BUUCTF-Crypto系列[65-80]WP
Published in：2021-08-07 | category： 密码学 BUUCTF

浪里淘沙

题目如下：

tonightsuccessnoticenoticewewesuccesstonightweexamplecryptoshouldwebackspacetonightbackspaceexamplelearnwesublimlearnbackspacetheshouldwelearnfoundsublimsystemexamplesublimfoundlearnshouldmorningsublimsystemuserlearnthecryptomorningexamplenoticetonightlearntonightlearntonightsublimenterusermorningfoundtonightweenterfoundnoticethecryptomorningthebackspacelearntonightlearnsublimtonightlearnfoundenterfoundsuccesstonightsuccessuserfoundmorningtonighttheshouldsublimentertonightenterbackspacelearnexamplenoticeexamplefoundsystemsuccesssublimsuccessshouldtonightcryptowelearncryptofoundshouldsublimsublimweentertonightsuccessshouldentertheentercryptouserbackspaceshouldentersystemsuccesssystementerfoundenterlearnexampletonightnoticemorningusertonightlearnmorningtonightfoundfoundsuccessnoticesystementerlearnexamplebackspaceshouldcryptocryptosublimweexampletonighttheshouldthemorningbackspacelearntonightsystemsuccesssuccessbackspacemorningnoticeuserfoundfoundtonightmorningenterenterthefoundbackspacelearnenterentershouldthesystemfounduserlearnlearnsystemnoticetonighttheshouldlearnuserbackspaceweusernoticeshouldthewefoundsystemwecryptocryptowethebackspacesystementershouldtonightsystemnoticemorningsystemweentermorningfoundsuccessusertonightsuccesstonightbackspaceshouldweenterthewesystemusernoticesystemthelearnexamplelearnfoundlearnnoticeexamplesystemthecryptocryptolearnsystemthecryptoenterlearnexamplemorningmorningweenterentersuccessexampleuserthebackspacenoticesublimenterbackspacesuccessbackspacethesublimexamplesystemtheexamplecryptolearnuserexamplelearnsystemusersuccessenterentersuccesstheuserbackspacelearnsuccessbackspacethesublimshouldwebackspaceexamplesuccesssuccesstonightweusershouldsuccessmorningcryptomorningfoundbackspacesublimshouldentershouldnoticesuccessmorningsuccessexamplelearnshouldsublimlearntonightshoulduserbackspacesublimlearncryptosuccessenternoticetonightmorningtonightwesuccessweuserbackspaceexamplewesystemnoticemorningsystemmorningcryptolearnsystemthethefoundcryptouserlearnusersystemwemorningenterexampleshouldlearncryptofoundenterbackspacelearnenterenterbackspaceshouldbackspacetheshouldthesystemshouldshouldsuccessmorningthefoundsystementersystemtonightcryptowelearnexampleexamplesystementerbackspaceshouldtheentersublimtonightfoundfoundsuccesssuccesssystemsublimcryptoshouldentersublimmorninglearnfoundtonightcryptobackspacesuccesscryptowebackspacefoundshouldnoticeshouldmorningnoticesystemcryptosystemlearnsystemnoticemorningsystementerwemorninglearnsuccessfoundwesuccesswetheusercryptousernoticebackspacesuccessshouldtonightmorningentermorninguserenternoticefoundmorningwetonightsystemthecryptotonightcryptosystemuserthefoundexampletonightusersystemcryptosublimmorninguserthefoundbackspaceshouldsuccesscryptotonightsystemnoticebackspaceusershouldenterthecryptomorningwesublimnoticesuccessnoticeusersuccesstonightlearnweuserenterfounduserexampleshouldshouldtonightwelearnthenoticethewefoundmorningexampleshouldexamplethesuccessnoticeenterfoundthecryptonoticeuserlearnuserweenterfoundmorningsystemweexamplenoticethebackspaceexamplesublimtheusermorningtonightthesuccesscryptosuccessusersuccesstonighttonightwelearnenterenterthemorningentersystemcryptobackspacemorningsystemexamplecryptouserexamplelearnsublimsuccessusersystemfoundmorningshouldcryptotonightsublimtheexamplemorningsystemuserexampleweexamplenoticesuccesssublimnoticecryptoshouldbackspaceshouldthetonightfoundsublimbackspacebackspacetonightshouldbackspacesuccesstonightbackspacesuccessmorningsystemcryptobackspaceentertonighttonightnoticelearnshoulduserfoundexamplesystemthesuccessweusertonightcryptousernoticeenternoticebackspaceusersystemfoundusernoticeshouldlearnuserfoundexampleusermorningshouldsuccessmorningmorningexampleexamplefoundsublimfoundenterbackspacenoticelearnfoundmorningcryptonoticecryptoshouldweshouldtonightcryptobackspacesublimcryptosublimenterentersublimentercryptonoticethethesublimexampleenterentershouldlearncryptoentershouldmorninglearnnoticeuserexamplesublimtonightshouldfoundtonightsuccessshouldmorningfoundtheweuserlearnsublimsystembackspacecryptotheusertonightcryptosublimmorningmorningexamplenoticetheenterlearnshouldmorningsublimfoundtonightsublimsublimexamplefounduserexamplethefoundwemorningnoticefoundcryptosuccesssublimsublimexamplethesuccessexamplenoticesuccessbackspacesublimlearnuserexamplesuccesssuccesssystemsuccessmorningmorninglearnexamplemorningtonightfoundbackspaceenternoticemorningentersuccessmorningusermorningbackspacelearncryptoenteruserenteruserthetonighttonightsuccesslearnenterfoundsuccesssystemfoundbackspaceenterlearnsystemsublimcryptoentermorningwetonightshouldlearnenterfoundcryptonoticelearnlearnshouldfoundsuccessexampletonightthesuccessfoundusertonightenterfoundsuccessshouldmorningusernoticemorningsystemsystemsuccessshouldwelearnenterfoundexamplewethefoundweshouldsystemsystemmorningmorningbackspaceshouldentersublimentertonightsuccesssystemsystemcryptousershouldsublimfoundwetonightnoticeexamplewewesuccessfoundusertonightfoundsystemexamplecryptofoundshouldshouldsuccessenterbackspaceexampletonightthelearnnoticeuserlearnsystemsublimfoundlearnsuccesssystemshouldsublimnoticelearnsystemnoticetonightexamplefoundusernoticeenterlearnnoticecryptousersystemmorningthewesystemfoundfoundshouldsystementerenterbackspacesystemsublimcryptousermorninglearnlearntonightsublimlearnenterenterbackspacesystemuserusercryptoentershouldtheusersublimnoticeexamplemorningexamplesublimsublimbackspacesystemexampleshouldsublimlearnfoundenterbackspacelearnmorningmorningfoundthetonightmorningnoticeenterlearnusersystemtonightbackspaceexamplelearntonightbackspaceweshouldcryptosuccessbackspaceexamplesuccesstheshouldmorninguserbackspacelearnthetheshouldcryptocryptotonightbackspacecryptocryptobackspacebackspacenoticeusertonightentermorningfoundweenterexampleenterfoundusersublimsystemtheexampleexamplesystemsuccessusersublimentermorningbackspacesystemfoundlearnsystemshouldsublimsublimentershouldtheusershouldexampleexampleshouldsuccesswelearnfoundsublimshoulduserweentertonightwenoticesublimsystemlearnshouldfoundsuccessuserentersuccessmorningcryptoenteruserfoundexampletonightlearnexampleexamplefoundlearnsuccesssystembackspacecryptonoticethefoundbackspacelearncryptothelearnlearnexamplesuccessnoticenoticesystemmorningcryptotonightnoticenoticeentersuccesscryptoenterbackspacesublimexampleenterfoundtonightcryptotonightsublimnoticesuccesssublimtheentertonighttheshouldthefoundsystemtonightuserbackspacesuccessshouldwebackspacenoticebackspacebackspacenoticewecryptobackspacebackspaceusertonightlearnsuccessmorningusertonightsuccessshouldbackspacecryptoenterentershouldsublimsystemexamplemorningcryptonoticethesuccessthebackspacenoticelearnsublimlearnsuccesscryptothesuccessenternoticecryptosystemsublimsuccesswebackspaceuserenterlearnuserwewemorningsuccesslearncryptobackspacewecryptosystemlearnenterenteruserexamplefoundsystemcryptousernoticefoundusersublimbackspacewesublimnoticemorningshouldexamplenoticecryptoshouldtonightmorningthefoundsystementerentersystemthecryptobackspacesublimlearnsuccessmorningsublimsystemcryptousersublimwesuccessmorningsublimbackspacecryptobackspacesublimthelearnsuccesssublimlearncryptoweweexamplecryptowenoticelearnfoundbackspacesystemsystemexampleshouldlearnsuccesssublimcryptobackspacetonightbackspacemorningmorningnoticeshouldnoticefoundthetheshouldtheshouldfoundfoundcryptosuccessbackspacesuccessshouldweenternoticeweweshouldmorningfoundusersuccessbackspacewenoticeusersuccessenterenterexamplelearnfoundwetonightusercryptothesublimsublimtonightsuccesslearnbackspacetonightentertonightthesublimnoticewefoundcryptobackspaceenterenterlearnlearntonightexamplesystementersublimnoticecryptoshoulduseruserbackspaceuserwesublimmorningwesystemshouldtonighttheusershouldnoticefoundusernoticeentersublimwethewefoundfoundlearnfoundwecryptosystemexamplemorningcryptocryptosublimtheexamplenoticefoundlearnwelearnmorningtheenterthesystemsublimtonightsuccesssystemlearnshouldenterbackspaceentersuccesssuccessbackspaceexamplenoticeentershouldsublimlearnbackspacetheshouldexamplelearnsystemusersublimbackspacebackspacesuccesswelearntonightexamplewecryptoenterwesystemsystemsublimexamplecryptolearnmorningsublimfoundsublimfoundbackspacefoundtonighttonightnoticesuccesssuccessexampleusersuccesstonightsublimcryptosystemweenterexamplesystemthethenoticesublimtonightbackspacenoticesystemexamplethesuccesstonightmorningsuccesstonightwenoticesublimtonightwelearntonightmorningsublimbackspaceenterthetonightenterwecryptofoundtheenternoticebackspacesuccesswesystemuserexamplebackspaceentersuccesstonightsublimwemorningsuccesssuccesswesublimsuccessnoticesublimfoundlearnlearnweexamplecryptonoticelearnweusershoulduserfoundcryptolearnfoundmorningtonightmorningmorningnoticewecryptowewesuccessfoundsublimweuserentershouldshouldshouldsublimbackspacetonightenterwesublimsuccessshouldfoundthethetonightwecryptoweenterfoundcryptoshouldcryptouseruserfoundentersublimsublimthelearntheshouldnoticebackspacefoundsuccessshouldtonightentermorningsystemmorningtonightwenoticelearnbackspaceexampleusershouldnoticesublimsublimexamplethesuccessnoticesystemmorningnoticecryptosystemsublimcryptosystemsuccessshouldmorningbackspaceshouldmorninglearnnoticenoticeshouldthewewesublimsublimnoticeusersuccessentersystemfoundshouldshouldcryptobackspaceusermorningsystemshouldshouldtonightwesublimuserfoundlearnbackspacethetonightmorningexampleuserthefoundbackspaceshouldtonightcryptocryptofounduserexamplenoticecryptousernoticethenoticeshouldweshouldfoundwemorningcryptosuccesslearnfoundtonightsublimnoticenoticewefoundwewesuccesssublimsublimcryptoweexampletonightsuccessfoundshouldsuccesstonightbackspacesystemshouldwesystemnoticebackspaceusersystembackspacewenoticelearnnoticenoticesuccesslearntonightuserlearnsuccessbackspacesuccesswesystemusercryptonoticethesystemusernoticewethesuccessweshouldfoundshouldcryptomorningtonightwethewesuccesslearntheshouldweexampletonightsuccessnoticenoticemorningfoundmorningfoundusersublimsystemsuccessbackspacesuccessmorninguserthefoundweexamplemorningsublimlearnfoundfoundnoticemorningshouldweuserwemorningexamplesuccesssuccessfoundthetheshouldweusershouldtheshouldexamplenoticefoundsuccesssystemfoundshouldsublimbackspacetonightshouldsystemtonightsuccesslearntonightsystemsublimsuccesscryptobackspacesystemsublimmorningmorningshouldmorninglearnsuccesslearnmorningusermorninglearnexamplecryptoshouldbackspacesublimshouldfoundbackspacesystemsystemweexamplesystemtonightsublimmorningmorninguserfoundcryptolearnbackspaceshouldbackspacenoticesublimfoundthecryptousershouldsuccesssystemsuccessshouldsystembackspacesublimshouldsublimsystembackspaceexampleshouldbackspacesublimnoticelearnsublimuserbackspaceusersublimsuccesssublimuserusernoticeshouldsuccessnoticenoticelearnexamplesystemweexamplesublimbackspacebackspacecryptoshouldusercryptosublimbackspacesublimshouldsystemnoticenoticethesuccesssuccesslearnsystemsublimwenoticelearnusersublimsystemusernoticeuserthesuccesslearnwelearnwenoticecryptolearncryptonoticenoticebackspacecryptothecryptousercryptobackspacesuccesslearnthesystemsuccessthesystemsystemcryptosuccessbackspacesublimlearnsublimcryptobackspacelearnsublimusersublimexamplecryptosublimsystemnoticecryptocryptousertheusernoticebackspacenoticenoticethecryptocryptosystembackspacesublimbackspacecryptocryptobackspacesystemuserthenoticesystemsystemsystemusernoticethecryptouserusersystemtheusercryptoexamplenoticecryptoexamplenoticetheexampleexamplethecryptotheusernoticetheexampleexamplecryptotheexampleexamplethenoticethecryptocryptoexampletheexamplecryptocryptothenoticeexamplecryptonoticetheexampleexampleexamplecryptocryptoexampleexamplethenoticethecryptothethethethethetheexampleexamplethetheexampletheexampletheexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexampleexample

正如题目标题所说的那样—-浪里淘沙，思路是统计单词在题目中的个数，然后按照其出现频率进行排序，答案是根据他给出的那串数字进行映射到排序表。

[AFCTF2018]Vigenère

题目如下：

Encode.c:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int main()
{
	freopen("flag.txt","r",stdin);
	freopen("flag_encode.txt","w",stdout);
	char key[] = /*SADLY SAYING! Key is eaten by Monster!*/;
	int len = strlen(key);
	char ch;
	int index = 0;
	while((ch = getchar()) != EOF){
		if(ch>='a'&&ch<='z'){
			putchar((ch-'a'+key[index%len]-'a')%26+'a');
			++index;
		}else if(ch>='A'&&ch<='Z'){
			putchar((ch-'A'+key[index%len]-'a')%26+'A');
			++index;
		}else{
			putchar(ch);
		}
	}
	return 0;
}
------------------------------------------------------------------------------------------------------------
flag_encode.txt
Yzyj ia zqm Cbatky kf uavin rbgfno ig hnkozku fyyefyjzy sut gha pruyte gu famooybn bhr vqdcpipgu jaaju obecu njde pupfyytrj cpez cklb wnbzqmr ntf li wsfavm azupy nde cufmrf uh lba enxcp, tuk uwjwrnzn inq ksmuh sggcqoa zq obecu zqm Lncu gz Jagaam aaj qx Hwthxn'a Gbj gfnetyk cpez, g fwwang xnapriv li phr uyqnvupk ib mnttqnq xgioerry cpag zjws ohbaul drinsla tuk liufku obecu ovxey zjwg po gnn aecgtsneoa.

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Hr nck lafhynl hvy Ckmang zx Tajy, vzy iofz fpoykugga aaj wmcryuslu fbx cpe caddcy gbum.

Pe ugu xinbvjmmn uou Yireetxzs gu rsmo Lncb wf vsowxeagk jvd cxgkment ovxoezcfwa, uarnas fauhyjdrj rv tukkj ileegcqoa zkdf dif Gbaeaz uziqlq hn wbggkfyz; aaj fpea yq kooprtmmd, uk jsm qtgkaty akidyytrj cw agzgfx po gnnu.

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Jw bws qobaoybgv Lapekbmnggvapa Hbabms ekrwupeqrh, noe urhioiam fqtu scffu fvxvvefy jam enigbqoay qf nde eopptf uh lba pruyte.

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Hr nja eajgspkuekm bo cxgnyjt gnn xocansneoa uo bhryg Knwtry; owr gncl jqrcubm ooyvjoytvtp bhr Rcom boe Tjbuegnatwtvuw wf Sutwccnrxb; zesauahc tb vjas bzjwlo tb kwkohxcyy phroa uitxclcknf nrbhrx, cfx navyrvg gng uijdvzrwnf uh fys Acvawpeoclcknf uo Taaju.

Zy daf ukateaelyz tuk Jlmvtkknnagoxv os Pwknecr hh zesauahc hvy Jasrtv li Hajy owr ryvsvhifnrvg Wafaweaee Ywwrxu.

Zy daf sjle Wafyyo drvnvdrtv gh dif Crtl nrqfy boe zqm trtwjy kf gnnqr blhawas, ntm bhr gogojt ntm xalsgfn kf gnnqr fgnsleef.

luig vy cxwpf{Jnxwobuqg_O_Cogiqi!}

Hr nck ynepznl a zanlcpuqk xn Nrc Qxzecry, jvd fkpl betuka awnxok ib Oslrkeey vg bwrnyb wue vggjhe ntm mag uwl ndevx bcbfzcfwa.

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Uk jsm wfsklbeq zq jyjdrx cpe Zonanwrl owleckpvyjt bl jvd farwleoe zx bhr Iknch Pbcnz.

