7.23 密码

SSSS

p=e4fa76fb77c30f889bbab54d7d7a3e7edbd7ae6c42a1a443f657e95c5708fa15
(1,9768db2b62cf5e892fd6a0fc26c1387974eefd609998d04a5db830cd85e334bd)
(3,7452014a456e0545c248617169f6426edd53d4f65ae9dced2ecc0c2911651410)
(5,ed09d852d4789d46891970459a710db6df790be773a9ba7f51528ac508fc74a)
(7,a2637283075c9494ac78f8d2dc0dd2e171ae93f03b0e29e84b785f552ffae39c)
(9,a54a9f5ada141d245eff3817ac40c80ed2120efcf6e90cdfffef8a3fcd34a3da)
AES-ECB '8a609296eae64a015148d512a6207fee4feed180dce7aed733d827428d0e0f19705151490bc4b488eaee469ac25a90463b78eb6bdbad0adaae4bb45ed66ceb38'
buu之前做过shamir加密 一下子就想到了
shamir
参考一下
大佬的脚本

p = 0xe4fa76fb77c30f889bbab54d7d7a3e7edbd7ae6c42a1a443f657e95c5708fa15

ss = [(1, 0x9768db2b62cf5e892fd6a0fc26c1387974eefd609998d04a5db830cd85e334bd),
(3, 0x7452014a456e0545c248617169f6426edd53d4f65ae9dced2ecc0c2911651410),
(5, 0xed09d852d4789d46891970459a710db6df790be773a9ba7f51528ac508fc74a),
(7, 0xa2637283075c9494ac78f8d2dc0dd2e171ae93f03b0e29e84b785f552ffae39c),
(9, 0xa54a9f5ada141d245eff3817ac40c80ed2120efcf6e90cdfffef8a3fcd34a3da)]

# shamir
Zp = Zmod(p)
A = Matrix(Zp, [[x^i for i in range(5)] for x, y in ss])
b = vector(Zp, [y for x, y in ss])
print(A)

x = A.solve_right(b)
print(x)

from Crypto.Util.number import *
a=[153421376468734975731045175614702317458439896683272740513610079, 13199192800511101628139098831052436664195660827240283169771088569098253935251, 61088415672732934161626778412472480144291218142728169239148393851672254005488, 18497361290777669420831397571339137872527963545071031656684459823492858013376, 79269397547729156551447820375139597186819542624125750350693624198297473706800]
for i in a:
    print(long_to_bytes(i))

第一个就是key aes解密

小二难度R54

给出了n，c，e
还有(2020p+501q)^(p*q-p-q) mod n
phi(n)=p*q-p-q+1
令a=(2020p+501q)
a^(phi(n)-1)=a^phi(n)*a^-1 mod n=a^-1 mod n #因为欧拉定理 a^phi(n)=1 mod n
所以我们可以求出a
然后解方程

var('p q')
a=327448806398192933801037652433354723329495079501058233693653746267433227317180518594815962904326861755308042692131522024714735549979411375294632496223054730601269843657930822421155465866751222008925869490328766285317785703175108703277038344625470471149420270843632927027865488878595360005766919889118978591175398
n=20947495659013288660808536751393787394664606045798093048128257278988208709333248671898749660848208653968668634891579612784633367362177864996602736258460476691940723323282467207875842974409286563660436709535601954405015261428106261369927836045794170912665351432105165546188591486357060490032334793140396757102052999128194027485573053073959574695808224922102635888141991154365047911830780778957642166757152369955362399379720841279167832886144458760347392316082994786119404006382441787685301119197529946566027319295285387108473752590621030421978808950305190250697199878929419723511578437404000924310974770501204226510397
solve([p*q==n,2021*p+501*q==a],[p,q])

