比赛记录

记录一下以前的一些比赛，这里感谢striving大佬，好多都是问的他哈哈哈哈。

杭师大

一步到喂

佩尔方程求解

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from gmpy2 import invert

def solve_pell(N):
    cf = continued_fraction(sqrt(N))
    for i in range(10000):
        denom = cf.denominator(i)
        numer = cf.numerator(i)
        if numer*numer - N * denom*denom == 1 :
            return numer, denom
    return None, None
d=105279230912868770223946474836383391725923
a=1293023064232431070902426583269468463
x,y=solve_pell(d//a)
print(x)
print(y)
assert 1293023064232431070902426583269468463*pow(x,2)==105279230912868770223946474836383391725923*pow(y,2)+1293023064232431070902426583269468463
p=9850212100620338486478343799784951721581757936621772924971218807300819701941457605399099898870264241518769370682330612103782092302148525012450902351701339
q=10749429992014823019923966246896511618886613763258781706004694804949547801668777655988055847885755337127548775758133804022361510427909703124161450470578543
c=66847321508502017574023222490247591875197038421108556106531660421662626233521063441647157067220450229816184622038471812597874582613385516838920632450015292570673816423432903604941781889308906893966588610214614726388822851471695742453496232748358301888465563812947038856742838097152549971517159475947566599664
phi=(p-1)*(q-1)
d=invert(x,phi)
print(pow(c,d,p*q))
#HZNUCTF{D0_y0u_know_p3ll_3qu4tion!!!}

checkin

连续24000次开方

quadratic_residue和cipolla

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import tqdm
from Crypto.Util.number import *
from sympy.ntheory.modular import crt

def square_root_of_quadratic_residue(n, modulo):
    """Square root of quadratic residue

    Solve the square root of quadratic residue using Cipolla's algorithm with Legendre symbol
    Returns:
        int -- if n is a quadratic residue,
                   return x, such that x^{2} = n (mod modulo)
               otherwise, return -1
    """
    if modulo == 2:
        return 1
    if n % modulo == 0:
        return 0
    Legendre = lambda n: pow(n, modulo - 1 >> 1, modulo)
    if Legendre(n) == modulo - 1:
        return -1
    t = 0
    while Legendre(t ** 2 - n) != modulo - 1:
        t += 1
    w = (t ** 2 - n) % modulo
    return (generate_quadratic_field(w, modulo)(t, 1) ** (modulo + 1 >> 1)).x

def generate_quadratic_field(d, modulo=0):
    """Generate quadratic field number class

    Returns:
        class -- quadratic field number class
    """
    assert (isinstance(modulo, int) and modulo >= 0)

    class QuadraticFieldNumber:
        def __init__(self, x, y):
            self.x = x % modulo
            self.y = y % modulo

        def __mul__(self, another):
            x = self.x * another.x + d * self.y * another.y
            y = self.x * another.y + self.y * another.x
            return self.__class__(x, y)

        def __pow__(self, exponent):
            result = self.__class__(1, 0)
            if exponent:
                temporary = self.__class__(self.x, self.y)
                while exponent:
                    if exponent & 1:
                        result *= temporary
                    temporary *= temporary
                    exponent >>= 1
            return result

        def __str__(self):
            return '({}, {} \\sqrt({}))'.format(self.x, self.y, d)

    return QuadraticFieldNumber

n = 22114546923564420607945063747927422619774890007937503484905798897563036431278699718161460968350749338680452479484253816646632515078192048118109272532310403715802657061990320170724360874028667484527150185159662403573637809180151665727445208585725264186578429094937215068881079399747998551453944363665401263
c = 7274219309267176700435453490636404568410293850833252412471984274955007037941820465929958008672185817002749418809077052781794306899476543760452010370102811167685901654480233874375880047900499814304539829706786470978714629827690730256369200773772396109793338097451559255985268375731804819829315168807228186
h = 1463929459818798711929811606552273520156490689917243949474579232718301828387871678397965433435537694532920957475947201372279363554005600100100224291888130
hint = 5610276127312766429915480651516095336201056367031530733662980757514427535721885723009367286870294772595629284861923351543396909892645845139050780691701736
# b'HZNUCTF{80f937af-6542-4142-b957-09534839da4d}'

e = 1 << 24_000
q = hint // 2024 + 2022
p = n // q
res1 = [c]
total = []
for _ in tqdm.tqdm(range(24_000)):
    tmp, res1 = res1, []
    for ct in tmp:
        mm = square_root_of_quadratic_residue(ct, p)
        if mm not in total:
            total.append(mm)
        else:
            continue
        if p - mm not in total:
            total.append(p - mm)
        else:
            continue
        if mm == 1 or mm == p - 1 or mm == -1:
            continue
        res1 += [mm, p - mm]
    res1 = list(set(res1))
res1 = res1[-2:]

res2 = c
for i in tqdm.tqdm(range(24000)):
    res2 = pow(res2, (q + 1) // 4, q)
res2 = [res2, p - res2]

for i in res1:
    for j in res2:
        m = crt([p, q], [i, j])[0]
        m = long_to_bytes(int(m))
        if b'HZNUCTF' in m:
            print(m)
            break
# b'HZNUCTF{80f937af-6542-4142-b957-09534839da4d}\x97\x86w@b\x1e\xa36\xc7\xe4H5g5\x13\xca\x0f,<\xa6\xb5\x16\x9ey\x92$\xd9\xd6\xa2~\x9f\xdagO\xbd\\3\x14\xfc\x05\xb8\x1f\r\xae2S\x1eA\x90\xaf'

