UNCTF复现赛-Crypto部分做题记录

【 不仅仅是RSA 】

通过这个网站得到C1和C2的值

C1:4314251881242803343641258350847424240197348270934376293792054938860756265727535163218661012756264314717591117355736219880127534927494986120542485721347351

C2:485162209351525800948941613977942416744737316759516157292410960531475083863663017229882430859161458909478412418639172249660818299099618143918080867132349

利用openssl对两个公钥进行解密，得到n1n2和e1e2：

import gmpy2
import libnum


n1 = gmpy2.mpz(10285341668836655607404515118077620322010982612318568968318582049362470680277495816958090140659605052252686941748392508264340665515203620965012407552377979)
n2 = gmpy2.mpz(8559553750267902714590519131072264773684562647813990967245740601834411107597211544789303614222336972768348959206728010238189976768204432286391096419456339)
e1 = 41221
e2 = 41221
p = gmpy2.gcd(n1,n2)
q1 = n1 / p
q2 = n2 / p
c1 = 4314251881242803343641258350847424240197348270934376293792054938860756265727535163218661012756264314717591117355736219880127534927494986120542485721347351
c2 = 485162209351525800948941613977942416744737316759516157292410960531475083863663017229882430859161458909478412418639172249660818299099618143918080867132349
phin1 = (p - 1)*(q1 - 1)
phin2 = (p - 1)*(q2 - 1)
d1 = gmpy2.invert(e1,phin1)
d2 = gmpy2.invert(e2,phin2)

print(libnum.n2s(pow(c1,d1,n1))+libnum.n2s(pow(c2,d2,n2)))

运行得到flag：UNCTF{ac01dff95336aa470e3b55d3fe43e9f6}

【 一句话加密 】

下载下来有两个东西：

encode.py里仅仅有三行代码：

c = pow(int(m.encode('hex'), 16),e,n)

c1:62501276588435548378091741866858001847904773180843384150570636252430662080263

c2:72510845991687063707663748783701000040760576923237697638580153046559809128516

得找到e和n
当时比赛的时候就没看懂这个e.png是个什么神仙东西QAQ...
把e.png用010editor打开，在结尾找到可疑的16进制字符串，应该是n，
和一个疑似hint的东西：

得知这个密码是kobe code：

这样就得到了n和e，分解n得到p和q，而且e为2，这个叫Rabin加密，修改C，运行两次脚本得到flag。

import gmpy2,libnum
n=0xc2636ae5c3d8e43ffb97ab09028f1aac6c0bf6cd3d70ebca281bffe97fbe30dd
p=275127860351348928173285174381581152299
q=319576316814478949870590164193048041239
e=2
c=62501276588435548378091741866858001847904773180843384150570636252430662080263
mp=pow(c,(p+1)/4,p)
mq=pow(c,(q+1)/4,q)
yp=gmpy2.invert(p,q)
yq=gmpy2.invert(q,p)
r=(yp*p*mq+yq*q*mp)%n
rr=n-r
s=(yp*p*mq-yq*q*mp)%n
ss=n-s
print libnum.n2s(r)
print libnum.n2s(rr)
print libnum.n2s(s)
print libnum.n2s(ss)

【 BABYRSA 】

我们亲爱的EDS师傅出的题（斜眼笑
开局的脚本里已经给出了n和e：

flag = open('flag.txt','r').read()
N = 221
e = 5
enc = b''

for i in flag:
    enc += bytes([pow(ord(i),e,N)])

encrypt = open('encrypt','wb')
encrypt.write(enc)
encrypt.close()

求出p和q为13和17，万事俱备，只差密文啦。

a = open("encrypt",'rb')
b = a.read()
for i in b:
    print(i,end=",")

得到：

102,91,84,67,83,106,121,29,29,33,51,52,109,51,85,85,218,191,54,109,85,33,52,175,186,115,29,109,172,52,85,66,191,51,186,29,175,29,177

对密文逐个进行解密：

import gmpy2
import libnum
p = 13
q = 17
e = 5
s = (p- 1) * (q - 1)
d = long(gmpy2.invert(e, s))
n = p *q
b = ''
a = [102,91,84,67,83,106,121,29,29,33,51,52,109,51,85,85,218,191,54,109,85,33,52,175,186,115,29,109,172,52,85,66,191,51,186,29,175,29,177]
for i in a:
    b += libnum.n2s(pow(i,d,n))
print b

运行得到flag：UNCTF{10023493ff87a9f246eb09d4f573e060}