moectf_crypto_wp
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CTF-CRYPTO,密码
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52

1385 字
|
46 分钟

题目

现代密码学指北

task:

from Crypto.Util.number import bytes_to_long, getPrime
from secret import flag
p = getPrime(128)
q = getPrime(128)
n = p*q
e = 65537
m = bytes_to_long(flag)
c = pow(m, e, n)
print(f"n = {n}")
print(f"p = {p}")
print(f"q = {q}")
print(f"c = {c}")
'''
n = 40600296529065757616876034307502386207424439675894291036278463517602256790833
p = 197380555956482914197022424175976066223
q = 205695522197318297682903544013139543071
c = 36450632910287169149899281952743051320560762944710752155402435752196566406306
'''

正常解密即可
exp

n = 40600296529065757616876034307502386207424439675894291036278463517602256790833
p = 197380555956482914197022424175976066223
q = 205695522197318297682903544013139543071
c = 36450632910287169149899281952743051320560762944710752155402435752196566406306
e=65537

from Crypto.Util.number import*

phi = (p-1)*(q-1)
d = inverse(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

baby_equation

task

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

l = len(flag)
m1, m2 = flag[:l//2], flag[l//2:]
a = bytes_to_long(m1)
b = bytes_to_long(m2)
k = 0x2227e398fc6ffcf5159863a345df85ba50d6845f8c06747769fee78f598e7cb1bcf875fb9e5a69ddd39da950f21cb49581c3487c29b7c61da0f584c32ea21ce1edda7f09a6e4c3ae3b4c8c12002bb2dfd0951037d3773a216e209900e51c7d78a0066aa9a387b068acbd4fb3168e915f306ba40
assert ((a**2 + 1)*(b**2 + 1) - 2*(a - b)*(a*b - 1)) == 4*(k + a*b)

思路
其实就是因式分解会得到$(ab-1-a-b)^2=4k$,那么先开根号，再因式分解，再对因子进行排列组合就好

exp

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

k = 0x2227e398fc6ffcf5159863a345df85ba50d6845f8c06747769fee78f598e7cb1bcf875fb9e5a69ddd39da950f21cb49581c3487c29b7c61da0f584c32ea21ce1edda7f09a6e4c3ae3b4c8c12002bb2dfd0951037d3773a216e209900e51c7d78a0066aa9a387b068acbd4fb3168e915f306ba40

x = iroot(4*k,2)
print(x)

# factor(x)
factors = [2,2,2,2,3,3,31,61,223,4013,281317,4151351,5404604441993,
           26798471753993,25866088332911027256931479223,64889106213996537255229963986303510188999911,370523737,339386329]

T = prod(factors)
t = T ** 0.5
n = len(factors)
dp = {1: []}  

for i in factors:
    for j in list(dp.keys()):
        cnt = j * i
        if cnt not in dp:
            dp[cnt] = dp[j] + [i]

min = min(dp.keys(), key=lambda x: abs(x - t))

factor1 = min
factor2 = T // min

print(factor1)
print(factor2)

print(long_to_bytes(2948658764987911698882278955266869405625105496652193856947916257370492+1))
print(long_to_bytes(2950365559902224963252311699604518749050102395042841254385700637274676-1))

big and small

task

from secret import flag
from Crypto.Util.number import*
m = long_to_bytes(flag)
p = getPrime(1024)
q = getPrime(1024)
n = p*q
e = 3
c = pow(m,e,n)
'''
c = 150409620528288093947185249913242033500530715593845912018225648212915478065982806112747164334970339684262757
e = 3
n = 20279309983698966932589436610174513524888616098014944133902125993694471293062261713076591251054086174169670848598415548609375570643330808663804049384020949389856831520202461767497906977295453545771698220639545101966866003886108320987081153619862170206953817850993602202650467676163476075276351519648193219850062278314841385459627485588891326899019745457679891867632849975694274064320723175687748633644074614068978098629566677125696150343248924059801632081514235975357906763251498042129457546586971828204136347260818828746304688911632041538714834683709493303900837361850396599138626509382069186433843547745480160634787
'''

思路
小明文攻击

exp

c = 150409620528288093947185249913242033500530715593845912018225648212915478065982806112747164334970339684262757
e = 3
n = 20279309983698966932589436610174513524888616098014944133902125993694471293062261713076591251054086174169670848598415548609375570643330808663804049384020949389856831520202461767497906977295453545771698220639545101966866003886108320987081153619862170206953817850993602202650467676163476075276351519648193219850062278314841385459627485588891326899019745457679891867632849975694274064320723175687748633644074614068978098629566677125696150343248924059801632081514235975357906763251498042129457546586971828204136347260818828746304688911632041538714834683709493303900837361850396599138626509382069186433843547745480160634787

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

flag=iroot(c,e)[0]
print(long_to_bytes(flag))

ez_hash

task

from hashlib import sha256
from secret import flag, secrets

assert flag == b'moectf{' + secrets + b'}'
assert secrets[:4] == b'2100' and len(secrets) == 10
hash_value = sha256(secrets).hexdigest()
print(f"{hash_value = }")
# hash_value = '3a5137149f705e4da1bf6742e62c018e3f7a1784ceebcb0030656a2b42f50b6a'

思路：
已知前四位，后六位未知，猜测都是数字，直接爆破

exp

import hashlib

target_hash = '3a5137149f705e4da1bf6742e62c018e3f7a1784ceebcb0030656a2b42f50b6a'

for i in range(100000000):
    secrets_candidate = f'2100{i:06}'.encode()  # 生成 '2100XXXXXX' 格式的字节串
    if hashlib.sha256(secrets_candidate).hexdigest() == target_hash:
        flag = b'moectf{' + secrets_candidate + b'}'
        print(flag)
        break

signin

task

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

m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
n = p*q
e = 65537
c = pow(m,e,n)
pq = (p-1)*(q-2)
qp = (q-1)*(p-2)
p_q = p + q