nothing

sm2签名

hnp问题

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from string import *
from hashlib import *
from gmssl import sm2
alp=ascii_letters+digits
print(alp)
def proof_of_work(end,sha):
    for a in alp:
        for b in alp:
            for c in alp:
                for d in alp:
                    start=a+b+c+d
                    s=(start+end).encode()
                    Sha=sha256(s).hexdigest()
                    if Sha==sha:
                        return start
from pwn import *

io=remote('43.156.230.30',10001)
context.log_level='debug'
io.recvuntil('sha256(XXXX+')
end=io.recvuntil(') == ')[:-5].decode()
print(end)
sha=io.recvuntil('\n')[:-1].decode()
print(sha)
xxxx=proof_of_work(end,sha)
io.recvuntil('[+] Plz tell me XXXX: ')
io.sendline(xxxx)
io.recvuntil('ready?[y/n]\n[*] ')
io.sendline('y')
R=[]
S=[]
n=0xFFFFFFFEFFFFFFFFFFFFFFFFFFFFFFFF7203DF6B21C6052B53BBF40939D54123
for _ in range(55):
    io.sendline('1')
    io.recvuntil('[+] U can get my signature\n')
    io.recvuntil('[+] ')
    rs=io.recvuntil('\n')[:-1].decode().zfill(128)
    r,s=int(rs[:64],16),int(rs[64:],16)
    R.append(r)
    S.append(s)  #ri,si
    #s(1+d)=k-r*d mod n
    #s+(s+r)*d=k mod n
    
M=Matrix(55+2,55+2)
M=Matrix(QQ,M)
for i in range(55):
    M[i,i]=n
    M[-2,i]=S[i]+R[i]
    M[-1,i]=S[i]
M[-2,-2]=2**245/n
M[-1,-1]=2**245
v=M.LLL()[1]
sk=int(v[-2]*n/(2**245))
sk=n-sk  #sk  #格子

sm2_easy = sm2.CryptSM2('1', '1')  #sm2
pk = sm2_easy._kg(sk, sm2_easy.ecc_table['g']) #
io.sendline('3')
io.recvuntil('I will give you the flag you want!(hexadecimal format)\n[*] ')
print(pk)
io.sendline(hex(sk))
io.recvall()

NKCTF招新赛

baby_RSA

考点：

  1. 已知e,dp,ne,dp,n分解nn
  2. 费马小定理，数论推导
  3. coppersmith攻击

第一步：已知dp=dmod  (p−1)d_p=d \mod {(p-1)}，所以e∗dp=e∗dmod  (p−1)e*d_p=e*d \mod {(p-1)}。

根据RSA参数之间的关系有：e∗d=1mod  (p−1)∗(q−1)e*d=1\mod{(p-1)*(q-1)}，那么e∗d=1mod  p−1e*d=1\mod{p-1}。

所以e∗dp=1mod  p−1e*d_p=1\mod{p-1}，即存在kk使得e∗dp−1=k∗(p−1)e*d_p-1=k*(p-1)，注意到dp<p−1d_p<p-1，那么k<ek<e。

那么通过枚举kk可以得到p=1+(e∗dp−1)//kp=1+(e*d_p-1)//k。

第二步：有c1=mpmod  n,c2=mqmod  nc_1=m^p\mod{n},c_2=m^q\mod{n}

根据同余以及费马小定理(ap−1=1mod  p)(a^{p-1}=1\mod{p})，有：c1=mmod  p,c2=mmod  qc_1=m\mod{p},c_2=m\mod{q}。

构造多项式：有f(m)=m2−(c1+c2)∗m+c1∗c2mod  nf(m)=m^2-(c_1+c_2)*m+c_1*c_2\mod{n}

第三步：coppersmith攻击可以帮助求解f(x0)=y0mod  Nf(x_0)=y_0\mod{N},其中x0<Xx_0<X，XX是一个上界。

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n =  114101396033690088275999670914803472451228154227614098210572767821433470213124900655723605426526569384342101959232900145334500170690603208327913698128445002527020347955300595384752458477749198178791196660625870659540794807018881780680683388008090434114437818447523471527878292741702348454486217652394664664641
N =  1159977299277711167607914893426674454199208605107323826176606074354449015203832606569051328721360397610665453513201486235549374869954501563523028914285006850687275382822302821825953121223999268058107278346499657597050468069712686559045712946025472616754027552629008516489090871415609098178522863027127254404804829735621706042266140637592206366042515190385496909533329383212542170504864473944657824502882014292528444918055958758310544435120502872883857209880723535754528096143707324179005292445100655695427777453144657819474805882956064292780031599790769618615908501966912635232746588639924772530057835864082951499028
dP =  33967356791272818610254738927769774016289590226681637441101504040121743937150259930712897925893431093938385216227201268238374281750681609796883676743311872905933219290266120756315613501614208779063819499785817502677885240656957036398336462000771885589364702443157120609506628895933862241269347200444629283263
from gmpy2 import *
e=65537
#已知e,dp,n分解n
#dp =d mod p-1
#e*dp =1 +k*(p-1)
for i in range(1,e):
    if (e*dP-1)%i==0:
        P=1+ (e*dP-1)//i
        if N%P==0:
            print(P)
'''
P=37269067352457630263351774206178494913957088859822110344334922741582406663357663275001777826744534556652993452577088773275825539553907027527722045884489297259984687894496505265384077983882580247333972954704644517469999916574893996324149548980338301147983367163067556434434982470623587914250041142380667816331
assert N%P==0
Q=N//P
R.<x>=Zmod(n)[]
f=x^2-(P+Q)*x+P*Q
flag=f.small_roots(X=2^300)
print(flag)
'''
#NKCTF{Th1S_a_babyRSA_y0u_are_tql!!!}

eZ_Math

考点：

  1. 数论推导，欧拉定理，rsa参数之间的关系等

题目给了很多aiximod  n=ci{a_i}^{x_i}\mod{n}={c_i}

注意到ai,cia_i,c_i都很小，而且有的aia_i是幂指关系：如4=22,8=234=2^2,8=2^3,在不同的底数下，也存在相同的cic_i。

我们可以找到其中一组幂指关系的：a1,a2=a1ba_1,a_2={a_1}^b，并且cic_i相同的。

那么此时a1x1=c1mod  na_1^{x_1}=c_1\mod{n}，a2x2=a1b∗x2=c1mod  n{a_2}^{x_2}={a_1}^{b*x_2}=c_1\mod{n}

