(2**512,c,m)
print(long_to_bytes(flag))

也可以用sage自带库：

#sage
from Crypto.Util.number import *
import gmpy2
m = 5321153468370294351697008906248782883193902636120413346203705810525086437271585682015110123362488732193020749380395419994982400888011862076022065339666193
c = 7383779796712259466884236308066760158536557371789388054326630574611014773044467468610300619865230550443643660647968413988480055366698747395046400909922513
p=2**512
flag = discrete_log(Mod(c,p),Mod(m,p))
print(long_to_bytes(flag))

小蓝鲨的方程

task:

from Cryptodome.Util.number import *
from random import *
from gmpy2 import *
import uuid
flag1='ISCTF{'+str(uuid.uuid4())+'}'

m1=bytes_to_long(flag1.encode())
def get_p():
    BITS = 256
    bits = 777
    oder = 4
    a = randint(1 << bits, 1 << bits + 1)
    p=getPrime(BITS)
    p1 = p**oder+a
    return p,p1
p,p1=get_p()
s=getPrime(1024)
q=getPrime(512)
n=p*q**4
e=65537
c1=pow(s,e,n)
c=pow(s**3+1,m1,s**5)
print("c1=",c1)
print("c =",c)
print("n =",n)
print("p1 =",p1)

'''
c1= 671390498592586008552998377599101093977542184109077889081448730480869018650843045119891777468161631085086340705902115332025675787789530562679603254577287153918966364523848382506106179394235772395029788721306186952016420794804145631124905952103136061076643266886961178241381892015555099638200222249447194504082451341122502519637821695210573997670753981061458264118355417889153180841281073262935937836447460470926729282834006229571453935760593644658459098721652426154970766417292435960463905367868753821950303919781798234432998272038029063155193184039985018137026245365188171178677898869374676546799536208952198558258306460302868688355653022725288744014143221560882404431652751343944983442109327
c = 8641190030376811670503537177719719233418166235794962118828671236836174132083208517733734760455990850156371205118391537919769888760384574011411232571257192285256730733174399297826587479261381970232162702657952399683882650083181048279650913795429823628186888540572704055008102853692060360140858142686334722286525699998854566609078547487420929457446776757558492454916447188774943818970599916514467335772992690805247630814156710861067503956707301402347944233660194395192354000788262111000900574820275786269075882923600474781645848712157460135387134196156906258218217831988828360827613420801773911833194097791649069743116686685667300622630909231822986237104627385544169938138006242341269672868611269202418482629393372933567053272565557137741441902377611003983050084491513897727856173625922194300103448148829004025229567101761111396110940066254801762424343522707712480796358754008120503317686600144600226149617189681233392693738216138797012278242152852923361635415564580582002132107424154426980566696622448291815571736676562214017436
n = 1076246859437269645898003764327104347852443049519429833372038915264009774423737482018987571807662568251485615769880354898666799006772572239466617428164721157850526408878346223839884319846641438292436373441749602341461361190584638190903978829024853974880636148520803145113551453821058269641304504880310836801494499720662704717315748614372503735165114899680682056477494953525794354656896362929510309669119173103242509398650608116835276076364248473952717811633756784397347121601006659623317417388283638159905288128181587304367489096254611610975352096229116491567502061775862811850081040850421151385474249060884479729988512713640536139010928836126719149031115182144744359297169350288886555784650111
p1 = 145356063641618996012874664536921616978986640263438210169671010403677822239343590475177543891188656103067696467174379510912427160232486984044862545338401652910975162942038201716552753723984593267892098222213049269335313670049037479410635628460505327693176152061750827570561482918795206276991967169087371403553
'''

分析题目：

我们先看a,p和p1的生成，

a = randint(1 << bits, 1 << bits + 1)
p=getPrime(BITS)
p1 = p**oder+a

p1-a=p**oder

左右两边开4次方，a也就会很好爆破了。然后再用next_prime逼进近似值p

这里求phi是用的欧拉定理的性质，具体可去看一下原理。

然后就是用二项式定理求出m1=(c-1)/s³%s⁶

exp:

