2023ctfshow愚人杯-Toy模板网

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easy_base

4C455A5645334C44474A55484D5A42544F5132574956525A50464E464F4E4C474D4656454D334359474A554751564B4949493255535532464E42544643504A35

丢进cyberchef网站直接梭

大牛的密码

from Crypto.Util.number import *
from flag import flag
from Crypto.Util.Padding import pad
from random import *
def s_box(a):
    box=[i for i in range(a)]
    shuffle(box)  #shuffle() 方法将序列的所有元素随机排序
    return box
BLOCK=16
flag=pad(flag,BLOCK)
S_BOX=s_box(len(flag))
m=[i for i in flag]
def swap(a,b):
    tmp = a
    a = b
    b = tmp
def encrypt1(m):
    enc=[m[i:i+BLOCK] for i in range(0,len(m),BLOCK)]
    for i in enc:
        for j in range(BLOCK):
            aa=j*7%BLOCK
            swap(i[j],i[aa])
def encrypt2(m):
    for i in range(16):
        m=[m[i] for i in S_BOX]
    return m
encrypt1(m)
c=encrypt2(m)
print(S_BOX)
print(c)
'''
[9, 31, 32, 38, 20, 1, 22, 4, 8, 2, 11, 21, 7, 18, 46, 23, 34, 3, 19, 12, 45, 30, 27, 37, 5, 47, 28, 36, 0, 43, 39, 10, 29, 14, 40, 24, 33, 16, 17, 6, 42, 15, 26, 41, 44, 25, 35, 13]
[99, 111, 102, 11, 107, 49, 11, 53, 121, 48, 114, 117, 11, 95, 112, 95, 109, 115, 11, 95, 101, 95, 119, 117, 79, 123, 111, 48, 110, 95, 121, 116, 121, 125, 116, 11, 119, 11, 97, 67, 11, 11, 11, 11, 11, 99, 110, 104]
'''

分析一下代码，其实encrypt1函数对m无任何影响，那就只需要弄清楚encrypt2，这里对m进行了16次加密，且都是指定位置替换，直接写脚本即可

S_BOX=[9, 31, 32, 38, 20, 1, 22, 4, 8, 2, 11, 21, 7, 18, 46, 23, 34, 3, 19, 12, 45, 30, 27, 37, 5, 47, 28, 36, 0, 43, 39, 10, 29, 14, 40, 24, 33, 16, 17, 6, 42, 15, 26, 41, 44, 25, 35, 13]
c=[99, 111, 102, 11, 107, 49, 11, 53, 121, 48, 114, 117, 11, 95, 112, 95, 109, 115, 11, 95, 101, 95, 119, 117, 79, 123, 111, 48, 110, 95, 121, 116, 121, 125, 116, 11, 119, 11, 97, 67, 11, 11, 11, 11, 11, 99, 110, 104]
for i in range(16):
    m = [0] * 48
    for j in range(48):
        m[S_BOX[j]]=c[j]
    c=m
for i in c:
    if i ==11:
        continue
    else:
        print(chr(i),end='')

easy_xor

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

assert len(flag[8:-1])==23  #说明m比特位是256位
m = bytes_to_long(flag)
p = getPrime(1024)
q = getPrime(1024)
n = p*q
e = 65537
c1 = m^p
c2 = pow(m,e,n)
print(f'c1 = {c1}')
print(f'c2 = {c2}')
print(f'n = {n}')
'''
c1 = 151198307301713399973545627808177783191262282577048906899567665485020342464366268384613589477129150406859219553325982275344405383612415523342568367197935454935162234419239807109194526080836070453102172720442102673200212658553214847476648456720629906051324248179394810385918370092764118401652990951968387233220
c2 = 7894512574379281106340582833782408137686355961537832816105517328532111343730615739255485918919146012721446905489729048235088965936700563973759759039693443386542070451737445467143517377017890468837697907596398070608179281207203217576205857817411996178441661371846647602166663752324880657668362355493701482869858528298247422875427747085642627978367348931707497113936723122393282697211257939351221141536029828744507560524637999804394951722319070365576391442828074457050403771353328835153787572457070779602728359333021922987279454923820866436212282592764768470608545881718922440010751845730974331917142224339664090863915
n = 20873587976264698212013861921447267548758723109929620330136081844796427967720295581580927324390713931549639540337285515365487607593546367886570408812338077846317206794057714877394609181224434104303259411081376607299962306250984285173463537669954845497211859940191392861121877814873939865829555350848523691546006073264112091406848179785659505299775196062799482197712761744192962658799557108701192680225134300686608396391566674966897700511638643429161735764600752699251493599533703928135311599575989253347234975026924804433742500175666009324057320386262109587593814197687132304704244158862263859846356497849518103755981
'''

c1=m^p，异或运算是针对二进制的逐一异或，这里告诉了m的比特位是256位，p的比特位1024位，那么c1前1024-256位是等于p的1024-256位的，那么只需要求p后面的256位即可，那么就可以联想到p的高位攻击，需要对c1进行变形成(c1>>256)<<256，那么自需要在sagemath进行p高位攻击求解p