Hr nck wkmoowmd jovz iphrxb bo fadbyyt hy cw a watamzipzrwn sutwccn gu xcr pupknethzrwn, ntf mhwcxtxelrjiwx xy baa tajy; iapent nra Afygfn po gnnqr Nivk ib pekcmnqkf Dycifrjbibt:

Hgl munxcmrvti dungr hxliry qx unmrj czobvu sgknt ab:

Noe vtgnacgowo tuko, ts w mbit Brvgn xlkm cawqsusgfn boe gwg Mhxfwlo wuolp tuka kbkuyj lwmzov gh phr Owpaoovshps bl cpefk Ulupef:

Lxz chzvahc osl xcr Gxcvy sign jtl cgtlm kf gnn eoerf:

Xin izvxaiam Vsras bt da wvzjgop ohx Lwnfkpl:

Zkr qkyziiopy oo ia sjvy pguwm, kf gnn jeakhan kf Gxril oe Lmlu:

Fbx czaayrglpiam da breqfx Oeny cw br ztayz fbx yzegkpvyz oslnvcry:

Hgl wbbrrahvti lba fekn Ayfzge ib Eamuqsu Rcom en n tnqguhqmlent Vawvvtew, yotnhuqsuopy ndeekrv aa Gttcprnxh ooiktfgang, gwl earcjaent oca Bbapvuniry bw af zq jyjdrx rb ag upuy wn rdjupyk cfx big owateaowhp fbx rvteufmwent zqm snsg svooyacm rhrg ahpo gnnae Pungheef

Lxz tnqkfa wwne xcr Pncjnarf, gkwlvyjahc ohx vwsg bcdowbyk Uiwf gpv uhtrxrvg sapvuieazjtll zjw Zkrzy xn ohx Igparasnvtf:

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Hr nja cbtullwiakm wue lgdfkw Pocqzrtu lugea Ijxtvbg gh phr nroh Fkck nk brga Irzy cyuenfz cpevx Egojtee, cw briqey phr kgmchzkgharf uo bhrot xleeajb inq Htwndrrt, xz tb lcdf phrsbmliku ts phroa Paaju.

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Aut zure Jk kmea ccfnent ow itgkplcknf zx wue Htanesu hamtuxgf. Qa hnbn eaetgv ndez lawm goow nk tvsn wf nzvwgltf hh bhrot dycifrjbuek vg yttrtm in htyslnaazjjlr pwjcodvicqoa uxwl qs. Jk qivr xgecjdrj cpez uh lba cvxlcmfzcfwas bl xcr rskylwtvuw inq yglnhezkwb hrxg. Oy daik jxprgnwx po gnnqr agvapa jhycqcr gpv gwgagwqmvza, shz wr njde pupboneq zqmm oe vzy piry xn ohx eggioa qrvdekf li zifgeww gngky qshxyitvupk, qdipn fwuyj kfyriggkty vtvwlnucz xcr pupfyytvuwa aaj eglnefvxvdrtew. Ndel zxw hnbg tyan qkjn tb zjw pkipk xn jhyvawa aaj xn cbtushcuvtrby. Jk ommp, tukamfbxg, swmuvkbke vt vzy jepkbaige, yzcyh qkwwuaigk iqr Fkyirnzkgh, wnq nxtd gnge, uo wr nxtd gng jyot bl vinxopv, Yjezona ia Ccj, cj Prglm Feogfxo.

Wr, zqmrrlqjy, phr Xnxrrygfnwtvbna os zjw ojigkm Atnzgk ib Azkaqcn, op Yyjeegu Koamtwmo, Afynubykf, sjlenrrvg gu vzy Oucxnue Wafyy kf gnn eoerf xin tuk amcgovmxa os udz iazgfneoay, mw, ia zjw Hwmr, gwl bl Gwlbkrvzh wf gng yikd Ckxxlr uh lbasr Ixtoaogk, mklrswty caddcoh ntm leprcjy, Phnz cpefk wfcpeq Ixtoaogk une, ntm wf Eoizn kutnc bo ok Hjya aaj Rvdrvgfxang Ycitry, vzup tukh irr Gdkihvrj ozoz gnd Uhlrmrinpk vg nde Oxrbifn Ejisn, ntm bhnz cdf loyocqcnr eghjepzrwn okvoyan gnnu aaj vzy Otnzn wf Txgsn Xrvzjqn, vy cfx kutnc bo ok vgnwlye mqsfunnyz; aaj cpag gu Xlae ntm Qnqkrwhzeaz Bbagku, lbay ugem fhrn Hisee zx teie Ysl, yoaiucdr Vgswa, cbtczapz Cdfeaaina, efzctfesu Ixumrxew, ujd gu mw ayr qlbar Nica aaj Vzcjgf cqqcu Opvyleajnvt Fzclyo mne xn rvmjl xk. — Aaj owr gng kolpbxc wf gnkk Xacygaitvup, ocph n lrzm eknaujcr uw bhr vtgnacgoxv os Jkncje Cxxdiqkpuy, se zaccayra hfadtk cw enij gndee udz Lvbgk, iqr Suabuaku, shz ohx bicekf Zijoe.

通过在线工具进行解密：

flag is afctf{Whooooooo_U_Gotcha!}

[NPUCTF2020]这是什么觅🐎

下载附件，使用winhex查看：

是一个压缩包，我们修改完后缀后，里面有一张图片：

根据图片中右下角的便签上的内容，找到日历中对应的数字，里面有两个’T’和’S’。记第一个T是星期二，第二个T是星期四；第一个S是星期五，第二个S是星期六。接着再根据数字找到字母表中的字母，这就是答案了。

flag{calendar}

[BJDCTF2020]easyrsa

题目如下，是一个py文件：

from Crypto.Util.number import getPrime,bytes_to_long
from sympy import Derivative
from fractions import Fraction
from secret import flag

p=getPrime(1024)
q=getPrime(1024)
e=65537
n=p*q
z=Fraction(1,Derivative(arctan(p),p))-Fraction(1,Derivative(arth(q),q))
m=bytes_to_long(flag)
c=pow(m,e,n)
print(c,z,n)
'''
output:
7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441