from Crypto.Util.number import *
import gmpy2
c1= 671390498592586008552998377599101093977542184109077889081448730480869018650843045119891777468161631085086340705902115332025675787789530562679603254577287153918966364523848382506106179394235772395029788721306186952016420794804145631124905952103136061076643266886961178241381892015555099638200222249447194504082451341122502519637821695210573997670753981061458264118355417889153180841281073262935937836447460470926729282834006229571453935760593644658459098721652426154970766417292435960463905367868753821950303919781798234432998272038029063155193184039985018137026245365188171178677898869374676546799536208952198558258306460302868688355653022725288744014143221560882404431652751343944983442109327
c = 8641190030376811670503537177719719233418166235794962118828671236836174132083208517733734760455990850156371205118391537919769888760384574011411232571257192285256730733174399297826587479261381970232162702657952399683882650083181048279650913795429823628186888540572704055008102853692060360140858142686334722286525699998854566609078547487420929457446776757558492454916447188774943818970599916514467335772992690805247630814156710861067503956707301402347944233660194395192354000788262111000900574820275786269075882923600474781645848712157460135387134196156906258218217831988828360827613420801773911833194097791649069743116686685667300622630909231822986237104627385544169938138006242341269672868611269202418482629393372933567053272565557137741441902377611003983050084491513897727856173625922194300103448148829004025229567101761111396110940066254801762424343522707712480796358754008120503317686600144600226149617189681233392693738216138797012278242152852923361635415564580582002132107424154426980566696622448291815571736676562214017436
n = 1076246859437269645898003764327104347852443049519429833372038915264009774423737482018987571807662568251485615769880354898666799006772572239466617428164721157850526408878346223839884319846641438292436373441749602341461361190584638190903978829024853974880636148520803145113551453821058269641304504880310836801494499720662704717315748614372503735165114899680682056477494953525794354656896362929510309669119173103242509398650608116835276076364248473952717811633756784397347121601006659623317417388283638159905288128181587304367489096254611610975352096229116491567502061775862811850081040850421151385474249060884479729988512713640536139010928836126719149031115182144744359297169350288886555784650111
p1 = 145356063641618996012874664536921616978986640263438210169671010403677822239343590475177543891188656103067696467174379510912427160232486984044862545338401652910975162942038201716552753723984593267892098222213049269335313670049037479410635628460505327693176152061750827570561482918795206276991967169087371403553

pp=gmpy2.iroot(p1,4)[0]

for a in range(3000):
    b=pp-a
    p=gmpy2.next_prime(b)
    if n%p==0:
        q=gmpy2.iroot((n // p), 4)[0]
        phi=(p-1)*(q-1)*(q**3)
        d=inverse(65537,phi)
        s=int(pow(c1,d,n))
        m=((c-1)//s**3)%s**6
        print(long_to_bytes(m))
        break

小蓝鲨的密码

打开附件发现有一个txt和图片还有个压缩包，压缩包要用密码打开，所以题目意思很明确了，用密码打开压缩包，具体密码怎么获得先看txt文件：

U2FsdGVkX1/spa9kXpumHMXclw7hNOVc6ySFWYESHjfM5igIuKER30eSBp5NS0vMjkBa4Za6NgQbHVW/hJRiGA==

这一串u2Fsd开头一眼是AES或者rabbit加密，然后现在就是去找密码，大概是在图片上找，在图片上试了一番什么图片影像，图片分离后无果，注意到题目名字，就试着去打开压缩包，没想到真的成功了，打开后里面是密码字典，找到最显眼的 isctf2024 试着去解码AES即可

蓝鲨的RSA

task:

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

def gethint(h,p):
    decimal.getcontext().prec = 1024
    H = decimal.Decimal(int(h))
    P = decimal.Decimal(int(p))
    leak = decimal.Decimal((8*H*P - 1) / (16*P*P))
    return leak
p = getPrime(512)
q = getPrime(512)
f = getPrime(512)
g = getPrime(128)
h = gmpy2.invert(f, p) * g % p

n = f*q
e = 65537
m = bytes_to_long(flag)

c = pow(m,e,n)
print('c =', c)
print('hint =', gethint(h,p))
print('n =',n)

#c = 587245179027322480379581200283415189810421958968516831191660631552695197401940961725169763339428980298128692606951200581483431566182271569207988054537414289564013883171160614196522169980339024564884190765084419167938640701193928669
#hint = 0.2427542737153618793334900104191212626446625872340179613972610728976081994921862517310186626304527115125924716035632505287111236596234811779375148657365336957626454491865164520834975233144235103885081268955448330597818844340656652982593545877449810282619387305007246499089258519062093814083383071737897364213169497762760797899310673216754376885295598952272100016962368762532805864796748393317534908268379601445004775495237901072144236328105526403608646831124542336002540011176406194984370372589752234640498423911217119220030242197564695880261480071310815379681250975672935544404797155655708441222387631967447088319826137200280810029390387418159394276760100487636516708987579464183208860911063948902432948269805493252899815187044807603000344378890835564906163242023600624338694473573763088471321731611077227112205396909637906507673367598721218000123789690455125909411309668615810240938664264212370815385282488986625554704015828254539339719586211726300858711328516487805251366293457402531199532556110786048074755505680210260049
#n = 839799159583571337450826982895478997157381520448790705455708438948150905361244823725400304016136863419723271227616684280477524669207590477657886623628732394537008838314015048569652202355464477680540884654473950183135276735347866051