print(f"{c = }")
print(f"{pq = }")
print(f"{qp = }")
print(f"{n = }")
print(f"{p_q = }")
'''
c = 5654386228732582062836480859915557858019553457231956237167652323191768422394980061906028416785155458721240012614551996577092521454960121688179565370052222983096211611352630963027300416387011219744891121506834201808533675072141450111382372702075488292867077512403293072053681315714857246273046785264966933854754543533442866929316042885151966997466549713023923528666038905359773392516627983694351534177829247262148749867874156066768643169675380054673701641774814655290118723774060082161615682005335103074445205806731112430609256580951996554318845128022415956933291151825345962528562570998777860222407032989708801549746
pq = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687154230787854196153067547938936776488741864214499155892870610823979739278296501074632962069426593691194105670021035337609896886690049677222778251559566664735419100459953672218523709852732976706321086266274840999100037702428847290063111455101343033924136386513077951516363739936487970952511422443500922412450462
qp = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687077087914198877794354459669808240133383828356379423767736753506794441545506312066344576298453957064590180141648690226266236642320508613544047037110363523129966437840660693885863331837516125853621802358973786440314619135781324447765480391038912783714312479080029167695447650048419230865326299964671353746764860
n = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687534959910892789661065614807265825078942931717855566686073463382398417205648946713373617006449901977718981043020664616841303517708207413215548110294271101267236070252015782044263961319221848136717220979435486850254298686692230935985442120369913666939804135884857831857184001072678312992442792825575636200505903
p_q = 279533706577501791569740668595544511920056954944184570513187478007551195831693428589898548339751066551225424790534556602157835468618845221423643972870671556362200734472399328046960316064864571163851111207448753697980178391430044714097464866523838747053135392202848167518870720149808055682621080992998747265496
'''

思路
解方程

exp

from sympy import symbols, Eq, solve

p, q = symbols('p q')

c = 5654386228732582062836480859915557858019553457231956237167652323191768422394980061906028416785155458721240012614551996577092521454960121688179565370052222983096211611352630963027300416387011219744891121506834201808533675072141450111382372702075488292867077512403293072053681315714857246273046785264966933854754543533442866929316042885151966997466549713023923528666038905359773392516627983694351534177829247262148749867874156066768643169675380054673701641774814655290118723774060082161615682005335103074445205806731112430609256580951996554318845128022415956933291151825345962528562570998777860222407032989708801549746
pq = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687154230787854196153067547938936776488741864214499155892870610823979739278296501074632962069426593691194105670021035337609896886690049677222778251559566664735419100459953672218523709852732976706321086266274840999100037702428847290063111455101343033924136386513077951516363739936487970952511422443500922412450462
qp = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687077087914198877794354459669808240133383828356379423767736753506794441545506312066344576298453957064590180141648690226266236642320508613544047037110363523129966437840660693885863331837516125853621802358973786440314619135781324447765480391038912783714312479080029167695447650048419230865326299964671353746764860
n = 18047017539289114275195019384090026530425758236625347121394903879980914618669633902668100353788910470141976640337675700570573127020693081175961988571621759711122062452192526924744760561788625702044632350319245961013430665853071569777307047934247268954386678746085438134169871118814865536503043639618655569687534959910892789661065614807265825078942931717855566686073463382398417205648946713373617006449901977718981043020664616841303517708207413215548110294271101267236070252015782044263961319221848136717220979435486850254298686692230935985442120369913666939804135884857831857184001072678312992442792825575636200505903
p_q = 279533706577501791569740668595544511920056954944184570513187478007551195831693428589898548339751066551225424790534556602157835468618845221423643972870671556362200734472399328046960316064864571163851111207448753697980178391430044714097464866523838747053135392202848167518870720149808055682621080992998747265496
e=65537

eq1 = Eq(p * q, n)
eq2 = Eq((p - 1) * (q - 2), pq)
eq3 = Eq((q - 1) * (p - 2), qp)
eq4 = Eq(p + q, p_q)

solution = solve((eq1, eq2, eq3, eq4), (p, q))

print(solution)

from Crypto.Util.number import*

p=101195416461091716428326199733504078281010548412226222689665080411126731520752210150756388683557219973649948209094722629248795549538890771346214761833764975454769057589710497693291150424006859232283601953197097456280805871953601208233200402046794268614613979577032173301390416040533984248749301081715040789947
q=178338290116410075141414468862040433639046406531958347823522397596424464310941218439142159656193846577575476581439833972909039919079954450077429211036906580907431676882688830353669165640857711931567509254251656241699372519476443505864264464477044478438521412625815994217480304109274071433871779911283706475549

phi=(p-1)*(q-1)
d = inverse(e,phi)
m = pow(c,d,n)
print(long_to_bytes(m))

大白兔

task

from Crypto.Util.number import *

flag = b'moectf{xxxxxxxxxx}'
m = bytes_to_long(flag)

e1 = 12886657667389660800780796462970504910193928992888518978200029826975978624718627799215564700096007849924866627154987365059524315097631111242449314835868137
e2 = 12110586673991788415780355139635579057920926864887110308343229256046868242179445444897790171351302575188607117081580121488253540215781625598048021161675697

def encrypt(m , e1 , e2):
    p = getPrime(512)
    q = getPrime(512)
    N = p*q
    c1 = pow((3*p + 7*q),e1,N)
    c2 = pow((2*p + 5*q),e2,N)
    e = 65537
    c = pow(m , e , N)
    return c

print(encrypt(m ,e1 , e2))

'''
N = 107840121617107284699019090755767399009554361670188656102287857367092313896799727185137951450003247965287300048132826912467422962758914809476564079425779097585271563973653308788065070590668934509937791637166407147571226702362485442679293305752947015356987589781998813882776841558543311396327103000285832158267
c1 = 15278844009298149463236710060119404122281203585460351155794211733716186259289419248721909282013233358914974167205731639272302971369075321450669419689268407608888816060862821686659088366316321953682936422067632021137937376646898475874811704685412676289281874194427175778134400538795937306359483779509843470045
c2 = 21094604591001258468822028459854756976693597859353651781642590543104398882448014423389799438692388258400734914492082531343013931478752601777032815369293749155925484130072691903725072096643826915317436719353858305966176758359761523170683475946913692317028587403027415142211886317152812178943344234591487108474
c = 21770231043448943684137443679409353766384859347908158264676803189707943062309013723698099073818477179441395009450511276043831958306355425252049047563947202180509717848175083113955255931885159933086221453965914552773593606054520151827862155643433544585058451821992566091775233163599161774796561236063625305050
'''