#sagemath

from Crypto.Util.number import *
import gmpy2

c1 = 151198307301713399973545627808177783191262282577048906899567665485020342464366268384613589477129150406859219553325982275344405383612415523342568367197935454935162234419239807109194526080836070453102172720442102673200212658553214847476648456720629906051324248179394810385918370092764118401652990951968387233220
c2 = 7894512574379281106340582833782408137686355961537832816105517328532111343730615739255485918919146012721446905489729048235088965936700563973759759039693443386542070451737445467143517377017890468837697907596398070608179281207203217576205857817411996178441661371846647602166663752324880657668362355493701482869858528298247422875427747085642627978367348931707497113936723122393282697211257939351221141536029828744507560524637999804394951722319070365576391442828074457050403771353328835153787572457070779602728359333021922987279454923820866436212282592764768470608545881718922440010751845730974331917142224339664090863915
n = 20873587976264698212013861921447267548758723109929620330136081844796427967720295581580927324390713931549639540337285515365487607593546367886570408812338077846317206794057714877394609181224434104303259411081376607299962306250984285173463537669954845497211859940191392861121877814873939865829555350848523691546006073264112091406848179785659505299775196062799482197712761744192962658799557108701192680225134300686608396391566674966897700511638643429161735764600752699251493599533703928135311599575989253347234975026924804433742500175666009324057320386262109587593814197687132304704244158862263859846356497849518103755981
e = 65537
R.<x> = PolynomialRing(Zmod(n), implementation='NTL')
c3=(c1>>256)<<256  #相当于p高位攻击
p = c3 + x
x0 = p.small_roots(X = 2^256, beta = 0.1)[0]
P = int(p(x0))
Q = n//P
phi=(P-1)*(Q-1)
d=inverse(e,phi)
m=gmpy2.powmod(c2,d,n)
print(long_to_bytes(m))

这里求两种方法求m，m=c1^p或正常的RSA求解

ecc_mini

#sage
from Crypto.Util.number import *
from secret import flag
flag=bytes_to_long(flag)
a =getPrime(256)
b =getPrime(256)
p =getPrime(256)
m1=int(str(flag)[:5])-4585
m2=int(str(flag)[5:])
#EllipticCurve([a1, a2, a3, a4, a6]) -- y^2+(a1)xy+(a3)y=x^3+(a2)x^2+(a4)x+(a6)
E = EllipticCurve(GF(p), [a, b])  #定义椭圆曲线
X=E.lift_x(m1)
Y=7*X
m = E.random_point()
G = E.random_point()
k = getPrime(256)
K = k * G
r = getPrime(256)
c1 = m + r * K
c2 = r * G
w2=m[0]*m2
print(f"p = {p}")
print(f"a = {a}")
print(f"b = {b}")
print(f"k = {k}")
print(f"E = {E}")
print(f'Y = {Y}')
print(f"c1 = {c1}")
print(f"c2 = {c2}")
print(f"w2 = {w2}")
'''
p = 71397796933602469825964946338224836258949974632540581233301840806613437378503
a = 106105288190268015217241182934677375171023341761047638573248022053052499733117
b = 76170541771321874396004434442157725545076211607587599314450304327736999807927
k = 58155941823118858940343657716409231510854647214870891375273032214774400828217
E = Elliptic Curve defined by y^2 = x^3 + 34707491256665545391276236596452538912073367128507057339946181246439062354614*x + 4772744837719404570039488103932889286126236975047018081148463521123562429424 over Finite Field of size 71397796933602469825964946338224836258949974632540581233301840806613437378503
Y = (33237936857741483513705672980652927705102229733798436323453609986072499230366 : 52619411226266177137991318059937693955038910547834999771526408984808553907338 : 1)
c1 = (37414446283406201193977113266234367761786780230360175925999700345196415953455 : 17037724145039910971426670298726906655653040365428438334942732090559637519851 : 1)
c2 = (60560423732267272277570046154733119097475794979191838027420415113112056962844 : 54372226143125971429691267751299496959531971082475860532181772357190222938465 : 1)
w2 = 16315249811700998894876359855091105114973337718373913477026230968747515636405
'''