显然此时有：x1=b∗x2mod  phix_1=b*x_2\mod{phi}，（欧拉定理、或该循环群的阶为phi)

此时可以得到phi的一个倍数kphi=x1−b∗x2kphi=x_1-b*x_2

得到d=e−1mod  kphid=e^{-1}\mod{kphi} (用phi的倍数求私钥与phi等价，可由欧拉定理得)

然后解rsa即可。

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a1=21606807968454766520529084107175158062623274703294799705984925156349373997035743632649715867216648159433730630939058861636582120303338699113498645592338621228070481894445156769437351313971471061779425442129487317917630554305634797690296919992201174927956121524640438775183413288101169580705360562693274958339417
b1=68789042322037899901399142739922213531190892214327212247633649397602243027562329029767224931385807659445647908974735092145336602662873465734923450809783198772444106655438821950560058413359621316189435942839212247873870560114926459781136160541556773230183543715576244986800296759341282346581691226676298068904581
n = 369520637995317866367336688225182965061898803879373674073832046072914710171302486913303917853881549637806426191970292829598855375370563396182543413674021955181862907847280705741114636854238746612618069619482248639049407507041667720977392421249242597197448360531895206645794505182208390084734779667749657408715621
c = 324131338592233305486487416176106472248153652884280898177125443926549710357763331715045582842045967830200123100144721322509500306940560917086108978796500145618443920020112366546853892387011738997522207752873944151628204886591075864677988865335625452099668804529484866900390927644093597772065285222172136374562043
from gmpy2 import *
phi=a1-3*b1
e=65537
d=invert(e,phi)
from Crypto.Util.number import *
print(long_to_bytes(pow(c,d,n)))

fake_MT、real_MT

考点：

  1. MT19937的预测、逆向，部件的逆向
  2. pwntools自动化脚本

1.可以参考https://www.anquanke.com/post/id/205861?display=mobile学习，很经典！很有意思

2.pwntools基本用法：接发数据等等

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from randcrack import *
from gmpy2 import invert
from random import Random


# right shift inverse
def inverse_right(res, shift):
    tmp = res
    bits = len(bin(res)[2:])
    for i in range(bits // shift):
        tmp = res ^ tmp >> shift
    return tmp


# right shift with mask inverse
def inverse_right_values(res, shift, mask):
    tmp = res
    bits = len(bin(res)[2:])
    for i in range(bits // shift):
        tmp = res ^ tmp >> shift & mask
    return tmp


# left shift inverse
def inverse_left(res, shift):
    tmp = res
    bits = len(bin(res)[2:])
    for i in range(bits // shift):
        tmp = res ^ tmp << shift
    return tmp


# left shift with mask inverse
def inverse_left_values(res, shift, mask):
    tmp = res
    bits = len(bin(res)[2:])
    for i in range(bits // shift):
        tmp = res ^ tmp << shift & mask
    return tmp

#extract_recover
def extract_recover(y):
    y = inverse_right(y, 18)
    y = inverse_left_values(y, 15, 4022730752)
    y = inverse_left_values(y, 7, 2636928640)
    y = inverse_right(y, 11)
    return y


def backtrace(cur):
    high = 0x80000000
    low = 0x7fffffff
    mask = 0x9908b0df
    state = cur
    for i in range(3, -1, -1):
        tmp = state[i + 624] ^ state[i + 397]
        # recover Y,tmp = Y
        if tmp & high == high:
            tmp ^= mask
            tmp <<= 1
            tmp |= 1
        else:
            tmp <<= 1
        # recover highest bit
        res = tmp & high
        # recover other 31 bits,when i =0,it just use the method again it so beautiful!!!!
        tmp = state[i - 1 + 624] ^ state[i + 396]
        # recover Y,tmp = Y
        if tmp & high == high:
            tmp ^= mask
            tmp <<= 1
            tmp |= 1
        else:
            tmp <<= 1
        res |= (tmp) & low
        state[i] = res
    return state


def recover_state(out):
    state = []
    for i in out:
        i = inverse_right(i, 18)
        i = inverse_left_values(i, 15, 4022730752)
        i = inverse_left_values(i, 7, 2636928640)
        i = inverse_right(i, 11)
        state.append(i)
    return state
#逆向


def _int32(x):
    return int(0xFFFFFFFF & x)


def init(seed):
    mt = [0] * 624
    mt[0] = seed
    for i in range(1, 624):
        mt[i] = _int32(1812433253 * (mt[i - 1] ^ mt[i - 1] >> 30) + i)
    return mt


def invert_right(res, shift):
    tmp = res
    bits = len(bin(res)[2:])
    for i in range(bits // shift):
        res = tmp ^ res >> shift
    return _int32(res)


def recover(last):
    n = 1 << 32
    inv = invert(1812433253, n)
    for i in range(623, 0, -1):
        last = ((last - i) * inv) % n
        last = invert_right(last, 30)
    return last
#逆向twist