思路
很经典的题目

$$
c1^e=(3p+7q)^{e1e2}=(3p)^{e1e2}+(7q)^{e1e2} mod N \
c2^e=(2p+5q)^{e1e2}=(2p)^{e1e2}+(5q)^{e1e2} mod N \
s=2^{e1e2}c1^e-3^{e1e2}c2^e=k*q^{e1e2} \
GCD(s,N)=q
$$

求出q之后p也就出来了

exp

from gmpy2 import*
from Crypto.Util.number import *
N = 107840121617107284699019090755767399009554361670188656102287857367092313896799727185137951450003247965287300048132826912467422962758914809476564079425779097585271563973653308788065070590668934509937791637166407147571226702362485442679293305752947015356987589781998813882776841558543311396327103000285832158267
c1 = 15278844009298149463236710060119404122281203585460351155794211733716186259289419248721909282013233358914974167205731639272302971369075321450669419689268407608888816060862821686659088366316321953682936422067632021137937376646898475874811704685412676289281874194427175778134400538795937306359483779509843470045
c2 = 21094604591001258468822028459854756976693597859353651781642590543104398882448014423389799438692388258400734914492082531343013931478752601777032815369293749155925484130072691903725072096643826915317436719353858305966176758359761523170683475946913692317028587403027415142211886317152812178943344234591487108474
c = 21770231043448943684137443679409353766384859347908158264676803189707943062309013723698099073818477179441395009450511276043831958306355425252049047563947202180509717848175083113955255931885159933086221453965914552773593606054520151827862155643433544585058451821992566091775233163599161774796561236063625305050
e1 = 12886657667389660800780796462970504910193928992888518978200029826975978624718627799215564700096007849924866627154987365059524315097631111242449314835868137
e2 = 12110586673991788415780355139635579057920926864887110308343229256046868242179445444897790171351302575188607117081580121488253540215781625598048021161675697

f1 = pow(2, e1*e2, N) * pow(c1, e2, N)
f2 = pow(3, e1*e2, N) * pow(c2, e1, N)
q = abs(gcd(N, f1-f2))
p = N//q
print(p)
print(q)

phi= (p-1)*(q-1)
e= 65537
d= inverse(e, phi)
m= pow(c, d, N)
print(long_to_bytes(m))

More_secure_RSA

task

from Crypto.Util.number import *

flag = b'moectf{xxxxxxxxxxxxxxxxx}'

m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
n = p * q 
e = 0x10001
c = pow(m, e, n)
print(f'c = {c}')
print(f'n = {n}')

'''
Oh,it isn't secure enough!
'''
r = getPrime(1024)
n = n * r
c = pow(m, e, n)
print(f'C = {c}')
print(f'N = {n}')

'''
c = 12992001402636687796268040906463852467529970619872166160007439409443075922491126428847990768804065656732371491774347799153093983118784555645908829567829548859716413703103209412482479508343241998746249393768508777622820076455330613128741381912099938105655018512573026861940845244466234378454245880629342180767100764598827416092526417994583641312226881576127632370028945947135323079587274787414572359073029332698851987672702157745794918609888672070493920551556186777642058518490585668611348975669471428437362746100320309846155934102756433753034162932191229328675448044938003423750406476228868496511462133634606503693079
n = 16760451201391024696418913179234861888113832949815649025201341186309388740780898642590379902259593220641452627925947802309781199156988046583854929589247527084026680464342103254634748964055033978328252761138909542146887482496813497896976832003216423447393810177016885992747522928136591835072195940398326424124029565251687167288485208146954678847038593953469848332815562187712001459140478020493313651426887636649268670397448218362549694265319848881027371779537447178555467759075683890711378208297971106626715743420508210599451447691532788685271412002723151323393995544873109062325826624960729007816102008198301645376867
C = 1227033973455439811038965425016278272592822512256148222404772464092642222302372689559402052996223110030680007093325025949747279355588869610656002059632685923872583886766517117583919384724629204452792737574445503481745695471566288752636639781636328540996436873887919128841538555313423836184797745537334236330889208413647074397092468650216303253820651869085588312638684722811238160039030594617522353067149762052873350299600889103069287265886917090425220904041840138118263873905802974197870859876987498993203027783705816687972808545961406313020500064095748870911561417904189058228917692021384088878397661756664374001122513267695267328164638124063984860445614300596622724681078873949436838102653185753255893379061574117715898417467680511056057317389854185497208849779847977169612242457941087161796645858881075586042016211743804958051233958262543770583176092221108309442538853893897999632683991081144231262128099816782478630830512
N = 1582486998399823540384313363363200260039711250093373548450892400684356890467422451159815746483347199068277830442685312502502514973605405506156013209395631708510855837597653498237290013890476973370263029834010665311042146273467094659451409034794827522542915103958741659248650774670557720668659089460310790788084368196624348469099001192897822358856214600885522908210687134137858300443670196386746010492684253036113022895437366747816728740885167967611021884779088402351311559013670949736441410139393856449468509407623330301946032314939458008738468741010360957434872591481558393042769373898724673597908686260890901656655294366875485821714239821243979564573095617073080807533166477233759321906588148907331569823186970816432053078415316559827307902239918504432915818595223579467402557885923581022810437311450172587275470923899187494633883841322542969792396699601487817033616266657366148353065324836976610554682254923012474470450197
'''