一看c1和c2就知道是ECC椭圆曲线加密，特意去看了ECC加密

C1 − k*C2 = M + r*K − k*( r*G ) = M + r*k*G − k*r*G = M

M为明文,r随机整数(r<n),k为密钥,G是基点,n是G的阶数

所以m=C1-k*c2，那么就只剩下m1，点与数的乘积实现逆过程很困难，又因为m1的位数只有5位，所以可以选择进行5位爆破，最终exp：

#sage
p = 71397796933602469825964946338224836258949974632540581233301840806613437378503
a = 106105288190268015217241182934677375171023341761047638573248022053052499733117
b = 76170541771321874396004434442157725545076211607587599314450304327736999807927
k = 58155941823118858940343657716409231510854647214870891375273032214774400828217
w2 = 16315249811700998894876359855091105114973337718373913477026230968747515636405
E = EllipticCurve(GF(p), [a, b])
c1=E(37414446283406201193977113266234367761786780230360175925999700345196415953455,17037724145039910971426670298726906655653040365428438334942732090559637519851)
c2=E(60560423732267272277570046154733119097475794979191838027420415113112056962844,54372226143125971429691267751299496959531971082475860532181772357190222938465)
m = c1-k*c2
m2=w2/m[0]
m3=str(m2)
u='0123456789'
for i in u:
    for j in u:
        for z in u:
            for x in u:
                for y in u:
                    m1=i+j+z+x+y
                    m=int(m1+m3)
                    m4=long_to_bytes(m)
                    if b'ctfshow' in m4:
                        print(m4)
                        break

Comedy

import gmpy2, libnum
from secret import flag1, flag2

m = libnum.s2n(flag1)
assert m.bit_length() < 200
B = gmpy2.next_prime(libnum.s2n(flag2))
A = (2022 - 2023 * m) % B
leak = pow(2, 2023, B)
print(A)
print(leak)
# 493275281479560936332761096886786925792234184811353209227551802099268192839677496844153534128991899414803550843408607188612593757622064753867565869035222715177143938385039508273050267347710495512806264863554858016145161165422812554800693811328453743229819656381224407015421235005940088439590887928051969351426291843586132741521121351667152673680122929827805479163871436776753859965413192837591532468372
# 238829196127128263156194898141748280130190920343265228257398802867203846004703877952990524473329125233083096275276064071930416561616135910190674099345267027039386328203653489152769309498199556401574021633071022874689081585677578010276529507102304828451681000682208089162940529052283763507244593173690786957816545746540436261888398732172965945762569416702401859253725696471593023885944262561159982327952

对A = (2022 - 2023 * m) % B进行化简->A+2023*m-2022=0(mod B)

pow(2,2023)-leak=0(mod B)因为m的比特位小于200，那么需要在kB这个整数环内求解m

#sage
from Crypto.Util.number import *
A = 493275281479560936332761096886786925792234184811353209227551802099268192839677496844153534128991899414803550843408607188612593757622064753867565869035222715177143938385039508273050267347710495512806264863554858016145161165422812554800693811328453743229819656381224407015421235005940088439590887928051969351426291843586132741521121351667152673680122929827805479163871436776753859965413192837591532468372
leak = 238829196127128263156194898141748280130190920343265228257398802867203846004703877952990524473329125233083096275276064071930416561616135910190674099345267027039386328203653489152769309498199556401574021633071022874689081585677578010276529507102304828451681000682208089162940529052283763507244593173690786957816545746540436261888398732172965945762569416702401859253725696471593023885944262561159982327952
#A+2023*m-2022 = 0 % B
KB = pow(2,2023)-leak
PR.<m> = PolynomialRing(Zmod(KB))
f = A+2023*m-2022
f = f.monic()
print(f)
x = f.small_roots(X=2^200, beta=0.4)
print(x)
print(long_to_bytes(int(x[0])))

又因为pow(2,2023)-leak与A+2023*m-2022存在共同的B因子，所以直接多俩求最小公约数文章来源地址https://www.toymoban.com/news/detail-436467.html

A=493275281479560936332761096886786925792234184811353209227551802099268192839677496844153534128991899414803550843408607188612593757622064753867565869035222715177143938385039508273050267347710495512806264863554858016145161165422812554800693811328453743229819656381224407015421235005940088439590887928051969351426291843586132741521121351667152673680122929827805479163871436776753859965413192837591532468372
leak=238829196127128263156194898141748280130190920343265228257398802867203846004703877952990524473329125233083096275276064071930416561616135910190674099345267027039386328203653489152769309498199556401574021633071022874689081585677578010276529507102304828451681000682208089162940529052283763507244593173690786957816545746540436261888398732172965945762569416702401859253725696471593023885944262561159982327952
m=2438621860802508754666419561610531898810985542251330229087
x=A+2023*m-2022
B=libnum.gcd(2**2023-leak,x)
print(long_to_bytes(B))

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