def guess_1(randoms):  # 预测
    rc = RandCrack()
    for i in randoms:
        temp = i
        while temp != 0:
            rc.submit(temp % pow(2, 32))
            temp >>= 32
    number = rc.predict_getrandbits(96)
    return number


def guess_2(randoms):
    partS = recover_state(randoms)
    state = backtrace([0] * 4 + partS)[:624]
    prng = Random()
    prng.setstate((3, tuple(state + [0]), None))
    nn = prng.getrandbits(32)
    number = prng.getrandbits(96)
    return number


def guess_3(last_number):
    seed = recover(last_number)
    return seed


def guess_4(ex_number):
    return extract_recover(ex_number)


from pwn import *

io = remote('node.yuzhian.com.cn',30279)
context.log_level='debug'

io.send('1')  #按任意键开始？

for i in range(20):
    io.recvuntil("Round: " + str(i + 1) + '\n')
    Line = io.recvuntil('\n')[:-1].decode()
    if 'randoms = ' in Line:
        randoms = eval(Line[9:].replace('L',''))  #去除L
        io.recvuntil('Guess ')
        tag = io.recvuntil(' number:').decode()
        print(tag)
        if 'after' in tag:
            io.sendline(str(guess_1(randoms)))
        if 'pre' in tag:
            io.sendline(str(guess_2(randoms)))
    if 'last number = ' in Line:
        last_number = int(Line[14:])
        io.recvuntil('Guess seed number:')
        io.sendline(str(guess_3(last_number)))
    if 'extract number = ' in Line:
        ex_number = int(Line[17:])
        io.recvuntil('Guess be extracted number:')
        io.sendline(str(guess_4(ex_number)))
io.recvall()

easyrsa

考点：

  1. 数论推导
  2. coppersmith攻击

第一问：n=p∗q∗s∗tn=p*q*s*t，phi=(p−1)∗(q−1)∗(s−1)∗(t−1)phi=(p-1)*(q-1)*(s-1)*(t-1)

首先可以求出d=e−1mod  phid=e^{-1}\mod{phi}，而c=memod  p∗qc=m^e\mod{p*q}

由欧拉定理有：cd=mmod  p∗qc^d=m\mod{p*q}

即：f(x)=cd−xmod  p∗qmod  nf(x)=c^d-x\mod{p*q}\mod{n}，mm较小。

利用coppersmith攻击求出x。

第二问同babyrsa。

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from gmpy2 import *

n = 8836130216343708623415307573630337110573363595188748983290313549413242332143945452914800845282478216810685733227137911630239808895196748125078747600505626165666334675100147790578546682128517668100858766784733351894480181877144793496927464058323582165412552970999921215333509253052644024478417393146000490808639363681195799826541558906527985336104761974023394438549055804234997654701266967731137282297623426318212701157416397999108259257077847307874122736921265599854976855949680133804464839768470200425669609996841568545945133611190979810786943246285103031363790663362165522662820344917056587244701635831061853354597
phi = 8836130216343708623415307573630337110573363595188748983290313549413242332143945452914800845282478216810685733227137911630239808895196748125078747600505622503351461565956106005118029537938273153581675065762015952483687057805462728186901563990429998916382820576211887477098611684072561849314986341226981300596338314989867731725668312057134075244816223120038573374383949718714549930261073576391501671722900294331289082826058292599838631513746370889828026039555245672195833927609280773258978856664434349221972568651378808050580665443131001632395175205804045958846124475183825589672204752895252723130454951830966138888560
c1 = 78327207863361017953496121356221173288422862370301396867341957979087627011991738176024643637029313969241151622985226595093079857523487726626882109114134910056673489916408854152274726721451884257677533593174371742411008169082367666168983943358876017521749198218529804830864940274185360506199116451280975188409
e = 65537
#c1 = m^e mod p*q
#m1 = c1^d mod p*q*r*t
#m1 = m mod p*q √
d = invert(e, phi)
c = int(pow(c1, d, n))
PR.< x > = PolynomialRing(Zmod(n))
f = c - x #c-x=k*p*q
f = f.monic()
x0 = f.small_roots(X=2^300,beta=0.4,epsilon=0.015)[0]
print(x0)
#NKCTF{it_i5_e45y_th4t_Kn0wn

N = 157202814866563156513184271957553223260772141845129283711146204376449001653397810781717934720804041916333174673656579086498762693983380365527400604554663873045166444369504886603233275868192688995284322277504050322927511160583280269073338415758019142878016084536129741435221345599028001581385308324407324725353
c2 = 63355788175487221030596314921407476078592001060627033831694843409637965350474955727383434406640075122932939559532216639739294413008164038257338675094324172634789610307227365830016457714456293397466445820352804725466971828172010276387616894829328491068298742711984800900411277550023220538443014162710037992032
c3 = 9266334096866207047544089419994475379619964393206968260875878305040712629590906330073542575719856965053269812924808810766674072615270535207284077081944428011398767330973702305174973148018082513467080087706443512285098600431136743009829009567065760786940706627087366702015319792328141978938111501345426931078