思路
用r作为模数进行求解

exp

from sympy import mod_inverse
from Crypto.Util.number import long_to_bytes

c = 12992001402636687796268040906463852467529970619872166160007439409443075922491126428847990768804065656732371491774347799153093983118784555645908829567829548859716413703103209412482479508343241998746249393768508777622820076455330613128741381912099938105655018512573026861940845244466234378454245880629342180767100764598827416092526417994583641312226881576127632370028945947135323079587274787414572359073029332698851987672702157745794918609888672070493920551556186777642058518490585668611348975669471428437362746100320309846155934102756433753034162932191229328675448044938003423750406476228868496511462133634606503693079
N = 1582486998399823540384313363363200260039711250093373548450892400684356890467422451159815746483347199068277830442685312502502514973605405506156013209395631708510855837597653498237290013890476973370263029834010665311042146273467094659451409034794827522542915103958741659248650774670557720668659089460310790788084368196624348469099001192897822358856214600885522908210687134137858300443670196386746010492684253036113022895437366747816728740885167967611021884779088402351311559013670949736441410139393856449468509407623330301946032314939458008738468741010360957434872591481558393042769373898724673597908686260890901656655294366875485821714239821243979564573095617073080807533166477233759321906588148907331569823186970816432053078415316559827307902239918504432915818595223579467402557885923581022810437311450172587275470923899187494633883841322542969792396699601487817033616266657366148353065324836976610554682254923012474470450197
e = 0x10001
n = 16760451201391024696418913179234861888113832949815649025201341186309388740780898642590379902259593220641452627925947802309781199156988046583854929589247527084026680464342103254634748964055033978328252761138909542146887482496813497896976832003216423447393810177016885992747522928136591835072195940398326424124029565251687167288485208146954678847038593953469848332815562187712001459140478020493313651426887636649268670397448218362549694265319848881027371779537447178555467759075683890711378208297971106626715743420508210599451447691532788685271412002723151323393995544873109062325826624960729007816102008198301645376867
C = 1227033973455439811038965425016278272592822512256148222404772464092642222302372689559402052996223110030680007093325025949747279355588869610656002059632685923872583886766517117583919384724629204452792737574445503481745695471566288752636639781636328540996436873887919128841538555313423836184797745537334236330889208413647074397092468650216303253820651869085588312638684722811238160039030594617522353067149762052873350299600889103069287265886917090425220904041840138118263873905802974197870859876987498993203027783705816687972808545961406313020500064095748870911561417904189058228917692021384088878397661756664374001122513267695267328164638124063984860445614300596622724681078873949436838102653185753255893379061574117715898417467680511056057317389854185497208849779847977169612242457941087161796645858881075586042016211743804958051233958262543770583176092221108309442538853893897999632683991081144231262128099816782478630830512

r=N//n
phi=r-1
d=mod_inverse(e,phi)
m=pow(C,d,r)
print(long_to_bytes(m))

ezlegendre

task

from sympy import *
from Crypto.Util.number import *

p = getPrime(128)
e = randprime(2, p)

FLAG = b'm'

def encrypt_flag(flag):
    ciphertext = []
    plaintext = ''.join([bin(i)[2:].zfill(8) for i in flag])
    print(plaintext)
    for bit in plaintext:
        n = pow(int(bit) + e, e , p)
        ciphertext.append(n)
    return ciphertext

print(f"p = {p}")
print(encrypt_flag(FLAG))

思路
只有两个数字，把83185897643827119919760550833655486588当作0，171219394072263643527316538070481587611当作1，就是01字符串，然后转字符就好了,数据太大，就不写了

exp

conversion_dict = {
    83185897643827119919760550833655486588: 0,
    171219394072263643527316538070481587611: 1
}

converted_c = [conversion_dict.get(x, x) for x in c]

converted_c_str = ''.join(map(str, converted_c))

print(converted_c_str)

s=b'01101101011011110110010101100011011101000110011001111011011011010110100101101110011101010111001101011111011011110110111001100101010111110011000101110011010111110110111000110000011101000101111101110001011101010011010001100100011100100011010001110100011010010110001101011111011100100011010001110011011010010110010001110101001101000101111101110111011010000110010101101110010111110111000001011111011011010110111101100100010111110110011000110000011101010111001001011111011001010111000101110101001101000011000101011111011101000110111101011111011101000110100001110010001100110011001101111101'

binary_string = b'01101101011011110110010101100011011101000110011001111011011011010110100101101110011101010111001101011111011011110110111001100101010111110011000101110011010111110110111000110000011101000101111101110001011101010011010001100100011100100011010001110100011010010110001101011111011100100011010001110011011010010110010001110101001101000101111101110111011010000110010101101110010111110111000001011111011011010110111101100100010111110110011000110000011101010111001001011111011001010111000101110101001101000011000101011111011101000110111101011111011101000110100001110010001100110011001101111101'

text = ''.join(chr(int(binary_string[i:i+8], 2)) for i in range(0, len(binary_string), 8))

print(text)

new_system

task

from random import randint
from Crypto.Util.number import getPrime,bytes_to_long 

flag = b'moectf{???????????????}'
gift = bytes_to_long(flag)

def parametergenerate():
    q = getPrime(256)
    gift1 = randint(1, q)
    gift2 = (gift - gift1) % q
    x =  randint(1, q)
    assert gift == (gift1 + gift2) % q
    return q , x , gift1, gift2

def encrypt(m , q , x):
    a = randint(1, q)
    c = (a*x + m) % q
    return [a , c]

q , x , gift1 , gift2 = parametergenerate()
print(encrypt(gift1 , q , x))
print(encrypt(gift2 , q , x))
print(encrypt(gift , q , x))
print(f'q = {q}')

'''
[48152794364522745851371693618734308982941622286593286738834529420565211572487, 21052760152946883017126800753094180159601684210961525956716021776156447417961]
[48649737427609115586886970515713274413023152700099032993736004585718157300141, 6060718815088072976566240336428486321776540407635735983986746493811330309844]
[30099883325957937700435284907440664781247503171217717818782838808179889651361, 85333708281128255260940125642017184300901184334842582132090488518099650581761]
q = 105482865285555225519947662900872028851795846950902311343782163147659668129411
'''

思路

$$
c_i = a_ix+g \mod q \
C = (a_1+a_2)x+g \mod q \
d = (a_1+a_2-a_3)*x \mod q
$$

带入求解就行了

exp

from Crypto.Util.number import inverse, long_to_bytes

a1, c1 = 48152794364522745851371693618734308982941622286593286738834529420565211572487, 21052760152946883017126800753094180159601684210961525956716021776156447417961
a2, c2 = 48649737427609115586886970515713274413023152700099032993736004585718157300141, 6060718815088072976566240336428486321776540407635735983986746493811330309844
a3, c3 = 30099883325957937700435284907440664781247503171217717818782838808179889651361, 85333708281128255260940125642017184300901184334842582132090488518099650581761
q = 105482865285555225519947662900872028851795846950902311343782163147659668129411

x = ((c3 - c2 - c1) * inverse(a3 - a2 - a1, q)) % q

gift1 = (c1 - a1 * x) % q
gift2 = (c2 - a2 * x) % q
gift = (gift1 + gift2) % q

flag = long_to_bytes(gift)
print(flag)