'''

题目中的n可以分解成两个素数，接下来就是普通的RSA算法了，脚本如下：

from Crypto.Util.number import getPrime,bytes_to_long, long_to_bytes
from sympy import Derivative
import fractions
import gmpy2

e = 65537
c = 7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035
z = 32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482
n = 15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441
p = 105909195259921349656664570904199242969110902804477734660927330311460997899731622163728968380757294196277263615386525795293086103142131020215128282050307177125962302515483190468569376643751587606016315185736245896434947691528567696271911398179288329609207435393579332931583829355558784305002360873458907029141
q = 144564833334456076455156647979862690498796694770100520405218930055633597500009574663803955456004439398699669751249623406199542605271188909145969364476344963078599240058180033000440459281558347909876143313940657252737586803051935392596519226965519859474501391969755712097119163926672753588797180811711004203301

phin = (p - 1) * (q - 1)
d = gmpy2.invert(e, phin)
m = pow(c, d, n)
print(long_to_bytes(m))

运行脚本可以得到flag。

[NCTF2019]babyRSA

题目是一个py文件：

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

def nextPrime(n):
    n += 2 if n & 1 else 1
    while not isPrime(n):
        n += 2
    return n

p = getPrime(1024)
q = nextPrime(p)
n = p * q
e = 0x10001
d = inverse(e, (p-1) * (q-1))
c = pow(bytes_to_long(flag.encode()), e, n)

# d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
# c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804

我们可以从e d ≡ 1 mod Φ(n)入手，p和q的数据规模相当，假设都是1024位，那么Φ(n)就是2048位，e d是一个2064位的数字，，所以k是一个不超过16位的数字（e d - 1 = k Φ(n)），我们可以遍历k求Φ(n)。脚本如下：

from Crypto.Util.number import long_to_bytes
import gmpy2
import sympy

e = 0x10001
c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804
d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
ed_1 = e * d - 1
p = 0
q = 0

for k in range(2 ** 15, 2 ** 16):
    if(ed_1 % k == 0):
        p = sympy.prevprime(gmpy2.iroot(ed_1 // k, 2)[0])
        q = sympy.nextprime(p)
        if((p - 1) * (q - 1) * k == ed_1):
            print(p, q)
            break

n = p * q
m = pow(c, d, n)
print(long_to_bytes(m))

#d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
#c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804

[ACTF新生赛2020]crypto-rsa3

题目如下：

rsa3.py:
from flag import FLAG
from Cryptodome.Util.number import *
import gmpy2
import random

e=65537
p = getPrime(512)
q = int(gmpy2.next_prime(p))
n = p*q
m = bytes_to_long(FLAG)
c = pow(m,e,n)
print(n)
print(c)
------------------------------------------------------------------------------------------------------------
output.txt:
177606504836499246970959030226871608885969321778211051080524634084516973331441644993898029573612290095853069264036530459253652875586267946877831055147546910227100566496658148381834683037366134553848011903251252726474047661274223137727688689535823533046778793131902143444408735610821167838717488859902242863683
1457390378511382354771000540945361168984775052693073641682375071407490851289703070905749525830483035988737117653971428424612332020925926617395558868160380601912498299922825914229510166957910451841730028919883807634489834128830801407228447221775264711349928156290102782374379406719292116047581560530382210049

RSA基本题，直接分解n，最后就是常规操作。脚本如下：

from Crypto.Util.number import long_to_bytes
import gmpy2
import sympy

e=65537
p = 13326909050357447643526585836833969378078147057723054701432842192988717649385731430095055622303549577233495793715580004801634268505725255565021519817179231
q = 13326909050357447643526585836833969378078147057723054701432842192988717649385731430095055622303549577233495793715580004801634268505725255565021519817179293
n = 177606504836499246970959030226871608885969321778211051080524634084516973331441644993898029573612290095853069264036530459253652875586267946877831055147546910227100566496658148381834683037366134553848011903251252726474047661274223137727688689535823533046778793131902143444408735610821167838717488859902242863683
c = 1457390378511382354771000540945361168984775052693073641682375071407490851289703070905749525830483035988737117653971428424612332020925926617395558868160380601912498299922825914229510166957910451841730028919883807634489834128830801407228447221775264711349928156290102782374379406719292116047581560530382210049
phin = (p - 1) * (q - 1)
d = gmpy2.invert(e, phin)
m = pow(c, d, n)
print(long_to_bytes(m))

运行脚本可得flag。

[AFCTF2018]你能看出这是什么加密么

题目如下：

p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f

q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061

e=0x10001

c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6

RSA基础题，脚本如下：

from Crypto.Util.number import long_to_bytes
import gmpy2
import sympy

p=0x928fb6aa9d813b6c3270131818a7c54edb18e3806942b88670106c1821e0326364194a8c49392849432b37632f0abe3f3c52e909b939c91c50e41a7b8cd00c67d6743b4f
q=0xec301417ccdffa679a8dcc4027dd0d75baf9d441625ed8930472165717f4732884c33f25d4ee6a6c9ae6c44aedad039b0b72cf42cab7f80d32b74061
e=0x10001
c=0x70c9133e1647e95c3cb99bd998a9028b5bf492929725a9e8e6d2e277fa0f37205580b196e5f121a2e83bc80a8204c99f5036a07c8cf6f96c420369b4161d2654a7eccbdaf583204b645e137b3bd15c5ce865298416fd5831cba0d947113ed5be5426b708b89451934d11f9aed9085b48b729449e461ff0863552149b965e22b6
n = p * q
phin = (p - 1) * (q - 1)
d = gmpy2.invert(e, phin)
m = pow(c, d, n)
print(long_to_bytes(m))