PR.< x > = PolynomialRing(Zmod(N))
f = x^2-(c2+c3)*x+c2*c3
f = f.monic()
x0 = f.small_roots(X=2^440,beta=0.4,epsilon=0.015)[0]
print(x0)
#_phi_4nd_N_dec0mp0ses_N_w1th_th3_s4m3_c0mm0n_n_but_pq}

eZ_Bl⊕ck

考点

  1. Feistel结构
  2. 异或

可以参考：http://striving.mapleice.top/index.php/archives/150/第三题学习

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from Crypto.Util.strxor import *

r=b"t\xf7\xaa\xac\x9d\x88\xa4\x8b\x1f+pA\x84\xacHg'\x07{\xcc\x06\xc4i\xdd)\xda\xc9\xad\xa9\xe8\x1fi"
c1=b"'{<z}\x91\xda\xc5\xd5S\x8b\xfa\x9f~]J\x0f\xf4\x9a\x1e\xe0\xef\x129N\xe7a\x928+\xe0\xee"
c2=b'8\x1f"\x83B4\x86)\xce\xebq3\x06\xa0w\x16U\x04M/w\xa1\x8f;)M\xdd~\x11:\xe3\xb3'

kn1=strxor(r[16:],c1[:16])

l1=strxor(kn1,c2[:16])
print(l1)

kn2=strxor(strxor(r[16:],r[:16]),c1[16:])
r1=strxor(strxor(kn2,c2[16:]),l1)
print(r1)

#NKCTF{1ccd5cee-c96d-4caf-8ce5-9a512b3d0655}

ez_polynomial

考点：

  1. 典型的多项式rsa

可以参考:https://blog.csdn.net/m0_57291352/article/details/124850426学习

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p=40031
R.<y> = PolynomialRing(GF(p))
N=24096*y^93 + 38785*y^92 + 17489*y^91 + 9067*y^90 + 1034*y^89 + 6534*y^88 + 35818*y^87 + 22046*y^86 + 12887*y^85 + 445*y^84 + 26322*y^83 + 37045*y^82 + 4486*y^81 + 3503*y^80 + 1184*y^79 + 38471*y^78 + 8012*y^77 + 36561*y^76 + 19429*y^75 + 35227*y^74 + 10813*y^73 + 26341*y^72 + 29474*y^71 + 2059*y^70 + 16068*y^69 + 31597*y^68 + 14685*y^67 + 9266*y^66 + 31019*y^65 + 6171*y^64 + 385*y^63 + 28986*y^62 + 9912*y^61 + 10632*y^60 + 33741*y^59 + 12634*y^58 + 21179*y^57 + 35548*y^56 + 17894*y^55 + 7152*y^54 + 9440*y^53 + 4004*y^52 + 2600*y^51 + 12281*y^50 + 22*y^49 + 17314*y^48 + 32694*y^47 + 7693*y^46 + 6567*y^45 + 19897*y^44 + 27329*y^43 + 8799*y^42 + 36348*y^41 + 33963*y^40 + 23730*y^39 + 27685*y^38 + 29037*y^37 + 14622*y^36 + 29608*y^35 + 39588*y^34 + 23294*y^33 + 757*y^32 + 20140*y^31 + 19511*y^30 + 1469*y^29 + 3898*y^28 + 6630*y^27 + 19610*y^26 + 11631*y^25 + 7188*y^24 + 11683*y^23 + 35611*y^22 + 37286*y^21 + 32139*y^20 + 20296*y^19 + 36426*y^18 + 25340*y^17 + 36204*y^16 + 37787*y^15 + 31256*y^14 + 505*y^13 + 27508*y^12 + 20885*y^11 + 32037*y^10 + 31236*y^9 + 7929*y^8 + 27195*y^7 + 28980*y^6 + 11863*y^5 + 16025*y^4 + 16389*y^3 + 570*y^2 + 36547*y + 10451
e = 65537
q1,q2=N.factor()
q1,q2=q1[0],q2[0]
phi = (p**q1.degree() - 1) * (p**q2.degree() - 1)
assert gcd(e, phi) == 1
d = inverse_mod(e, phi)
S.<x> = R.quotient(N)
C=3552*x^92 + 6082*x^91 + 25295*x^90 + 35988*x^89 + 26052*x^88 + 16987*x^87 + 12854*x^86 + 25117*x^85 + 25800*x^84 + 30297*x^83 + 5589*x^82 + 23233*x^81 + 14449*x^80 + 4712*x^79 + 35719*x^78 + 1696*x^77 + 35653*x^76 + 13995*x^75 + 13715*x^74 + 4578*x^73 + 37366*x^72 + 25260*x^71 + 28865*x^70 + 36120*x^69 + 7047*x^68 + 10497*x^67 + 19160*x^66 + 17939*x^65 + 14850*x^64 + 6705*x^63 + 17805*x^62 + 30083*x^61 + 2400*x^60 + 10685*x^59 + 15272*x^58 + 2225*x^57 + 13194*x^56 + 14251*x^55 + 31016*x^54 + 10189*x^53 + 35040*x^52 + 7042*x^51 + 29206*x^50 + 39363*x^49 + 32608*x^48 + 38614*x^47 + 5528*x^46 + 20119*x^45 + 13439*x^44 + 25468*x^43 + 30056*x^42 + 19720*x^41 + 21808*x^40 + 3712*x^39 + 25243*x^38 + 10606*x^37 + 16247*x^36 + 36106*x^35 + 17287*x^34 + 36276*x^33 + 1407*x^32 + 28839*x^31 + 8459*x^30 + 38863*x^29 + 435*x^28 + 913*x^27 + 36619*x^26 + 15572*x^25 + 9363*x^24 + 36837*x^23 + 17925*x^22 + 38567*x^21 + 38709*x^20 + 13582*x^19 + 35038*x^18 + 31121*x^17 + 8933*x^16 + 1666*x^15 + 21940*x^14 + 25585*x^13 + 840*x^12 + 21938*x^11 + 20143*x^10 + 28507*x^9 + 5947*x^8 + 20289*x^7 + 32196*x^6 + 924*x^5 + 370*x^4 + 14849*x^3 + 10780*x^2 + 14035*x + 15327
m=C^d
for i in m:
    print(chr(int(i)),end='')