得到p和q

import gmpy2
from Crypto.Util.number import *
c=13390681135321846035598057088735733735860895610541899486616159864716324918810264721878447895634342127744578566110322466944217562868186608760962032192994397783118528288276520451944892998435079744244578731427626946331165523865930693902700790185275273534104979885060728696532991031786741950704918951536399577118136416956670893081637730646528913282395731901667720418372650030593319596584787752412110672058692368924987360383096340538971725402687347195347344826404005229912821371282465882351660619944919637382790572267512735645269618163597227604601321699186335016345484182059187046972681187078878556533926780789183784240737
a=12911378830212711575909332427930495830030418987483519620282504671823307660633472092466534392403086505995560725428252134905285658936113891795434303336259751169171583600394870893708505805256284455729584616439559184469715186920464999723861722097244025658190194027561300165723184060071016117033960821040587421503448139025974851980482004179865110864844573575034406782936965166402665401330436229441569042660851847498727291447251591027480750458209012729510702196684303778564353025395186191064801000127420683298000173389589468742142444444759536629401472836827952997758216526858512433131954439154124668711408079361172485321041
n=20947495659013288660808536751393787394664606045798093048128257278988208709333248671898749660848208653968668634891579612784633367362177864996602736258460476691940723323282467207875842974409286563660436709535601954405015261428106261369927836045794170912665351432105165546188591486357060490032334793140396757102052999128194027485573053073959574695808224922102635888141991154365047911830780778957642166757152369955362399379720841279167832886144458760347392316082994786119404006382441787685301119197529946566027319295285387108473752590621030421978808950305190250697199878929419723511578437404000924310974770501204226510397
a=gmpy2.invert(a,n)

p =118025808013874567502972067396154378205921004159004592806833496003217679639841107629090128959992942958292926872941356148283178987927403535617729247751812776636333788905610726461255956234207606642302396722821445909489968841256943549868105578713493559907241758893208485830735235619636112619927867287431575647321
q =177482332140024816122816575300851746457741976238942019223639223243373845838446387378093437677008231609976122718397287922224412805145965728166071230572137942154170172214953182121671014605624049869925600226560127948579957435119412752482229481128742488177414523194528098131635883615390771259266866469500527361157
phi=(p-1)*(q-1)
e=0o200001
d=gmpy2.invert(e,phi)
m=pow(c,d,n)
print(long_to_bytes(m))

emkkocy{lnyu_wp_kxs_kc_ahz_dvjg_by?}
一看就是被加密过的 不是凯撒 只能自己分析

a='emkkocy{lnyu_wp_kxs_kc_ahz_dvjg_by?}'
b='ctfshow{'
n=-1
for i in a:
    if(i>='a' and i<='z'):
        n=n+1
        if(n%6==0):
            print(chr((ord(i)-97-2)%26+97),end='')
        if(n%6==1):
            print(chr((ord(i)-97+7)%26+97),end='')
        if(n%6==2):
            print(chr((ord(i)-97-5)%26+97),end='')
        if(n%6==3):
            print(chr((ord(i)-97+8)%26+97),end='')
        if(n%6==4):
            print(chr((ord(i)-97-7)%26+97),end='')
        if(n%6==5):
            print(chr((ord(i)-97+12)%26+97),end='')
    else:
        print(i,end='')

抱我

cstring = 'abcdefghijklmnopqrstuvwxyz{}_0123456789'
key = 'flag{********}'
length = 300

def encode():
    res = ''
    for i in range(1, length):
        c = random.randint(0, 36)
        res += cstring[c]
        for n in range(10):
            c = random.randint(0, len(key) - 1)
            res += key[c]
    return res