先用连分数逼进分数，然后求p的近似值,再恢复h1，再构造NTRU,参考链接：https://dexterjie.github.io/2023/07/29/%E6%A0%BC%E5%AF%86%E7%A0%81%E5%85%A5%E9%97%A8/

exp:

import gmpy2
from Crypto.Util.number import *
from fractions import Fraction
leak = 0.2427542737153618793334900104191212626446625872340179613972610728976081994921862517310186626304527115125924716035632505287111236596234811779375148657365336957626454491865164520834975233144235103885081268955448330597818844340656652982593545877449810282619387305007246499089258519062093814083383071737897364213169497762760797899310673216754376885295598952272100016962368762532805864796748393317534908268379601445004775495237901072144236328105526403608646831124542336002540011176406194984370372589752234640498423911217119220030242197564695880261480071310815379681250975672935544404797155655708441222387631967447088319826137200280810029390387418159394276760100487636516708987579464183208860911063948902432948269805493252899815187044807603000344378890835564906163242023600624338694473573763088471321731611077227112205396909637906507673367598721218000123789690455125909411309668615810240938664264212370815385282488986625554704015828254539339719586211726300858711328516487805251366293457402531199532556110786048074755505680210260049
c = 587245179027322480379581200283415189810421958968516831191660631552695197401940961725169763339428980298128692606951200581483431566182271569207988054537414289564013883171160614196522169980339024564884190765084419167938640701193928669
n = 839799159583571337450826982895478997157381520448790705455708438948150905361244823725400304016136863419723271227616684280477524669207590477657886623628732394537008838314015048569652202355464477680540884654473950183135276735347866051

cf = continued_fraction(leak)   

list_numden=cf.convergents()    
numerator_list=[]
denominator_list=[]

for i in list_numden:
    fraction=Fraction(i)
    numerator = fraction.numerator()  
    denominator = fraction.denominator() 
    denominator_list.append(denominator)
for i,p in enumerate(denominator_list):

    if gmpy2.iroot(p//16,2)[1]==True:
        p1= int(gmpy2.iroot(p//16,2)[0])  

e=65537     
h=525006825733225408907597067990642584862207418097213674443701993658004487397784939081087821440376288434095962749583145952889046963712522108407121467693817733948555003104602864877809068405079168763660338489863597493767710662358756925070265792090561589011628760109484806572919266760895186597799962875710969125263

h1=(h+1)//(8*p1)
M=matrix(ZZ,[[h1,1],
            [p1,0]])
f=int(M.LLL()[0][1])
q=n//f
phi=(q-1)*(f-1)
d=gmpy2.invert(e,int(phi))
m=pow(c,d,n)
print(long_to_bytes(int(m)))

ezmath

task:

import random
import base64
from hashlib import md5
from secret import flag
from libnum import s2n
from Crypto.Cipher import AES


INF = 0xff
bigINF = 0xffffffff
# ---------------------error------------------------------

def sumFunc(func):
   def wapper(*args,start = 1,end=INF):
      sums = 0
      for i in trange(start, 0xffff * end):
         sums += function(*args,j/0xf) * (1/0xf)
      return sums
   return wapper

def limitFunc(func):
   def wapper(*args, approch = bigINF, pos = "+"):
      o = 1/bigINF
      return function(args, eval(f"{approch} {pos} {o}"))
   return wapper


# -------------------enderror-----------------------------


def pad(data):
   data=data.encode('utf8')
   while len(data) % 16 !=0:
      data+=b'\x00'
   return data


def encode(key,m):
   mode=AES.MODE_ECB
   aes=AES.new(pad(key),mode)
   en_m=aes.encrypt(pad(m))
   en_m=base64.encodebytes(en_m)
   en_m=en_m.decode('utf8')
   return en_m


def enc(msg, key):
   random.seed(key)
   new_key = md5(str(random.getrandbits(256)).encode('utf-8')).hexdigest()
   return encode(new_key, msg)


@sumFunc
def gamma(x,t):
   data = pow(t,x-1) * pow(magicNumber,-t)
   return data

@sumFunc
def common(t):
   data = pow(magicNumber,-pow(t,2))
   return data

@limitFunc
def getMagicNumber(t):
   data = pow(1+1/t,t)
   return data

magicNumber = getMagicNumber()
encKey1 = str(gamma(3/2))[2:6]
encKey2 = str(common())[2:6]
assert encKey1 == encKey2

key = int(str(gamma(5/2))[2:])
print(enc(flag, key))
# n2SQK64JMsXstCtZurBiz81pMr3ZmgMjhuyL67hssm3shqJGYGfS/mWubINeE5HZ

看到gamma ，sage自带gamma：

再求出他的值，取小数点前15位为：329340388179137

这样就可以求得key了，直接逆一下求得flag。

import random
import base64
from hashlib import md5
from Crypto.Cipher import AES

def pad(data):
    data = data.encode('utf8')
    while len(data) % 16 != 0:
        data += b'\x00'
    return data

def decode(key, en_m):
    mode = AES.MODE_ECB
    aes = AES.new(pad(key), mode)
    en_m = base64.decodebytes(en_m.encode('utf8'))
    de_m = aes.decrypt(en_m)
    return de_m.decode('utf8').rstrip('\x00')

def dec(encrypted_msg, key):
    random.seed(key)
    new_key = md5(str(random.getrandbits(256)).encode('utf-8')).hexdigest()
    return decode(new_key, encrypted_msg)


key = 329340388179137
encrypted_msg = "n2SQK64JMsXstCtZurBiz81pMr3ZmgMjhuyL67hssm3shqJGYGfS/mWubINeE5HZ"
flag = dec(encrypted_msg, key)
print(flag)