Raven

挺有意思的一个题目，多项式的系数都比较小。所以可以看作是一个6次的一元多项式。造格子出系数（不过不知道为啥d不能直接出，这里是先出a,b,c。再单独二元copper求d）。

涉及到格基规约和二元coppersimth，较为复杂，先自行研究。

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p = 
pairs = 

K=2**520
M=[[0 for i in range(12)] for j in range(12)]  #M=[[0]*12]*12
for i in range(4):
    for j in range(7):
        M[j][i]=p-pow(pairs[i][0],6-j,p)
    M[7+i][i]=p
    M[11][i]=pairs[i][1]
for i in range(7):
    M[i][4+i]=1
M[-1][-1]=K
M=Matrix(M)
L=M.LLL()
for i in L:
    if int(i[-1])==K:
        print(i)
        sol=i*M^-1
sol=[int(i) for i in sol[:7]]
print(sol)
from gmpy2 import *
a=int(iroot(sol[0],2)[0])
b=sol[1]//(2*a)
c=(sol[2]-b*b)//(2*a)
print(a,b,c)
#d=256836945952994885035467059118377791095

二元coppersmith恢复key

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import itertools


# 一种求多项式小根问题的方法
def small_roots(f, bounds, m=1, d=None):
    if not d:
        d = f.degree()

    R = f.base_ring()
    N = R.cardinality()

    f /= f.coefficients().pop(0)
    f = f.change_ring(ZZ)

    G = Sequence([], f.parent())
    for i in range(m + 1):
        base = N ^ (m - i) * f ^ i
        for shifts in itertools.product(range(d), repeat=f.nvariables()):
            g = base * prod(map(power, f.variables(), shifts))
            G.append(g)

    B, monomials = G.coefficient_matrix()
    monomials = vector(monomials)

    factors = [monomial(*bounds) for monomial in monomials]
    for i, factor in enumerate(factors):
        B.rescale_col(i, factor)

    B = B.dense_matrix().LLL()

    B = B.change_ring(QQ)
    for i, factor in enumerate(factors):
        B.rescale_col(i, 1 / factor)

    H = Sequence([], f.parent().change_ring(QQ))
    for h in filter(None, B * monomials):
        H.append(h)
        I = H.ideal()
        if I.dimension() == -1:
            H.pop()
        elif I.dimension() == 0:
            roots = []
            for root in I.variety(ring=ZZ):
                root = tuple(R(root[var]) for var in f.variables())
                roots.append(root)
            return roots

    return []

r3=38110026992499584203762396049303607649993098725357412914752750999593537061009
x3=68711183101845981499596464753252121346970486988311398916877579778110690480447199642602267233989256728822535174215153145632158860662954277116345331672194812126361911061449082917955000137698138358926301360506687271134873
y3=995428771589393162202488762223106955302099250561593105620410424291405842350539887383005328242236156038373244928147473800972534658018117705291472213770335998508454938607290279268848513727721410314612261163489156360908800
p=1018551160851728231474335384388576586031917743463656622083024684199383855595168341728561337234276243780407755294430553694832049089534855113774546001494743212076463713621965520780122783825100696968959866614846174188401153

a,b,c=(248509346369130391242170998803167584710,34130269465266010935033619912669159600,190347116220905792012773821811284452644)
A=a*x3*x3*x3+b*x3*x3+c*x3
# 设置一个基于整数环的多项式,一元或者多元
P.< x, y >= PolynomialRing(Zmod(p))
# f为多项式
f = (A+x)*(A+x)+y-y3
# bounds为根的最大值,粗略范围即可
bounds =(2^128,2^256)
# m,d选填
m = 3
d = 4
roots = small_roots(f, bounds, m, d)
print(roots)

解AES

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

key=256836945952994885035467059118377791095
cipher = AES.new(key=long_to_bytes(key), IV=bytes(range(16)), mode=AES.MODE_CBC)

ct = b"|2\xf0v7\x05Y\x89\r]\xe93s\rr)#3\xe9\x90%Z\x9a\xd9\x9ck\xba\xec]q\xb8\xf2'\xc8e~fL\xcf\x93\x00\xd6^s-\xc9\xd6M"
print(cipher.decrypt(ct))

easy_high

考点：

  1. 异或
  2. 已知p的部分比特的coppersmith攻击

参考学习coppersmith攻击的经典应用：已知p的部分比特，已知部分明文，已知部分私钥低位等。

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p0=149263925308155304734002881595820602641174737629551638146384199378753884153459661375931646716325020758837194837271581361322079811468970876532640273110966545339040194118880506352109559900553776706613338890047890747811129988585025948270181264314668772556874718178868209009192010129918138140332707080927643141811