RSA_revenge

task

from Crypto.Util.number import getPrime, isPrime, bytes_to_long
from secret import flag

def emirp(x):
    y = 0
    while x !=0:
        y = y*2 + x%2
        x = x//2
    return y

while True:
    p = getPrime(512)
    q = emirp(p)
    if isPrime(q):
        break

n = p*q
e = 65537
m = bytes_to_long(flag)
c = pow(m,e,n)
print(f"{n = }")
print(f"{c = }")

"""
n = 141326884939079067429645084585831428717383389026212274986490638181168709713585245213459139281395768330637635670530286514361666351728405851224861268366256203851725349214834643460959210675733248662738509224865058748116797242931605149244469367508052164539306170883496415576116236739853057847265650027628600443901
c = 47886145637416465474967586561554275347396273686722042112754589742652411190694422563845157055397690806283389102421131949492150512820301748529122456307491407924640312270962219946993529007414812671985960186335307490596107298906467618684990500775058344576523751336171093010950665199612378376864378029545530793597
"""

思路
可以看看https://kt.gy/blog/2015/10/asis-2015-finals-rsasr/

exp

n = 141326884939079067429645084585831428717383389026212274986490638181168709713585245213459139281395768330637635670530286514361666351728405851224861268366256203851725349214834643460959210675733248662738509224865058748116797242931605149244469367508052164539306170883496415576116236739853057847265650027628600443901
def t(a, b, k):
    if k == 256:
        if a*b == n:
            print(a, b)
        return
    for i in range(2):
        for j in range(2):
            a1 = a + i*(2**k) + j*(2**(511-k))
            b1 = b + j*(2**k) + i*(2**(511-k))
            if a1*b1 > n:
                continue
            if (a1+(2**(511-k)))*(b1+(2**(511-k))) < n:
                continue
            if ((a1*b1)%(2**(k+1))) != (n%(2**(k+1))):
                continue
            t(a1, b1, k+1)

for i in range(2):
    t(i*(2**256), i*(2**256), 0)

p=12119998731259483292178496920109290754181396164390285597126378297678818779092115139911720576157973310671490865211601201831597946479039132512609504866583931
q=11660635291534613230423193509391946961264539191735481147071890944740311229658362673314192872117237108949853531941630122241060679012089130178372253390640871
c=47886145637416465474967586561554275347396273686722042112754589742652411190694422563845157055397690806283389102421131949492150512820301748529122456307491407924640312270962219946993529007414812671985960186335307490596107298906467618684990500775058344576523751336171093010950665199612378376864378029545530793597
from Crypto.Util.number import *
n=p*q
phi=(p-1)*(q-1)
d=inverse(65537,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

One more bit

task

from Crypto.Util.number import getStrongPrime, bytes_to_long, GCD, inverse
from Crypto.Util.Padding import pad
from secret import flag
import random

def genKey(nbits,dbits):
    p = getStrongPrime(nbits//2)
    q = getStrongPrime(nbits//2)
    n = p*q
    phi = (p-1)*(q-1)
    while True:
        d = random.getrandbits(dbits)
        if d.bit_length() == dbits:
            if GCD(d, phi) == 1:
                e = inverse(d, phi)
                pk = (n, e)
                sk = (p, q, d)
                return pk, sk

nbits = 1024
dbits = 258
message = pad(flag,16)
msg = pad(message, 16)
m = bytes_to_long(msg)
pk= genKey(nbits, dbits)[0]
n, e = pk
ciphertext = pow(m, e, n)

with open("data.txt","w") as f:
    f.write(f"pk = {pk}\n")
    f.write(f"ciphertext = {ciphertext}\n")
    f.close()

思路
Boneh_Durfee攻击的板子题

exp
省略

EzPack

task

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

p = 2050446265000552948792079248541986570794560388346670845037360320379574792744856498763181701382659864976718683844252858211123523214530581897113968018397826268834076569364339813627884756499465068203125112750486486807221544715872861263738186430034771887175398652172387692870928081940083735448965507812844169983643977
assert len(flag) == 42

def encode(msg):
    return bin(bytes_to_long(msg))[2:].zfill(8*len(msg))

def genkey(len):
    sums = 0
    keys = []
    for i in range(len):
        k = random.randint(1,7777)
        x = sums + k
        keys.append(x)
        sums += x
    return keys

key = genkey(42*8)

def enc(m, keys):
    msg = encode(m)
    print(len(keys))
    print(len(msg))
    assert len(msg) == len(keys)
    s = sum((k if (int(p,2) == 1) else 1) for p, k in zip(msg, keys))
    print(msg)
    for p0,k in zip(msg,keys):
        print(int(p0,2))
    return pow(7,s,p)

cipher = enc(flag,key)

with open("output.txt", "w") as fs:
    fs.write(str(key)+'\n')
    fs.write(str(cipher))

思路
p-1是光滑的，直接DLP，然后背包,数据太大了，不写了

exp

from Crypto.Util.number import*
from sympy import discrete_log
#pow(7,s,p)
s=363965742933281351259442199216117822475210003294088371760914916341815880641228470807683148775152284520244

decoded_bits = []
for k in reversed(key):
    if s >= k:
        decoded_bits.append('1')
        s -= k
    else:
        decoded_bits.append('0')

decoded_bits = ''.join(reversed(decoded_bits))

flag = long_to_bytes(int(decoded_bits, 2))
print(flag)

EzMatrix

task

from Crypto.Util.number import *
from secret import FLAG,secrets,SECERT_T

assert len(secrets) == 16
assert FLAG == b'moectf{' + secrets + b'}'
assert len(SECERT_T) <= 127

class LFSR:
    def __init__(self):
        self._s = list(map(int,list("{:0128b}".format(bytes_to_long(secrets)))))
        for _ in range(8*len(secrets)):
            self.clock()

    def clock(self):
        b = self._s[0]
        c = 0
        for t in SECERT_T:c ^= self._s[t]
        self._s = self._s[1:] + [c]
        return b

    def stream(self, length):
        return [self.clock() for _ in range(length)]

c = LFSR()
stream = c.stream(256)
print("".join(map(str,stream))[:-5])
# 11111110011011010000110110100011110110110101111000101011001010110011110011000011110001101011001100000011011101110000111001100111011100010111001100111101010011000110110101011101100001010101011011101000110001111110100000011110010011010010100100000000110