#qdfl33{6{6gs3afa6{3}agf{}aagdf}6fl36d{dfl{6ay6gafddfg}{j3f}}6la{3}bfdf3}gla}65}lg6g6dflf0{dfgd3fdfgc{g6a}a3{6}mfa{}f}f}}}3363}}f6a6a7g{a}g66{d3xgfffg}a}3}_{lad}a33ga5fd33}{{dl}{}f{3da}g}3egfal{a3l}3f33}dfdda{3sa{d6g{ff}6vgl33d6g333h{gd{{lg6ldg{ad{3333a6oalf6a{33{de3{fa}ggl{abfd}6}6}}l33fa}f{{3{3fla}a6}af}{amg}{{d}}a6gallfg36{g3dh{{{a{lfg3{sll6g6gfaggid6d{3afl}3rff3gfad3d}1dlllff6}6}h3g66gla336b{6d3gf}f{30d63l}3dfl6a3llfgld3{}qg}gf}dg{6l}3gal}agdl6{lg{g}ddfaaealf{f3llgge3ad3{3adf{c}fllf6f}3at{aag}a66d3}ad{dfg{}dlz6gld}6{3flxgf{3g3ald}3g}g63f6ggf3}gfd}f3ga3efllf6}363fu6366fdlggfx6}6l3}}a{afg{{}}3fdaluaa}al{dg3dpfga}}l}d3l4afg}f{d{lgcfgffglal}dq6l}fgflldavdad6}df{}dw}l6g}}{l3gf6fdaa66aadt}f6lg{dg33h{fa3d{}laao3l6aal{lfdv{3dlf6af36bddg}3ggad3o}{}3g3fgddyffd3lddgdd6{gdfl{{la3ild}dg{g}dgef{a3{d6dfgq3adll{fdadt}66fdflg{3x{l3ll}3{{g4a3af6lag3gdaf66dadg6dfglaf66l3f{2}6{afaf}3l6all}{l}lfdla6{fgff{}g13dl{a{6{l6rd6}}l3dgg3_f{66gll{f6a3d3dga6{lg}{g}d{6{d36lll3dd6{3dg3afal}d}gff26}l}al}}{a6}g66gaaff}0fga{g6dfld{{}fglf{af}iddf6g6}l361{ag}{{dlfak}{d3fa{6{godgg{l36a{gmllgfa3fa{}f}}3{a6{a3{nafg{l3d6}g2lf6{gg}{g}sg{ga{63g{}la6g{g6{{63o6{l}{3}l3ag36{af33g3dw6d33f3lfdan{dddad{{6l6}}fad63lgd1ffaa}g}3flkg3d}aalf3lbgf{g}f}}d3agf{ld{dl3l4fl{{3fla}}r3g}{}gda{}_df3g}fa36gq}la{f{6l}66fgdg}6ag6feaal6all3{d}lfgl}}{{6lal}gf}}gfgd4d{g36daff}l6fd63ag6}f7}l3{{d}{al6lff66gda}f7dfaf6}fd3ldfgfl36gf337a6al663afd{dff}6}df{lt}66}ag6a3{na}3la{6daa}63fgldf3ggcl6dd{3fg{}}gfgaf{633lpfadalldgglg{l}{6}gf{agf6{3l3a366wa6l6}fdla}wfl}33}d{6d6aa}laldag}bgaa3gff}3db{gd}lfga3{}ffddd6}{la4}3{agdg3{}bf33adg3a632d}66f}dgd67}{333dfg}}mgg3all3l}fd6dd3{g}{}}v6}a6f6lgd3nfgg6aff3a}d3da{l3ldldz{}{}g3}6fdg6f{gd{g3adx{gll6{fg3dc63lf}6dl{d63f3g{3adda5f3dgfla3{6}gd{3{d6dlldal6g66}{ddp}lalafd}d{lgl}g6g33agjg}33dgf}lg0adlda6gfdlx{3g}{g3a{a76}gdf3la}lh}l{l{}}a6gm{gdd{agg}6xfgg}{336}d_a{df3}df33jgf}6d3}}f}h3l{6ga6fll2}dd{l36d66}ldafdlga3gbgd}d6df}ff1gf6a{ll3a3w{3g}allfafldal}aal}dlra33l3f}3dff{6{6}f}la}lgf}}}gd{f3z3l3{d3636dpl3fag3{faa1{3ga33l}6ll6{gg6}ddf}t6g}{gl6ggl{d}aafalf{lw6a{dad}}a3x{ada{fg6d}a3g{d{fggdawdfal{{3dlfndl636}36alv633ada6gf6hd{3l66ddlfpglda}{g3fdogdfa3}3g}3k3d3gda33}dvd}laa{fa{a{{}a}36}}}{r6d6{a6}}6{0laa36gd{36kf63a{3}gga4af6}f3gfgf0lf{6g}{{6}pafg6dg}g6{b}3d36ad6d{h6f3agff}63p}{l3ag3}lf1f3dgd{66a37}}}d6gglaftaf3l6a3{{a7{lgd3d}fl6tlfl663lgg3wa}33gl}d{3i6aaagl6{{}n3gd}l3l6}l7a{gf{a}l}f3al{alg63fln{{dd}3l{ll1}{3g}6{6}{u63{f3{g6lgf{3d}{636}{u3}{f6d{{d3lg{3l6aldf{i{f366{f3l{eg{d{gll{3dhgdgfgaf{}}g}{lg3{a{flm}fa3ldf{d32fagllf{{66q363}dl66gg2fa6af6d6g37lffl{d{3lltgl33}}{}d3o{lfld3d{}a6a663a66{fabfd6ld333g3rafa}}fddfgt{ggad3ag}lr63af6lgg}gy{6{{6}6dd626{gl6a{ad3b3df}alf3afdaf66ll}lf6jd}3{6dldfgg}f3lg63l{lr3ff3l{gafaa}f}agl6l33xglfggg{{{fq66}g6lfa3{736lllflalglf}{}gf{aggdg3{a}}da{fp6fglla3l}65gf36{l6dl}g}f{la6{l{fpf{}63{f6gdfaalf6{dffgdgf{lgaf{f{56}g6af63l6a}a}{lfa{3gblda}l}{fl{s{g}}6{g6la56g6g3{f}ddfaa{l}dg6g}0glda6{6d}ff}f{6laadd6zaag{l3l}6dc}f3gg}lffgsag}l3l6d3apd3gd3fd}}aga3ga}a3{6f1f3df{{d}}av3laf}6adf3_d{afa6f}adt{faf{d33aaol3}{l}ld}3yl3a6a{fa6}_d33gf3fll}of{6lad}}fdx}6d{f}ll63ugag66d{6f3}33}al6l{ffwf{}{fl3a36ogg3{}}g6}3hl}6dg6ld{digaa}g}{{l}da{ddg3{{d}w3}ld}adgg3m{lad{gd{a{7afff}{d6}fsf3{f}gflgavfldg6a6{ldqf}fd{f3f3}73ddad{666fz6}d{3{l36a1d6fal3fl6lrl{}aga{fdlsa}{6l6ag3gtgdg{6lgf3f