p1=p0%(2**444)  #低位
b=p0.bit_length()-250
p2=(p0>>b)<<b
N=17192509201635459965397076685948071839556595198733884616568925970608227408244870123644193452116734188924766414178232653941867668088060274364830452998991993756231372252367134508712447410029668020439498980619263308413952840568602285764163331028384281840387206878673090608323292785024372223569438874557728414737773416206032540038861064700108597448191546413236875600906013508022023794395360001242071569785940215873854748631691555516626235191098174739613181230094797844414203694879874212340812119576042962565179579136753839946922829803044355134086779223242080575811804564731938746051591474236147749401914216734714709281349
'''
PR.< x > = PolynomialRing(Zmod(N))
f = p1+p2+x*2^444
f=f.monic()
x0 = f.small_roots(X=2 ^ 340, beta=0.4)[0]
print(x0)
'''
x0=1414640121010865756473118885903739629697184780736973873076436168814438323477468170166972852648894898
p=p1+p2+x0*2**444
q=N//p
phi=(p-1)*(q-1)
from gmpy2 import *
from Crypto.Util.number import *
d=invert(65537,phi)
c=4881545863615247924697512170011400857004555681758106351259776881249360423774694437921554056529064037535796844084045263140567168171628832384672612945806728465127954937293787045302307135365408938448006548465000663247116917564500525499976139556325841597810084111303039525833367199565266613007333465332710833102978756654324956219855687611590278570749890543277201538208370370097424105751568285050703167350889953331829275262932104042040526209179357770495596739361176548337593674366015027648541293309465113202672923556991818236011769228078267484362980348613669012975963468592763463397575879215173972436831753615524193609612
print(long_to_bytes(pow(c,d,N)))

eZ_LargeCG

考点

  1. p+1光滑

  2. p-1光滑

  3. 线性递推->矩阵快速幂

可以参考http://striving.mapleice.top/index.php/archives/141/第四题学习

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'''
#p-1光滑
python primefac -vs -m=p-1 xxx
'''
p=427721675251610827084310512123962488210068003845592404231631542730839819224381
n1=39755206609675677517559022219519767646524455449142889144073217274247893104711318356648198334858966762944109142752432641040037415587397244438634301062818169
q=n1//p
q1=max(p,q)
p1=min(p,q)
'''
#p+1光滑
python primefac -vs -m=p+1 xxx
'''
p=106481130516780473105954611077340189174861459077145246394800660809527032990637
q=288551157776490110472645044398395422160196115791981535735903775378294599329633
q2=max(p,q)
p2=min(p,q)

#线性递推 快速幂
r = 7948275435515074902473978567170931671982245044864706132834233483354166398627204583162848756424199888842910697874390403881343013872330344844971750121043493
A = 6085327340671394838391386566774092636784105046872311226269065664501131836034666722102264842236327898770287752026397099940098916322051606027565395747098434
B = 1385551782355619987198268805270109182589006873371541520953112424858566073422289235930944613836387546298080386848159955053303343649615385527645536504580787
C = 2529291156468264643335767070801583140819639532551726975314270127875306069067016825677707064451364791677536138503947465612206191051563106705150921639560469

a,b,c,d=p1,q1,p2,q2
M=matrix(Zmod(r),[[c,b,a,1],
                  [1,0,0,0],
                  [0,1,0,0],
                  [0,0,0,1]])
N=matrix(Zmod(r),[[C,B,A,d]])
N=N.T
nn=6**666
M=M^nn
print(M^-1*N)

print(long_to_bytes(718268686893438084080386300782696916434605242000201123193568557324202508308322958959518435405441886593296740-2023))

complex_matrix

考点：

  1. 扩展维纳攻击（四维）

可以参考https://huangx607087.online/2021/03/01/LatticeNotes6/学习

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from Crypto.Util.number import *

n = 71841248095369087024928175623295380241516644434969868335504061065977014103487197287619667598363486210886674500469383623511906399909335989202774281795855975972913438448899231650449810696539722877903606541112937729384851506921675290984316325565141178015123381439392534417225128922398194700511937668809140024838070124095703585627058463137549632965723304713166804084673075651182998654091113119667582720831809458721072371364839503563819080226784026253
e = 65537
c = 39297018404565022956251803918747154798377576057123078716166221329195959669756819453426741569480551313085435037629493881038383709458043802420338889323233368852331387845200216275712388921820794980987541224782392553528127093154957890356084331463340193478391679540506421250562554424770350351514435220782124981277580072039637811543914983033300225131364246910828188727043248991987332274929827173923543187017105236008487756190002204169623313222748976369
e1, e2, e3, e4 = (
65128799196671634905309494529154568614228788035735808211836905142007976099865571126946706559109393187772126407982007858423859147772762638898854472065889939549916077695303157760259717113616428849798058080633047516455513870697383339784816006154279428812359241282979297285283850338964993773227397528608557211742425548651971558377656644211835094019462699301650412862894391885325969143805924684662849869947172175608502179438901337558870349697233790535,
58756559706647121529575085912021603170286163639572075337348109911506627489265537716060463072086480156516641723700802217411122982693536541892986623158818442274840863016647800896033363360822503445344748132842451806511693779600370832206455202293028402486647422212959763287987847280322100701242139127654031151565924132562837893975505159702015125483479126108892709063135006366792197127007229210558758401679638300464111782814561428899998471531067163715,
34828685390969672139784723764579499920301439564705391196519314224159563070870933754477650614819514127121146216049444888554338415587165719098661141454627820126445291802801256297252654045398330613075575527685542980264993711077876535643646746742646371967302159565887123638001580042027272379341650995728849759541960087953160211696369079708787543303742132161742979856720539914370868829868891655221361545648778590685232034703220732697083024449894197969,
26717968456600556973167180286909817773394160817933525240720067057464671317174201540556176814203780603153696663101158205367554829261808020426363683474848952397963507069306452835776851274959389849223566030857588019845781623271395012194869024566879791449466064832273531795430185178486425688475688634844530106740480643866537205900809400383304665727460014210405339697947582657505028211149470787536144302545259243549176816653560626044921521516818788487)