运行脚本可得flag。

[AFCTF2018]可怜的RSA

题目中包含了一个pub.key文件和flag.enc文件，通过在线工具可以获取公钥中的内容。

接下来就是读取enc文件中的内容，进行解密。脚本如下：

from Crypto.PublicKey import RSA
from Crypto.Cipher import PKCS1_OAEP
from base64 import b64decode

n = 79832181757332818552764610761349592984614744432279135328398999801627880283610900361281249973175805069916210179560506497075132524902086881120372213626641879468491936860976686933630869673826972619938321951599146744807653301076026577949579618331502776303983485566046485431039541708467141408260220098592761245010678592347501894176269580510459729633673468068467144199744563731826362102608811033400887813754780282628099443490170016087838606998017490456601315802448567772411623826281747245660954245413781519794295336197555688543537992197142258053220453757666537840276416475602759374950715283890232230741542737319569819793988431443
e = 65537
d=406853230956379689450620815713768871010712825839536410687962650677800895818003893712259622281477453292088146173840036827322518131453630576229976208523593618949818777897059256426591560532784635697190752924923710375949616954069804342573867253630978123632384795587951365482103468722384133084798614863870775897915929475258974188300927376911833763105616386167881813301748585233563049693794370642976326692672223638908164822104832415788577945314264232531947860576966629150456995512932232264881080618006698700677529111454508900582785420549466798020451488168615035256292977390692401388790460066327347700109341639992159475755036449
p = 3133337
q = 25478326064937419292200172136399497719081842914528228316455906211693118321971399936004729134841162974144246271486439695786036588117424611881955950996219646807378822278285638261582099108339438949573034101215141156156408742843820048066830863814362379885720395082318462850002901605689761876319151147352730090957556940842144299887394678743607766937828094478336401159449035878306853716216548374273462386508307367713112073004011383418967894930554067582453248981022011922883374442736848045920676341361871231787163441467533076890081721882179369168787287724769642665399992556052144845878600126283968890273067575342061776244939


key_info = RSA.construct((n, e, d, p, q))
key = RSA.importKey(key_info.exportKey())
key = PKCS1_OAEP.new(key)
f = open('flag1.enc', 'r').read()
c = b64decode(f)
flag = key.decrypt(c)
print(flag)

运行脚本可得flag。

鸡藕椒盐味

题目如下：

考察的是海明码，看了一篇关于海明码写得很好的博客：https://baijiahao.baidu.com/s?id=1598006039749022275&wfr=spider&for=pc。通过计算发现是第九位发生了变化，进行置位后再MD5加密就可以了。脚本如下：

import hashlib
cipher = "110110100000"
md = hashlib.md5()
md.update(cipher.encode("utf-8"))
flag = md.hexdigest()
print("flag{" + flag + "}")

运行脚本可得flag。

[ACTF新生赛2020]crypto-classic0

题目中有一个加密的howtoencrypto.zip文件、hint.txt、cipher文件，内容如下：

hint.txt:
哼，压缩包的密码？这是小Z童鞋的生日吧==
cipher:
Ygvdmq[lYate[elghqvakl}

根据提示盲猜密码是8位，通过ARCHPR软件进行暴力破解zip文件

接着压缩包中的内容是一个.c文件，内容如下：

#include<stdio.h>

char flag[25] = ***

int main()
{
	int i;
	for(i=0;i<25;i++)
	{
		flag[i] -= 3;
		flag[i] ^= 0x7;
		printf("%c",flag[i]);
	}
	return 0; 
}

解密就是加密的逆运算过程，脚本如下：

cipher = "Ygvdmq[lYate[elghqvakl}"
for i in cipher:
    i = ord(i)
    i ^= 0x7
    i += 3
    print(chr(i), end='')

运行脚本可得flag。

[RoarCTF2019]babyRSA

题目如下：

import sympy
import random

def myGetPrime():
    A= getPrime(513)
    print(A)
    B=A-random.randint(1e3,1e5)
    print(B)
    return sympy.nextPrime((B!)%A)
p=myGetPrime()
#A1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467234407
#B1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467140596

q=myGetPrime()
#A2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858418927
#B2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858351026

r=myGetPrime()

n=p*q*r
#n=85492663786275292159831603391083876175149354309327673008716627650718160585639723100793347534649628330416631255660901307533909900431413447524262332232659153047067908693481947121069070451562822417357656432171870951184673132554213690123308042697361969986360375060954702920656364144154145812838558365334172935931441424096270206140691814662318562696925767991937369782627908408239087358033165410020690152067715711112732252038588432896758405898709010342467882264362733
c=pow(flag,e,n)
#e=0x1001
#c=75700883021669577739329316795450706204502635802310731477156998834710820770245219468703245302009998932067080383977560299708060476222089630209972629755965140317526034680452483360917378812244365884527186056341888615564335560765053550155758362271622330017433403027261127561225585912484777829588501213961110690451987625502701331485141639684356427316905122995759825241133872734362716041819819948645662803292418802204430874521342108413623635150475963121220095236776428
#so,what is the flag?

碰到RSA的新题型，加密过程还是比较清晰的，但是不知道如何求解B! % A。还是去看看大佬的博客，大佬的博客中说道：威尔逊定理。

当且仅当p为素数时：( p -1 )! ≡ -1 ( mod p )，我们的加密过程x = B! % A。从代码块中我们知道，A是素数，那么就有：

(A - 1)! ≡ -1 (mod A)，又因为：

B=A-random.randint(1e3,1e5)

