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[CTF(x)]λ – 50pts

Description:

I used this program to encrypt a flag. The output was:

n1s4_t1An([email protected]_h3)m3lp3y__Eas

λ.py:

print (lambda j,m:(lambda f,t:t if len(t) <= 1 elsej([f(f,x)for x in m(j,m(reversed,(lambdas:zip(*[iter(s)]*(len(s)/2)))(t+"x01"*(len(t)%2))))]))(lambdaf,t:t if len(t) <= 1 else j([f(f,x)for x inm(j,m(reversed,(lambda s:zip(*[iter(s)]*(len(s)/2)))(t+"x01"*(len(t)%2))))]),raw_input("Plaintext:")))(''.join,map).replace("x01","")

Solution:

我看到這麼一堆東西是完全不想追算法的 , 那就不追好了 , 直接暴力破之

首先我使用以上的算法加密:

1234567890abcdfghijklmnopqrstuvwxyz

結果得出

cbd0aihfg435129867vuwstzxymlnjkrqop

看來只是把次序調了 , 所以得出以下算法

{CTF(x)}[Cryptography]λ - 50pts

運行後得出

{CTF(x)}[Cryptography]λ - 50pts

ctf([email protected]_1nsAn1ty_pl3asE_h3lp_m3)

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