[UTCTF2020]basic_crypto

[UTCTF2020]basic_crypto

题目:

01010101 01101000 00101101 01101111 01101000 00101100 00100000 01101100 01101111 01101111 01101011 01110011 00100000 01101100 01101001 01101011 01100101 00100000 01110111 01100101 00100000 01101000 01100001 01110110 01100101 00100000 01100001 01101110 01101111 01110100 01101000 01100101 01110010 00100000 01100010 01101100 01101111 01100011 (...)

分析：

是好久不见的古典密码题（密码学签到题

这道题套了好几次加密，很基础且友善但是感觉很有意思，感觉很适合把0基础的新人拐进这个大坑（。

首先上来先转码：

str="01010101 01101000 00101101 01101111 01101000 00101100 00100000 01101100 01101111 01101111 01101011 01110011 00100000 01101100 01101001 01101011 01100101 00100000 01110111 01100101 00100000 01101000 01100001 01110110 01100101 00100000 01100001 01101110 01101111 01110100 01101000 01100101 01110010 00100000 01100010 01101100 01101111 01100011 (...)"
li=str.split(" ")
flag=''
for i in li:
    flag+=chr(eval('0b'+i))		#'0b'不可省略
print(flag)

得到：

Uh-oh, looks like we have another block of text, with some sort of special encoding. Can you figure out what this encoding is? (hint: if you look carefully, you'll notice that there only characters present are A-Z, a-z, 0-9, and sometimes / and +. See if you can find an encoding that looks like this one.)
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

一眼base64。

import base64
list="TmV3IGNoYWxsZW5nZSEgQ2FuIHlvdSBmaWd1cmUgb3V0IHdoYXQncyBnb2luZyBvbiBoZXJlPyBJdCBsb29rcyBsaWtlIHRoZSBsZXR0ZXJzIGFyZSBzaGlmdGVkIGJ5IHNvbWUgY29uc3RhbnQuIChoaW50OiB5b3UgbWlnaHQgd2FudCB(...)"
flag=base64.b64decode(list).decode()
print(flag)

得到：

New challenge! Can you figure out what's going on here? It looks like the letters are shifted by some constant. (hint: you might want to start looking up Roman people).
kvbsqrd, iye'bo kvwycd drobo! Xyg pyb dro psxkv (kxn wkilo dro rkbnocd...) zkbd: k celcdsdedsyx mszrob. Sx dro pyvvygsxq dohd, S'fo dkuox wi wocckqo kxn bozvkmon ofobi kvzrklodsm mrkbkmdob gsdr k mybboczyxnoxmo dy k nsppoboxd mrkbkmdob - uxygx kc k celcdsdedsyx mszrob. Mkx iye psxn dro psxkv pvkq? rsxd: Go uxyg drkd dro pvkq sc qysxq dy lo yp dro pybwkd edpvkq{...} - grsmr wokxc drkd sp iye coo drkd zkddobx, iye uxyg grkd dro mybboczyxnoxmoc pyb e, d, p, v k, kxn q kbo. Iye mkx zbylklvi gybu yed dro bowksxsxq mrkbkmdobc li bozvkmsxq drow kxn sxpobbsxq mywwyx gybnc sx dro Oxqvscr vkxqekqo. Kxydrob qbokd wodryn sc dy eco pboaeoxmi kxkvicsc: go uxyg drkd 'o' crygc ez wycd ypdox sx dro kvzrklod, cy drkd'c zbylklvi dro wycd mywwyx mrkbkmdob sx dro dohd, pyvvygon li 'd', kxn cy yx. Yxmo iye uxyg k pog mrkbkmdobc, iye mkx sxpob dro bocd yp dro gybnc lkcon yx mywwyx gybnc drkd cryg ez sx dro Oxqvscr vkxqekqo.
rghnxsdfysdtghu! qgf isak cthtuike dik zknthhkx rxqldgnxsliq risyykhnk. ikxk tu s cysn cgx syy qgfx isxe kccgxdu: fdcysn{h0v_di4du_vi4d_t_r4yy_rxqld0}. qgf vtyy cthe disd s ygd gc rxqldgnxsliq tu pfud zftyethn gcc ditu ugxd gc zsutr bhgvykenk, she td xksyyq tu hgd ug zse scdkx syy. iglk qgf khpgqke dik risyykhnk!

可以看到密文中的符号都没有发生变换，所以应该是替换加密方式。

hint里说要往罗马人那方面想，凯撒密码就是罗马共和国时期产生的，基本可以确定就是凯撒密码加密了。

不过在一般情况下，如果知道是替换加密但不确定是凯撒密码时，我们需要词频分析看看。

凯撒密码偏移量最大也就25，进行一个穷举。

当偏移量为10时，可以看到出现了我们想要的结果。

alright, you're almost there! now for the final (and maybe the hardest...) part: a substitution cipher. in the following text, i've taken my message and replaced every alphabetic character with a correspondence to a different character - known as a substitution cipher. can you find the final flag? hint: we know that the flag is going to be of the format utflag{...} - which means that if you see that pattern, you know what the correspondences for u, t, f, l a, and g are. you can probably work out the remaining characters by replacing them and inferring common words in the english language. another great method is to use frequency analysis: we know that 'e' shows up most often in the alphabet, so that's probably the most common character in the text, followed by 't', and so on. once you know a few characters, you can infer the rest of the words based on common words that show up in the english language.
hwxdnitvoitjwxk! gwv yiqa sjxjkyau tya padjxxan hngbtwdnibyg hyiooaxda. yana jk i soid swn ioo gwvn yinu asswntk: vtsoid{x0l_ty4tk_ly4t_j_h4oo_hngbt0}. gwv ljoo sjxu tyit i owt ws hngbtwdnibyg jk fvkt pvjoujxd wss tyjk kwnt ws pikjh rxwloauda, ixu jt naioog jk xwt kw piu istan ioo. ywba gwv axfwgau tya hyiooaxda!

这提示真的非常贴心了，总结起来实际上就是我们上文所提到的词频分析。

首先我们知道这个比赛的flag长什么样：utflag{XXXXXXXXXXXXX}

因为是替换加密，所以我们看花括号就能在密文中找到对应的文本，即：vtsoid{x0l_ty4tk_ly4t_j_h4oo_hngbt0}

也就是utflag=vtsoid

那么我们就可以拿这个对应关系来进行词频分析：

显然输出结果中的第一个就是我们需要的最终答案：

	congratulations! you have finished the beginner cryptography challenge. here is a flag for all your hard efforts: utflag{n0w_th4ts_wh4t_i_c4ll_crypt0}. you will find that a lot of cryptography is just building off this sort of basic knowledge, and it really is not so bad after all. hope you enjoyed the challenge!

you will find that a lot of cryptography is just building off this sort of basic knowledge, and it really is not so bad after all. hope you enjoyed the challenge!

总结：

充满出题人善意的题目，确实很enjoy。