思路
lfsr每次生成一位都会形成一个方程

$$
mask*statei=state{i+1}
$$

根据条件，一共有123组方程，然后反解出mask

exp

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

def xor(a,b):
    res=0
    for i,j in zip(a,b):
        if j==1:
            res^=int(i)
    return res

def rev(stream,mask):
    temp = [0] + stream[:-1]
    if xor(temp, mask) == stream[-1]:
        return temp

    temp = [1] + stream[:-1]
    if xor(temp, mask) == stream[-1]:
        return temp
    return None

lfsr_stream='11111110011011010000110110100011110110110101111000101011001010110011110011000011110001101011001100000011011101110000111001100111011100010111001100111101010011000110110101011101100001010101011011101000110001111110100000011110010011010010100100000000110'

M=matrix(Zmod(2),0,128)

for i in range(len(lfsr_stream)-128):
    v=vector([int(x) for x in lfsr_stream[i:i+128]])
    M=M.stack(v)

res=vector(Zmod(2), [int(x) for x in lfsr_stream[128:]])

mask = M.solve_right(res)
for k in range(5):
    stream = [int(x) for x in lfsr_stream[:128]]
    for idx in range(16 * 8):
        tmp = rev(stream, list(mask))

        if tmp is None:
            break
        stream = tmp
    flag=int(''.join([str(x) for x in stream]), 2)
    print(long_to_bytes(flag))
    mask += M.right_kernel().basis()[k]

hidden_poly

task

from Crypto.Util.Padding import pad
from Crypto.Util.number import *
from Crypto.Cipher import AES
import os

q = 264273181570520944116363476632762225021
key = os.urandom(16)
iv = os.urandom(16)
root = 122536272320154909907460423807891938232
f = sum([a*root**i for i,a in enumerate(key)])
assert key.isascii()
assert f % q == 0

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

cipher = AES.new(key,AES.MODE_CBC, iv)
ciphertext = cipher.encrypt(pad(flag,16)).hex()

with open('output.txt','w') as f:
    f.write(f"{iv = }" + "\n")
    f.write(f"{ciphertext = }" + "\n")

思路

$$
\sum_{i} a_i*root^i=0\mod p
$$

其中$a_i$都是0到128，很小，那么就可以写成

$$
(a_1,···,a_15,1)L=(a_1,···,a_15,a_0) \
L=
\begin{pmatrix}
1 & & root \
&· &· \
&1& root^{15} \
&& q
\end{pmatrix}
$$

直接LLL就行了

exp

iv = b'Gc\xf2\xfd\x94\xdc\xc8\xbb\xf4\x84\xb1\xfd\x96\xcd6\\'
ciphertext ='d23eac665cdb57a8ae7764bb4497eb2f79729537e596600ded7a068c407e67ea75e6d76eb9e23e21634b84a96424130e'
q = 264273181570520944116363476632762225021
root = 122536272320154909907460423807891938232

from Crypto.Util.number import *
from Crypto.Cipher import AES

L = Matrix(ZZ, 16, 16)
for i in range(16 - 1):
    L[i, i] = 1
    L[i, 15] = root ** (i + 1)

L[-1, -1] = q
L = L.LLL()

for res in L:
    tmp = []
    for idx in res:
        tmp.append(int(abs(idx)))
    tmp = [tmp[-1]] + tmp[:-1]
    if all(0 < x < 128 for x in tmp):
        key = bytearray([x for x in tmp])
        cipher = AES.new(key,AES.MODE_CBC, iv)
        flag = cipher.decrypt(long_to_bytes(int(ciphertext, 16)))
        print(flag)

babe-Lifting

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

p = getPrime(512)
q = getPrime(512)
n = p*q
e = 0x1001
d = inverse(e, (p-1)*(q-1))
bit_leak = 400
d_leak = d & ((1<<bit_leak)-1)
msg = bytes_to_long(flag)
cipher = pow(msg,e,n)
pk = (n, e)

with open('output.txt','w') as f:
    f.write(f"pk = {pk}\n")
    f.write(f"cipher = {cipher}\n")
    f.write(f"hint = {d_leak}\n")
    f.close()

n,e = (53282434320648520638797489235916411774754088938038649364676595382708882567582074768467750091758871986943425295325684397148357683679972957390367050797096129400800737430005406586421368399203345142990796139798355888856700153024507788780229752591276439736039630358687617540130010809829171308760432760545372777123, 4097)
cipher = 14615370570055065930014711673507863471799103656443111041437374352195976523098242549568514149286911564703856030770733394303895224311305717058669800588144055600432004216871763513804811217695900972286301248213735105234803253084265599843829792871483051020532819945635641611821829176170902766901550045863639612054
hint = 1550452349150409256147460237724995145109078733341405037037945312861833198753379389784394833566301246926188176937280242129

思路
d的低位泄露板子

exp

from Crypto.Util.number import long_to_bytes,inverse
n,e = (53282434320648520638797489235916411774754088938038649364676595382708882567582074768467750091758871986943425295325684397148357683679972957390367050797096129400800737430005406586421368399203345142990796139798355888856700153024507788780229752591276439736039630358687617540130010809829171308760432760545372777123, 4097)
c = 14615370570055065930014711673507863471799103656443111041437374352195976523098242549568514149286911564703856030770733394303895224311305717058669800588144055600432004216871763513804811217695900972286301248213735105234803253084265599843829792871483051020532819945635641611821829176170902766901550045863639612054
dlow = 1550452349150409256147460237724995145109078733341405037037945312861833198753379389784394833566301246926188176937280242129

def get_full_p(p_low,n,pbits):
    kbits = p_low.bit_length()
    R.<x> = PolynomialRing(Zmod(n))
    f = x * 2^kbits + p_low
    f = f.monic()
    res = f.small_roots(X = 2^(pbits-kbits),beta=0.4)
    if res != []:
        p = int(res[0]) * 2^kbits + p_low
        return p