通过代码可以知道res每11位的第一位是cstring后10位位flag里的内容
去出这10位然后词频分析

res='qdfl33{6{6gs3afa6{3}agf{}aagdf}6fl36d{dfl{6ay6gafddfg}{j3f}}6la{3}bfdf3}gla}65}lg6g6dflf0{dfgd3fdfgc{g6a}a3{6}mfa{}f}f}}}3363}}f6a6a7g{a}g66{d3xgfffg}a}3}_{lad}a33ga5fd33}{{dl}{}f{3da}g}3egfal{a3l}3f33}dfdda{3sa{d6g{ff}6vgl33d6g333h{gd{{lg6ldg{ad{3333a6oalf6a{33{de3{fa}ggl{abfd}6}6}}l33fa}f{{3{3fla}a6}af}{amg}{{d}}a6gallfg36{g3dh{{{a{lfg3{sll6g6gfaggid6d{3afl}3rff3gfad3d}1dlllff6}6}h3g66gla336b{6d3gf}f{30d63l}3dfl6a3llfgld3{}qg}gf}dg{6l}3gal}agdl6{lg{g}ddfaaealf{f3llgge3ad3{3adf{c}fllf6f}3at{aag}a66d3}ad{dfg{}dlz6gld}6{3flxgf{3g3ald}3g}g63f6ggf3}gfd}f3ga3efllf6}363fu6366fdlggfx6}6l3}}a{afg{{}}3fdaluaa}al{dg3dpfga}}l}d3l4afg}f{d{lgcfgffglal}dq6l}fgflldavdad6}df{}dw}l6g}}{l3gf6fdaa66aadt}f6lg{dg33h{fa3d{}laao3l6aal{lfdv{3dlf6af36bddg}3ggad3o}{}3g3fgddyffd3lddgdd6{gdfl{{la3ild}dg{g}dgef{a3{d6dfgq3adll{fdadt}66fdflg{3x{l3ll}3{{g4a3af6lag3gdaf66dadg6dfglaf66l3f{2}6{afaf}3l6all}{l}lfdla6{fgff{}g13dl{a{6{l6rd6}}l3dgg3_f{66gll{f6a3d3dga6{lg}{g}d{6{d36lll3dd6{3dg3afal}d}gff26}l}al}}{a6}g66gaaff}0fga{g6dfld{{}fglf{af}iddf6g6}l361{ag}{{dlfak}{d3fa{6{godgg{l36a{gmllgfa3fa{}f}}3{a6{a3{nafg{l3d6}g2lf6{gg}{g}sg{ga{63g{}la6g{g6{{63o6{l}{3}l3ag36{af33g3dw6d33f3lfdan{dddad{{6l6}}fad63lgd1ffaa}g}3flkg3d}aalf3lbgf{g}f}}d3agf{ld{dl3l4fl{{3fla}}r3g}{}gda{}_df3g}fa36gq}la{f{6l}66fgdg}6ag6feaal6all3{d}lfgl}}{{6lal}gf}}gfgd4d{g36daff}l6fd63ag6}f7}l3{{d}{al6lff66gda}f7dfaf6}fd3ldfgfl36gf337a6al663afd{dff}6}df{lt}66}ag6a3{na}3la{6daa}63fgldf3ggcl6dd{3fg{}}gfgaf{633lpfadalldgglg{l}{6}gf{agf6{3l3a366wa6l6}fdla}wfl}33}d{6d6aa}laldag}bgaa3gff}3db{gd}lfga3{}ffddd6}{la4}3{agdg3{}bf33adg3a632d}66f}dgd67}{333dfg}}mgg3all3l}fd6dd3{g}{}}v6}a6f6lgd3nfgg6aff3a}d