持续创作，加速成长！这是我参与「掘金日新计划 · 10 月更文挑战」的第2天，点击查看活动详情
今天课比较多，就少做一些题目吧。
[BJDCTF2020]easyrsa
下载附件得到代码：
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
'''
根据以往RSA题目，我们先找有用的值，根据代码得到c,z,n的值，然后可以通过z的算式推导出p,q。
(p+q)2=z+2n
(p-q)2=z-2n
根据上面的推导求出P,Q，我们就可以写解题脚本了。
// python2
from gmpy2 import *
from Crypto.Util.number import *
c=mpz(7922547866857761459807491502654216283012776177789511549350672958101810281348402284098310147796549430689253803510994877420135537268549410652654479620858691324110367182025648788407041599943091386227543182157746202947099572389676084392706406084307657000104665696654409155006313203957292885743791715198781974205578654792123191584957665293208390453748369182333152809882312453359706147808198922916762773721726681588977103877454119043744889164529383188077499194932909643918696646876907327364751380953182517883134591810800848971719184808713694342985458103006676013451912221080252735948993692674899399826084848622145815461035)
z=mpz(32115748677623209667471622872185275070257924766015020072805267359839059393284316595882933372289732127274076434587519333300142473010344694803885168557548801202495933226215437763329280242113556524498457559562872900811602056944423967403777623306961880757613246328729616643032628964072931272085866928045973799374711846825157781056965164178505232524245809179235607571567174228822561697888645968559343608375331988097157145264357626738141646556353500994924115875748198318036296898604097000938272195903056733565880150540275369239637793975923329598716003350308259321436752579291000355560431542229699759955141152914708362494482)
n=mpz(15310745161336895413406690009324766200789179248896951942047235448901612351128459309145825547569298479821101249094161867207686537607047447968708758990950136380924747359052570549594098569970632854351825950729752563502284849263730127586382522703959893392329333760927637353052250274195821469023401443841395096410231843592101426591882573405934188675124326997277775238287928403743324297705151732524641213516306585297722190780088180705070359469719869343939106529204798285957516860774384001892777525916167743272419958572055332232056095979448155082465977781482598371994798871917514767508394730447974770329967681767625495394441)
e=65537
pqplus=iroot(z+2*n,2)[0]
pqminus=iroot(z-2*n,2)[0]
p=(pqminus+pqplus)//2
q=(pqplus-pqminus)//2
phi=(p-1)*(q-1)
d=invert(e,phi)
m=pow(c,d,n)
flag=long_to_bytes(m)
print flag
#BJD{Advanced_mathematics_is_too_hard!!!}
#flag{Advanced_mathematics_is_too_hard!!!}
[NPUCTF2020]这是什么觅🐎
比较有趣的一个题目，下载后发现是一个压缩包，我们解压后得到图片：
一个奇怪的日历，纸条上的内容大概就是加密内容：
F1 W1 S22 S21 T12 S11 W1 S13
这个很简单就能想到包含的字母都是星期英文首字母。其中，由于T和S开头的都有两个，所以按先后顺序有T1=TUE，T2=THU，S1=SAT，S2=SUN。
我们对照字母表，就可以得到flag:
flag{calendar}
总体来说这个题目还是挺有趣的，比较适合刚入门的新生去做，也能锻炼一下思维。
[NCTF2019]babyRSA
下载后得到代码：
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
一眼要求我们求p,q，q是1024位的二进制数，getprime(n)返回随机的n个bit位的素数， q是p的下一个素数。
∵e*d=1mod(p-1)(q-1)
∴e*d-1=k(p-1)(q-1)
但这之后就不太明白了，两个1024位的数乘起来至少是2047位，再测量一下e*d-1是2064位，相差17位，所以猜测k应该是个16-18位的数，我们爆破k值就可以了：
e = 0x10001
d = 19275778946037899718035455438175509175723911466127462154506916564101519923603308900331427601983476886255849200332374081996442976307058597390881168155862238533018621944733299208108185814179466844504468163200369996564265921022888670062554504758512453217434777820468049494313818291727050400752551716550403647148197148884408264686846693842118387217753516963449753809860354047619256787869400297858568139700396567519469825398575103885487624463424429913017729585620877168171603444111464692841379661112075123399343270610272287865200880398193573260848268633461983435015031227070217852728240847398084414687146397303110709214913
c = 5382723168073828110696168558294206681757991149022777821127563301413483223874527233300721180839298617076705685041174247415826157096583055069337393987892262764211225227035880754417457056723909135525244957935906902665679777101130111392780237502928656225705262431431953003520093932924375902111280077255205118217436744112064069429678632923259898627997145803892753989255615273140300021040654505901442787810653626524305706316663169341797205752938755590056568986738227803487467274114398257187962140796551136220532809687606867385639367743705527511680719955380746377631156468689844150878381460560990755652899449340045313521804
import sympy.crypto
import gmpy2
from Crypto.Util.number import *
ed1=e*d-1
for k in range(pow(2,15),pow(2,16)):
if ed1%k==0:
p=sympy.prevprime(gmpy2.iroot(ed1//k,2)[0])
q=sympy.nextprime(p)
if (p-1)*(q-1)*k==ed1:
break
n=p*q
print(n)
m=gmpy2.powmod(c,d,n)
print(m)
print(long_to_bytes(m))
运行得到flag
flag{70u2_nn47h_14_v3ry_gOO0000000d}
[UTCTF2020]basic-crypto
下载附件是一串二进制(由于太多了没有写全)
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
转成字符串
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，我们随便找一个网站去解码。
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!
语句不太通顺，感觉像是被位移了，我们用凯撒密码解一下，偏移量为10时：
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!
词频分析得到flag
flag{n0w_th4ts_wh4t_i_c4ll_crypt0}