for k in range(e):
    var('p')
    f1 = e*dlow*p - (k*n*p - k*p^2 - k*n + (k+1)*p)
    roots = solve_mod(f1,2^400)
    if roots != []:
        for root in roots:
            if int(root[0]).bit_length() == 400:
                p = get_full_p(int(root[0]),n,512)
                if p:
                    q = n // p
                    d = inverse(e,(p-1)*(q-1))
                    m = pow(c,d,n)
                    print(long_to_bytes(m))
                    break

ezLCG

task

from sage.all import *
from random import getrandbits, randint
from secrets import randbelow
from Crypto.Util.number import getPrime,isPrime,inverse
from Crypto.Util.Padding import pad
from Crypto.Cipher import AES
from secret import priKey, flag
from hashlib import sha1
import os

q = getPrime(160)
while True:
    t0 = q*getrandbits(864)
    if isPrime(t0+1):
        p = t0 + 1
        break

x = priKey
assert p % q == 1
h = randint(1,p-1)
g = pow(h,(p-1)//q,p)
y = pow(g,x,p)

def sign(z, k):
    r = pow(g,k,p) % q
    s = (inverse(k,q)*(z+r*priKey)) % q
    return (r,s)

def verify(m,s,r):
    z = int.from_bytes(sha1(m).digest(), 'big')
    u1 = (inverse(s,q)*z) % q
    u2 = (inverse(s,q)*r) % q
    r0 = ((pow(g,u1,p)*pow(y,u2,p)) % p) % q
    return r0 == r

def lcg(a, b, q, x):
    while True:
        x = (a * x + b) % q
        yield x

msg = [os.urandom(16) for i in range(5)]

a, b, x = [randbelow(q) for _ in range(3)]
prng = lcg(a, b, q, x)
sigs = []
for m, k in zip(msg,prng):
    z = int.from_bytes(sha1(m).digest(), "big") % q
    r, s = sign(z, k)
    assert verify(m, s, r)
    sigs.append((r,s))

print(f"{g = }")
print(f"{h = }")
print(f"{q = }")
print(f"{p = }")
print(f"{msg = }")
print(f"{sigs = }")
key = sha1(str(priKey).encode()).digest()[:16]
iv = os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC,iv)
ct = cipher.encrypt(pad(flag,16))
print(f"{iv = }")
print(f"{ct = }")

'''
g = 81569684196645348869992756399797937971436996812346070571468655785762437078898141875334855024163673443340626854915520114728947696423441493858938345078236621180324085934092037313264170158390556505922997447268262289413542862021771393535087410035145796654466502374252061871227164352744675750669230756678480403551
h = 13360659280755238232904342818943446234394025788199830559222919690197648501739683227053179022521444870802363019867146013415532648906174842607370958566866152133141600828695657346665923432059572078189013989803088047702130843109809724983853650634669946823993666248096402349533564966478014376877154404963309438891
q = 1303803697251710037027345981217373884089065173721
p = 135386571420682237420633670579115261427110680959831458510661651985522155814624783887385220768310381778722922186771694358185961218902544998325115481951071052630790578356532158887162956411742570802131927372034113509208643043526086803989709252621829703679985669846412125110620244866047891680775125948940542426381
msg = [b'I\xf0\xccy\xd5~\xed\xf8A\xe4\xdf\x91+\xd4_$', b'~\xa0\x9bCB\xef\xc3SY4W\xf9Aa\rO', b'\xe6\x96\xf4\xac\n9\xa7\xc4\xef\x82S\xe9 XpJ', b'3,\xbb\xe2-\xcc\xa1o\xe6\x93+\xe8\xea=\x17\xd1', b'\x8c\x19PHN\xa8\xbc\xfc\xa20r\xe5\x0bMwJ']
sigs = [(913082810060387697659458045074628688804323008021, 601727298768376770098471394299356176250915124698), (406607720394287512952923256499351875907319590223, 946312910102100744958283218486828279657252761118), (1053968308548067185640057861411672512429603583019, 1284314986796793233060997182105901455285337520635), (878633001726272206179866067197006713383715110096, 1117986485818472813081237963762660460310066865326), (144589405182012718667990046652227725217611617110, 1028458755419859011294952635587376476938670485840)]
iv = b'M\xdf\x0e\x7f\xeaj\x17PE\x97\x8e\xee\xaf:\xa0\xc7'
ct = b"\xa8a\xff\xf1[(\x7f\xf9\x93\xeb0J\xc43\x99\xb25:\xf5>\x1c?\xbd\x8a\xcd)i)\xdd\x87l1\xf5L\xc5\xc5'N\x18\x8d\xa5\x9e\x84\xfe\x80\x9dm\xcc"
'''
from sage.all import *
from random import getrandbits, randint
from secrets import randbelow
from Crypto.Util.number import getPrime,isPrime,inverse
from Crypto.Util.Padding import pad
from Crypto.Cipher import AES
from secret import priKey, flag
from hashlib import sha1
import os

q = getPrime(160)
while True:
    t0 = q*getrandbits(864)
    if isPrime(t0+1):
        p = t0 + 1
        break

x = priKey
assert p % q == 1
h = randint(1,p-1)
g = pow(h,(p-1)//q,p)
y = pow(g,x,p)

def sign(z, k):
    r = pow(g,k,p) % q
    s = (inverse(k,q)*(z+r*priKey)) % q
    return (r,s)

def verify(m,s,r):
    z = int.from_bytes(sha1(m).digest(), 'big')
    u1 = (inverse(s,q)*z) % q
    u2 = (inverse(s,q)*r) % q
    r0 = ((pow(g,u1,p)*pow(y,u2,p)) % p) % q
    return r0 == r

def lcg(a, b, q, x):
    while True:
        x = (a * x + b) % q
        yield x

msg = [os.urandom(16) for i in range(5)]

a, b, x = [randbelow(q) for _ in range(3)]
prng = lcg(a, b, q, x)
sigs = []
for m, k in zip(msg,prng):
    z = int.from_bytes(sha1(m).digest(), "big") % q
    r, s = sign(z, k)
    assert verify(m, s, r)
    sigs.append((r,s))

print(f"{g = }")
print(f"{h = }")
print(f"{q = }")
print(f"{p = }")
print(f"{msg = }")
print(f"{sigs = }")
key = sha1(str(priKey).encode()).digest()[:16]
iv = os.urandom(16)
cipher = AES.new(key, AES.MODE_CBC,iv)
ct = cipher.encrypt(pad(flag,16))
print(f"{iv = }")
print(f"{ct = }")