3da{l3ldldz{}{}g3}6fdg6f{gd{g3adx{gll6{fg3dc63lf}6dl{d63f3g{3adda5f3dgfla3{6}gd{3{d6dlldal6g66}{ddp}lalafd}d{lgl}g6g33agjg}33dgf}lg0adlda6gfdlx{3g}{g3a{a76}gdf3la}lh}l{l{}}a6gm{gdd{agg}6xfgg}{336}d_a{df3}df33jgf}6d3}}f}h3l{6ga6fll2}dd{l36d66}ldafdlga3gbgd}d6df}ff1gf6a{ll3a3w{3g}allfafldal}aal}dlra33l3f}3dff{6{6}f}la}lgf}}}gd{f3z3l3{d3636dpl3fag3{faa1{3ga33l}6ll6{gg6}ddf}t6g}{gl6ggl{d}aafalf{lw6a{dad}}a3x{ada{fg6d}a3g{d{fggdawdfal{{3dlfndl636}36alv633ada6gf6hd{3l66ddlfpglda}{g3fdogdfa3}3g}3k3d3gda33}dvd}laa{fa{a{{}a}36}}}{r6d6{a6}}6{0laa36gd{36kf63a{3}gga4af6}f3gfgf0lf{6g}{{6}pafg6dg}g6{b}3d36ad6d{h6f3agff}63p}{l3ag3}lf1f3dgd{66a37}}}d6gglaftaf3l6a3{{a7{lgd3d}fl6tlfl663lgg3wa}33gl}d{3i6aaagl6{{}n3gd}l3l6}l7a{gf{a}l}f3al{alg63fln{{dd}3l{ll1}{3g}6{6}{u63{f3{g6lgf{3d}{636}{u3}{f6d{{d3lg{3l6aldf{i{f366{f3l{eg{d{gll{3dhgdgfgaf{}}g}{lg3{a{flm}fa3ldf{d32fagllf{{66q363}dl66gg2fa6af6d6g37lffl{d{3lltgl33}}{}d3o{lfld3d{}a6a663a66{fabfd6ld333g3rafa}}fddfgt{ggad3ag}lr63af6lgg}gy{6{{6}6dd626{gl6a{ad3b3df}alf3afdaf66ll}lf6jd}3{6dldfgg}f3lg63l{lr3ff3l{gafaa}f}agl6l33xglfggg{{{fq66}g6lfa3{736lllflalglf}{}gf{aggdg3{a}}da{fp6fglla3l}65gf36{l6dl}g}f{la6{l{fpf{}63{f6gdfaalf6{dffgdgf{lgaf{f{56}g6af63l6a}a}{lfa{3gblda}l}{fl{s{g}}6{g6la56g6g3{f}ddfaa{l}dg6g}0glda6{6d}ff}f{6laadd6zaag{l3l}6dc}f3gg}lffgsag}l3l6d3apd3gd3fd}}aga3ga}a3{6f1f3df{{d}}av3laf}6adf3_d{afa6f}adt{faf{d33aaol3}{l}ld}3yl3a6a{fa6}_d33gf3fll}of{6lad}}fdx}6d{f}ll63ugag66d{6f3}33}al6l{ffwf{}{fl3a36ogg3{}}g6}3hl}6dg6ld{digaa}g}{{l}da{ddg3{{d}w3}ld}adgg3m{lad{gd{a{7afff}{d6}fsf3{f}gflgavfldg6a6{ldqf}fd{f3f3}73ddad{666fz6}d{3{l36a1d6fal3fl6lrl{}aga{fdlsa}{6l6ag3gtgdg{6lgf3f'
for i in range(1,300):
    c=res[(i-1)*11:i*11]
    print(c[1:])

盲猜36d