(A - 1) (A - 2) (A - 3) …… (B + 1) B (B - 1) …… 2 * 1 ≡ -1 (mod A)，又有：

(A - 1) (A - 2) (A - 3) …… (B + 1) B (B - 1) …… 2 * 1 ≡ A -1 (mod A)，所以：

(A - 2) (A - 3) …… (B + 1) B (B - 1) …… 2 1 ≡ 1 (mod A)。

所以要求B! % A，只要求(A - 2) (A - 3) …… * (B + 1) 对A的逆就好了。脚本如下：

import gmpy2
import sympy
from Crypto.Util.number import long_to_bytes

A1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467234407
B1=21856963452461630437348278434191434000066076750419027493852463513469865262064340836613831066602300959772632397773487317560339056658299954464169264467140596
A2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858418927
B2=16466113115839228119767887899308820025749260933863446888224167169857612178664139545726340867406790754560227516013796269941438076818194617030304851858351026
n=85492663786275292159831603391083876175149354309327673008716627650718160585639723100793347534649628330416631255660901307533909900431413447524262332232659153047067908693481947121069070451562822417357656432171870951184673132554213690123308042697361969986360375060954702920656364144154145812838558365334172935931441424096270206140691814662318562696925767991937369782627908408239087358033165410020690152067715711112732252038588432896758405898709010342467882264362733
e=0x1001
c=75700883021669577739329316795450706204502635802310731477156998834710820770245219468703245302009998932067080383977560299708060476222089630209972629755965140317526034680452483360917378812244365884527186056341888615564335560765053550155758362271622330017433403027261127561225585912484777829588501213961110690451987625502701331485141639684356427316905122995759825241133872734362716041819819948645662803292418802204430874521342108413623635150475963121220095236776428
re = 0
def decrypto(A, B):
    tmp = 1
    #注意不乘（A - 1）
    for i in range(B + 1, A - 1):
        tmp *= i
        tmp %= A
    B_A = gmpy2.invert(tmp, A)
    re = sympy.nextprime(B_A)
    return re

p = decrypto(A1, B1)
q = decrypto(A2, B2)
r = n // (p * q)
phin = (p - 1) * (q - 1) * (r - 1)
d = gmpy2.invert(e, phin)
m = pow(c, d, n)
print(long_to_bytes(m))

运行脚本可得flag。

[RoarCTF2019]RSA

题目如下：

A=(((y%x)**5)%(x%y))**2019+y**316+(y+1)/x
p=next_prime(z*x*y)
q=next_prime(z)
A =  2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724
n =  117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c =  41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128

首先在A的生成公式中存在着一个隐含条件：(y + 1) // x。那么就会有y = k * x - 1，当k=1时，y < x，有y % x = y，x % y = 1 —>(y^5)^2019 + y^316 + (y + 1)/x，其中最大次数的项决定了A的数据规模，用脚本查看A的二进制位数：

A = 2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724

str = str(bin(A))
print(len(str))

得到A的二进制数长度为2017，而上面推导出的(y^5)^2019显然可以将y < x排除在外了。所以x < y，此时有y % x = -1，x % y = x —>

(x - 1)^2019 + y^316 + (y + 1)/x，为了满足A的数据规模，此时x<=2。求y时可以把x=2代入，也可以粗略算一下：(x - 1) = 1，式子两边同乘以x —> A x = x y^316 + y + 1，（x + y + 1）可以忽略掉（因为并不会影响A的数据规模），A开316次方算得y=83。

再看看p和q的生成条件：p ≈ z x y；q ≈ z，以为n = p q = z x y z，所以：

x和y都已知了，我们可以进行爆破p和q，接下去就是RSA的基本操作了。脚本如下：

from Crypto.Util.number import long_to_bytes
import gmpy2
import itertools


#A=(((y%x)**5)%(x%y))**2019+y**316+(y+1)/x
#p=next_prime(z*x*y)
#q=next_prime(z)
A = 2683349182678714524247469512793476009861014781004924905484127480308161377768192868061561886577048646432382128960881487463427414176114486885830693959404989743229103516924432512724195654425703453612710310587164417035878308390676612592848750287387318129424195208623440294647817367740878211949147526287091298307480502897462279102572556822231669438279317474828479089719046386411971105448723910594710418093977044179949800373224354729179833393219827789389078869290217569511230868967647963089430594258815146362187250855166897553056073744582946148472068334167445499314471518357535261186318756327890016183228412253724
n = 117930806043507374325982291823027285148807239117987369609583515353889814856088099671454394340816761242974462268435911765045576377767711593100416932019831889059333166946263184861287975722954992219766493089630810876984781113645362450398009234556085330943125568377741065242183073882558834603430862598066786475299918395341014877416901185392905676043795425126968745185649565106322336954427505104906770493155723995382318346714944184577894150229037758434597242564815299174950147754426950251419204917376517360505024549691723683358170823416757973059354784142601436519500811159036795034676360028928301979780528294114933347127
c = 41971850275428383625653350824107291609587853887037624239544762751558838294718672159979929266922528917912189124713273673948051464226519605803745171340724343705832198554680196798623263806617998072496026019940476324971696928551159371970207365741517064295956376809297272541800647747885170905737868568000101029143923792003486793278197051326716680212726111099439262589341050943913401067673851885114314709706016622157285023272496793595281054074260451116213815934843317894898883215362289599366101018081513215120728297131352439066930452281829446586562062242527329672575620261776042653626411730955819001674118193293313612128
x = 2
y = 83
e = 65537

for q in itertools.count(gmpy2.iroot(n // 166, 2)[0]):
    if n % q == 0:
        p = n // q
        phin = (p - 1) * (q - 1)
        d = gmpy2.invert(e, phin)
        m = pow(c, d, n)
        print(long_to_bytes(m))
        break

运行脚本可得flag。

[网鼎杯 2020 青龙组]boom

使用cmd打开运行的文件，如下图所示：

按任意键试试：

通过在线工具破解MD5密文：

提交后：

使用sage解方程组：

提交后：

继续使用sage求解：

提交上去后就出现flag：

[WUSTCTF2020]B@se

题目如下：

密文：MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbvD==
JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/

oh holy shit, something is missing...