'''
g = 81569684196645348869992756399797937971436996812346070571468655785762437078898141875334855024163673443340626854915520114728947696423441493858938345078236621180324085934092037313264170158390556505922997447268262289413542862021771393535087410035145796654466502374252061871227164352744675750669230756678480403551
h = 13360659280755238232904342818943446234394025788199830559222919690197648501739683227053179022521444870802363019867146013415532648906174842607370958566866152133141600828695657346665923432059572078189013989803088047702130843109809724983853650634669946823993666248096402349533564966478014376877154404963309438891
q = 1303803697251710037027345981217373884089065173721
p = 135386571420682237420633670579115261427110680959831458510661651985522155814624783887385220768310381778722922186771694358185961218902544998325115481951071052630790578356532158887162956411742570802131927372034113509208643043526086803989709252621829703679985669846412125110620244866047891680775125948940542426381
msg = [b'I\xf0\xccy\xd5~\xed\xf8A\xe4\xdf\x91+\xd4_$', b'~\xa0\x9bCB\xef\xc3SY4W\xf9Aa\rO', b'\xe6\x96\xf4\xac\n9\xa7\xc4\xef\x82S\xe9 XpJ', b'3,\xbb\xe2-\xcc\xa1o\xe6\x93+\xe8\xea=\x17\xd1', b'\x8c\x19PHN\xa8\xbc\xfc\xa20r\xe5\x0bMwJ']
sigs = [(913082810060387697659458045074628688804323008021, 601727298768376770098471394299356176250915124698), (406607720394287512952923256499351875907319590223, 946312910102100744958283218486828279657252761118), (1053968308548067185640057861411672512429603583019, 1284314986796793233060997182105901455285337520635), (878633001726272206179866067197006713383715110096, 1117986485818472813081237963762660460310066865326), (144589405182012718667990046652227725217611617110, 1028458755419859011294952635587376476938670485840)]
iv = b'M\xdf\x0e\x7f\xeaj\x17PE\x97\x8e\xee\xaf:\xa0\xc7'
ct = b"\xa8a\xff\xf1[(\x7f\xf9\x93\xeb0J\xc43\x99\xb25:\xf5>\x1c?\xbd\x8a\xcd)i)\xdd\x87l1\xf5L\xc5\xc5'N\x18\x8d\xa5\x9e\x84\xfe\x80\x9dm\xcc"
'''

思路
可以发现我们有9组多项式方程，且度都不高。用groebner，即可解出未知元

exp

from Crypto.Util.number import *
from Crypto.Cipher import AES
from hashlib import sha1
g = 81569684196645348869992756399797937971436996812346070571468655785762437078898141875334855024163673443340626854915520114728947696423441493858938345078236621180324085934092037313264170158390556505922997447268262289413542862021771393535087410035145796654466502374252061871227164352744675750669230756678480403551
h = 13360659280755238232904342818943446234394025788199830559222919690197648501739683227053179022521444870802363019867146013415532648906174842607370958566866152133141600828695657346665923432059572078189013989803088047702130843109809724983853650634669946823993666248096402349533564966478014376877154404963309438891
q = 1303803697251710037027345981217373884089065173721
p = 135386571420682237420633670579115261427110680959831458510661651985522155814624783887385220768310381778722922186771694358185961218902544998325115481951071052630790578356532158887162956411742570802131927372034113509208643043526086803989709252621829703679985669846412125110620244866047891680775125948940542426381
msg = [b'I\xf0\xccy\xd5~\xed\xf8A\xe4\xdf\x91+\xd4_$', b'~\xa0\x9bCB\xef\xc3SY4W\xf9Aa\rO', b'\xe6\x96\xf4\xac\n9\xa7\xc4\xef\x82S\xe9 XpJ', b'3,\xbb\xe2-\xcc\xa1o\xe6\x93+\xe8\xea=\x17\xd1', b'\x8c\x19PHN\xa8\xbc\xfc\xa20r\xe5\x0bMwJ']
sigs = [(913082810060387697659458045074628688804323008021, 601727298768376770098471394299356176250915124698), (406607720394287512952923256499351875907319590223, 946312910102100744958283218486828279657252761118), (1053968308548067185640057861411672512429603583019, 1284314986796793233060997182105901455285337520635), (878633001726272206179866067197006713383715110096, 1117986485818472813081237963762660460310066865326), (144589405182012718667990046652227725217611617110, 1028458755419859011294952635587376476938670485840)]
iv = b'M\xdf\x0e\x7f\xeaj\x17PE\x97\x8e\xee\xaf:\xa0\xc7'
ct = b"\xa8a\xff\xf1[(\x7f\xf9\x93\xeb0J\xc43\x99\xb25:\xf5>\x1c?\xbd\x8a\xcd)i)\xdd\x87l1\xf5L\xc5\xc5'N\x18\x8d\xa5\x9e\x84\xfe\x80\x9dm\xcc"

msg = [bytes_to_long(sha1(x).digest()) for x in msg]
r = []
s = []

for i, j in sigs:
    r.append(i)
    s.append(j)

PR.<k1,k2,k3,k4,k5,a,b,x> = PolynomialRing(Zmod(q))
f1 = a * k1 + b - k2
f2 = a * k2 + b - k3
f3 = a * k3 + b - k4
f4 = a * k4 + b - k5
f5 = msg[0] + r[0] * x - s[0] * k1
f6 = msg[1] + r[1] * x - s[1] * k2
f7 = msg[2] + r[2] * x - s[2] * k3
f8 = msg[3] + r[3] * x - s[3] * k4
f9 = msg[4] + r[4] * x - s[4] * k5

Fs = [f1, f2, f3, f4, f5, f6, f7, f8, f9]
I = Ideal(Fs)
B = I.groebner_basis()

print(B)
prikey = -1144162652064701115049643134487732928553039124427

priKey = prikey % q
key = sha1(str(priKey).encode()).digest()[:16]

cipher = AES.new(key, AES.MODE_CBC,iv)
flag = cipher.decrypt(ct)
print(flag)
心如草木，向阳而生
wp密码
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