DASCTF 2020 安恒七月赛 |

admin 2021年10月11日22:38:16DASCTF 2020 安恒七月赛 |已关闭评论215 views字数 5095阅读16分59秒阅读模式

MISC

welcome to the misc world

解压得一张图red_blue.png和加密压缩包flag.rar,那么密码肯定在图里了。

由关键字red/blue想到lsb隐写,用Stegsolve工具查找无结果,放到zsteg命令下看,发现png图,提取出来:

image-20200725163904873

得到压缩包密码 /*///1258/*/@#

用360压缩打开直接能看到NTFS隐写文件: (23333~~)

image-20200725164537137

打开flag.txt,得到一串字符,由包含的字符猜测为base85编码,尝试解码得到flag。

CRYPTO

bullshit

给定脚本

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from flag import flag
def pairing(a,b):
shell = max(a, b)
step = min(a, b)
if step == b:
flag = 0
else:
flag = 1
return shell ** 2 + step * 2 + flag

def encrypt(message):
res = ''
for i in range(0,len(message),2):
res += str(pairing(message[i],message[i+1]))
return res

print(encrypt(flag))

加密脚本思路:

将明文message每两个字符分别作为pairing()函数的参数,返回的int结果值连接,得到输出密文。

因flag可能包含的ASCII字符的十进制值

48d126

,由pairing()函数返回值知,得到的int结果值范围为

482+482+0v1262+1262+12400v16129

按范围分割密文,再分别爆破每个的输入参数:

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import string
dic=string.printable[:-6]

def pairing(a,b):
shell = max(a, b)
step = min(a, b)
if step == b:
flag = 0
else:
flag = 1
return shell ** 2 + step * 2 + flag

c=[11869,10804,15229,10199,3354,10105,3241,10519,9908,2499,10505,10103,9519,9519,3232,9711,9520,3127,15722]
flag=''
for k in c:
for x in dic:
for y in dic:
if pairing(ord(x),ord(y))==k:
flag+=x+y
print(flag)

ezrsa

给定脚本

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from secret import flag
from Crypto.Util.number import getPrime,inverse,bytes_to_long,long_to_bytes
from sympy import isprime

m = bytes_to_long(flag)

i=0
p=getPrime(1024)
r=getPrime(1024)
while True:
i+=1
q = 5*p+i
if isprime(q) :
break

n=p*q*r
e = 65537
c = pow(m,e,n)
p3 = pow(p,3,n)
q3 = pow(q,3,n)
print c
print e
print n
print p3
print q3







由脚本知

p,q

为线性关系:

q=5p+k,kZ

又已知

p,q

e=3

得到的密文

p3,q3

,满足 Related Message Attack 条件,运行Sage脚本得到

p,q

,再求得

r

,按照常规RSA解法求出flag。

Related Message Attack Sage脚本:

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def short_pad_attack(c1, c2, e, n):
PRxy.<x,y> = PolynomialRing(Zmod(n))
PRx.<xn> = PolynomialRing(Zmod(n))
PRZZ.<xz,yz> = PolynomialRing(Zmod(n))
g1 = x^e - c1
g2 = (5*x+y)^e - c2
q1 = g1.change_ring(PRZZ)
q2 = g2.change_ring(PRZZ)
h = q2.resultant(q1)
h = h.univariate_polynomial()
h = h.change_ring(PRx).subs(y=xn)
h = h.monic()
kbits = n.nbits()//(2*e*e)
diff = h.small_roots(X=2^kbits, beta=0.4)[0]
return diff
def related_message_attack(c1, c2, diff, e, n):
PRx.<x> = PolynomialRing(Zmod(n))
g1 = x^e - c1
g2 = (5*x+diff)^e - c2
def gcd(g1, g2):
while g2:
g1, g2 = g2, g1 % g2
return g1.monic()
return -gcd(g1, g2)[0]
if __name__ == '__main__':
n = 20361372240024088786698455948788052559208001789410016096382703853157107986024860262721685000417719260611935731634077852127432140361792767202581631816544546972750034494061276779878409544779707914261679633764772575040304712361634318086289783951555842021028438799649252652041211341825451500751760872572402250747982495384263677669526575825183733353800694161425360299521143726681387485097281832219009682768523304737252763907939642212542959846630464628135025203489075698699980715986689341069964387779523254203021424865355054215122316160201073604105317768112281914334065349420946717116563634883368316247495042216330408372176714499012778410160478384503335610321108263706243329745785632599707740534386988945259578897614317582546751658480917188464178997026284336861027299289073045677754342746386408505695243800685323283852020325044649604548575089927541935884800327121875191739922436199496098842684301207745090701158839031935190703347091
e = 3
c1 = 3639847731266473012111996909765465259684540134584180368372338570948892196816095838781423020996407457408188225238520927483809091079993151555076781372882518810174687150067903870448436299501557380508793238254471833275507634732947964907461619182112787911133054275872120243558556697900528427679352181961312958660881800731678134481664074711076672290178389996403357076809805422591851145306425951725627843352207233693810474618882394140691334742086008967260117740486955640068190440609984095657695423536016475468229419187489359563800737261212975921663803729112420222039005478830477455592167092520074509241894829304209406713781082959299623674294927249556083486223036858674077173104518013601628447504500606447821540687465361616447631579976579754996021653630804073535352129315413118764836270751250405649683786487251823247828947202336680538849571498780353357272103697510910576879383751704763858882439578045020243015928994208017750848637513
c2 = 7030777127779173206633582847346001157991477456002191926122836599155148909465054067800807615361108442560942058865403188672629297039703065927801771646334817871335134889139894648729527452541098449842202838983982508551750669662540615534327150829869964429006130891731472099912937717406120443380283548571270317421722042835639732966975812764084015221255115940508456442279902250677665136380988902682370875602145833135937210740790528756301051981994351553247852018355526641012434670664732924491790949235519600899289515495046353559475806935200029321563549553167235419039924276406059858659476329718809657072997385947262654743181242885709558209249589482036673428723035300722280229192727192487772217518673838209646300548275957450994828221329299666216457961746189885356929698674294944243729739850927111231235060005119781652245234537583181232715964191675241206562888107252569566488402724441835466680342239244581162530424964324562530832713397
diff = short_pad_attack(c1, c2, e, n)
print("difference of two messages is %d" % diff)
p = related_message_attack(c1, c2, diff, e, n)
print("p:", p)
print("q:", 5*p + diff)

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  • 本文由 发表于 2021年10月11日22:38:16
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