A = [
    [1, -n, 0, n ^ 2, 0, 0, 0, -n ^ 3, 0, 0, 0, 0, 0, 0, 0, n ^ 4],
    [0, 1, -1, -n, -1, 0, n, n ^ 2, -1, 0, n, 0, n, 0, -n ^ 2, -n ^ 3],
    [0, 0, 1, -n, 0, n, 0, n ^ 2, 0, n, 0, 0, 0, -n ^ 2, 0, -n ^ 3],
    [0, 0, 0, 1, 0, -1, -1, -n, 0, -1, -1, 0, 0, n, n, n ^ 2],
    [0, 0, 0, 0, 1, -n, -n, n ^ 2, 0, 0, 0, -n ^ 2, 0, 0, 0, -n ^ 3],
    [0, 0, 0, 0, 0, 1, 0, -n, 0, 0, 0, n, -1, 0, n, n ^ 2],
    [0, 0, 0, 0, 0, 0, 1, -n, 0, 0, 0, n, 0, n, 0, n ^ 2],
    [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, -1, -1, -n],
    [0, 0, 0, 0, 0, 0, 0, 0, 1, -n, -n, n ^ 2, -n, n ^ 2, n ^ 2, -n ^ 3],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, -n, 0, -n, 0, n ^ 2],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -n, 0, 0, -n, n ^ 2],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -n],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -n, -n, n ^ 2],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, -n],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, -n],
    [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]]
Q = [n ^ 2, n ^ 1.5, n ^ 2.4, n, n ^ 2.4, n ^ 1.9, n ^ 1.9, n ^ 0.5, n ^ 2.4, n ^ 1.9, n ^ 1.9, n ^ 1.4, n ^ 1.9,
     n ^ 1.4, n ^ 1.4, 1]
P = [1, e1, e2, e1 * e2, e3, e1 * e3, e2 * e3, e1 * e2 * e3, e4, e1 * e4, e2 * e4, e1 * e2 * e4, e3 * e4, e1 * e3 * e4,
     e2 * e3 * e4, e1 * e2 * e3 * e4]
A, P, Q = matrix(ZZ, A), diagonal_matrix(ZZ, P), diagonal_matrix(ZZ, Q)
A = P * A * Q
w = vector(ZZ, A.LLL()[0])
v = w * A ^ (-1)
phi = int(e1 * v[1] // v[0])
print(hex(phi))
d = inverse(e, phi)
print(long_to_bytes(pow(c, d, n)))

baby_classical

看着复杂，把题目函数都理一遍，其实就一个矩阵运算，类维吉尼亚加密

考点：

  1. python
  2. 矩阵运算
  3. 维吉尼亚加密
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import string

dic = string.ascii_uppercase + string.ascii_lowercase + string.digits + '+/' #大写小写+/
f1nd = lambda x: dic.find(x)

enc_key='pVvRe/G08rLhfwa'

res=[f1nd(i) for i in enc_key]
res=[res[i:i+3] for i in range(0,len(res),3)]
print(res)
K= matrix(Zmod(64),[[13 ,37 ,10],
 [15 ,17 ,41],
 [13  ,0 ,10]])
long=[]
for i in res:
    r=matrix(Zmod(64),[i])
    l=r*K^-1
    print(l)
    for j in l[0]:
        long.append(int(j))
print(long)
dicn2s={i: dic[i] for i in range(64)}
dics2n = dict(zip(dicn2s.values(), dicn2s.keys()))
print(dics2n)
key=''.join([dicn2s[i] for i in long])[:-1]
print(key) #W3ar3N0wayBack

c='1k2Pe{24seBl4_a6Ot_fp7O1_eHk_Plg3EF_g/JtIonut4/}'
c=c.replace('_',' ')
flag=''
j=0
for i  in c:
    if i in dic:
        flag+=dic[ (f1nd(i)-f1nd(key[j%len(key)]) )%64 ]
    else:
        flag+=i
    j+=1
print(flag)
flag=flag.replace('}','_').replace('{','_').replace(' ','_').split('_')
flag='_'.join([i[::-1] for i in flag])
print(flag.swapcase())
#NKCTF{ClaSsic_c0de_d0l1s_aRe_r3a1ly_int3reSting}

N1CTF招新赛

ezezez

知识点：椭圆曲线，数论，Related Message Attack

语言：sagemath

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from Crypto.Util.number import *
a = 259416965926853287275013962386631097709
n = 8963244108103857378610478915798540027754205184635651624801332901515870452435977078824267979279218575027152343028911705254378598817260939339516835322386427
gy = [1896722249337161391193797083807118705857637064141055866260977430130447249097804719101078827699386498665026965515554644281163046869672493999194358926022865, 1971472878970365404484102034878792535366624572891039926216530789642810299559238977732710681752685090332485586910206690093655977703795335533365202958013518, 4074555112002157403114317690671003025383051142150876016519408756236925988856478235258960189636836867363451314784889114354880607055705753428890566240606935]
#y^2=x^3+ax+b mod n
magic = bytes_to_long(b'N1Junior_CheckIn')
e=24
gx1=magic
b=(gy[1]*gy[1]-gx1*gx1*gx1-a*gx1)%n #求出b

#fx=y0^2-(x^e)^3-a(x^e)-b
#gx=y2^2-((x+magic)^e)^3-a( (x+magic)^e )-b
def gcd(g1, g2):
    while g2:
        g1, g2 = g2, g1 % g2
    return g1.monic()
PR.<x>=PolynomialRing(Zmod(n))
fx=gy[0]*gy[0]-x^(3*e)-a*x^e-b
gx=gy[2]*gy[2]-(x+magic)^(3*e)-a*((x+magic)^e)-b
print(gcd(fx,gx))
比赛记录
http://yzsandw.com/2023/12/11/比赛记录/
作者
5Y2z
发布于
2023年12月11日
许可协议