思路：对其中未知的四位字符进行爆破，在按照题目给出的编码表进行正常的base64解密即可。脚本如下：

import string
import binascii
import itertools

str="JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs****kxyz012789+/"
ciper="MyLkTaP3FaA7KOWjTmKkVjWjVzKjdeNvTnAjoH9iZOIvTeHbVD"

lost_letters = ""
for i in string.ascii_letters+string.digits:
    if i not in str:
        lost_letters += i

for i in itertools.permutations(lost_letters,4):
    ss="JASGBWcQPRXEFLbCDIlmnHUVKTYZdMovwipatNOefghq56rs"+"".join(i)+"kxyz012789+/"
    flag = ""
    for j in ciper:
        flag+=bin(ss.index(j))[2:].zfill(6)
    print(binascii.unhexlify(hex(eval("0b"+flag))[2:-1]))

运行脚本进行flag的筛选即可。

救世捷径

题目如下：

附件内容：

1 2 100 FLAG{
2 3 87 AFQWE
2 4 57 ETKLS
2 5 50 WEIVK
2 6 51 AWEIW
3 7 94 QIECJF
3 8 78 QSXKE
3 9 85 QWEIH
4 13 54 WQOJF
4 14 47 KDNVE
4 15 98 QISNV
5 10 43 AEWJV
5 11 32 QWKXF
5 12 44 ASJVL
6 16 59 ASJXJ
6 17 92 QJXNV
6 18 39 SCJJF
6 23 99 SJVHF
7 19 99 WJCNF
8 20 96 SKCNG
9 20 86 SJXHF
10 21 60 SJJCH
11 21 57 SJHGG
12 22 47 SJCHF
14 10 55 EJFHG
16 17 59 ASJVH
18 12 53 SJFHG
18 24 93 SHFVG
21 22 33 SJFHB
19 25 88 ASHHF
20 25 96 SJVHG
22 25 23 SJVHJ
25 26 75 SDEV}

根据题目意思考察的是最短路径问题，使用Dijkstra算法进行解密。主要脚本如下：

list = []
with open("Encrypto.txt", "rb") as f:
    lines = f.readlines()
    for line in lines:
        line = line.strip('\r\n'.encode())
        list.append(line)

g = {}
for i in list:
    data = i.split(" ".encode())
    g[str(int(data[0])) + "-" + str(int(data[1]))] = int(data[2])

dist = {}
st = {}
for i in range(1, 27):
    dist[str(i)] = str(int(0x3f3f3f3f))
    st[str(i)] = False
#print(dist)
#print(st)

def Dijkstra(dist, st, g):
    dist['1'] = 0

    for i in range(1, 27):
        t = -1
        for j in range(1, 27):
            if (st[str(j)] != True and (t == -1 or int(dist[str(t)]) > int(dist[str(j)]))):
                t = j

        st[str(t)] = True

        for j in range(1, 27):
            if str(t)+"-"+str(j) in g:
                dist[str(j)] = min(int(dist[str(j)]), int(dist[str(t)]) + g[str(t)+"-"+str(j)])

    if (dist['26'] == 0x3f3f3f3f):
        return -1
    else:
        return dist['26']

print(Dijkstra(dist, st, g))

[AFCTF2018]Single

题目如下：

cipher.txt:
Jmqrida rva Lfmz (JRL) eu m uqajemf seny xl enlxdomrexn uajiderc jxoqarerexnu. Rvada mda rvdaa jxooxn rcqau xl JRLu: Paxqmdyc, Mrrmjs-Yalanja mny oekay.

Paxqmdyc-urcfa JRLu vmu m jxiqfa xl giaurexnu (rmusu) en dmnza xl jmrazxdeau. Lxd akmoqfa, Wab, Lxdanuej, Jdcqrx, Benmdc xd uxoarvenz afua. Ramo jmn zmen uxoa qxenru lxd atadc uxftay rmus. Oxda qxenru lxd oxda jxoqfejmray rmusu iuimffc. Rva nakr rmus en jvmen jmn ba xqanay xnfc mlrad uxoa ramo uxfta qdatexiu rmus. Rvan rva zmoa reoa eu xtad uio xl qxenru uvxwu cxi m JRL wenad. Lmoxiu akmoqfa xl uijv JRL eu Yaljxn JRL gimfu.

Waff, mrrmjs-yalanja eu mnxrvad enradaurenz seny xl jxoqarerexnu. Vada atadc ramo vmu xwn narwxds(xd xnfc xna vxur) werv tifnmdmbfa uadtejau. Cxid ramo vmu reoa lxd qmrjvenz cxid uadtejau mny yatafxqenz akqfxeru iuimffc. Ux, rvan xdzmnehadu jxnnajru qmdrejeqmnru xl jxoqarerexn mny rva wmdzmoa urmdru! Cxi uvxify qdxrajr xwn uadtejau lxd yalanja qxenru mny vmjs xqqxnanru lxd mrrmjs qxenru. Veurxdejmffc rveu eu m ledur rcqa xl JRLu, atadcbxyc snxwu mbxir YAL JXN JRL - uxoarvenz fesa m Wxdfy Jiq xl mff xrvad jxoqarerexnu.

Oekay jxoqarerexnu omc tmdc qxuuebfa lxdomru. Er omc ba uxoarvenz fesa wmdzmoa werv uqajemf reoa lxd rmus-bmuay afaoanru (a.z. IJUB eJRL).

JRL zmoau xlran rxijv xn omnc xrvad muqajru xl enlxdomrexn uajiderc: jdcqrxzdmqvc, urazx, benmdc mnmfcueu, datadua anzanaadenz, oxbefa uajiderc mny xrvadu. Zxxy ramou zanadmffc vmta urdxnz useffu mny akqadeanja en mff rvaua euuiau.

Iuimffc, lfmz eu uxoa urdenz xl dmnyxo ymrm xd rakr en uxoa lxdomr. Akmoqfa mljrl{Xv_I_lxiny_er_neja_rDc}
------------------------------------------------------------------------------------------------------------
Encode.cpp:
#include <bits/stdc++.h>
using namespace std;
int main()
{
	freopen("Plain.txt","r",stdin);
	freopen("Cipher.txt","w",stdout);
	map<char, char> f;
	int arr[26];
	for(int i=0;i<26;++i){
		arr[i]=i;
	}
	random_shuffle(arr,arr+26);
	for(int i=0;i<26;++i){
		f['a'+i]='a'+arr[i];
		f['A'+i]='A'+arr[i];
	}
	char ch;
	while((ch=getchar())!=EOF){
		if(f.count(ch)){
			putchar(f[ch]);
		}else{
			putchar(ch);
		}
	}
	return 0;
}

应该是一种简单的替换加密，直接爆破试试：

flag就在最后一句话了。

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