Buuctf Crpyto writeup --page1

发布于:2022-11-28 ⋅ 阅读:(712) ⋅ 点赞:(0)

1、md5码,md5网站解码

2、一眼就解密

3、 Url编码

%66%6c%61%67%7b%61%6e%64%20%31%3d%31%7d

4、 看我回旋踢

synt{5pq1004q-86n5-46q8-o720-oro5on0417r1}

5、 .. .-.. --- ...- . -.-- --- ..-

6、 password

姓名:张三 
生日:19900315

key格式为key{xxxxxxxxxx}

flag{zs19900315}

7、变异凯撒

afZ_r9VYfScOeO_UL^RWUc

8、Quoted-printable

=E9=82=A3=E4=BD=A0=E4=B9=9F=E5=BE=88=E6=A3=92=E5=93=A6

9、Rabbit

Rabbit在线网站解码

10、篱笆墙的影子

11、RSA

12、 丢失的MD5

import hashlib   
for i in range(32,127):
    for j in range(32,127):
        for k in range(32,127):
            m=hashlib.md5()
            m.update('TASC'+chr(i)+'O3RJMV'+chr(j)+'WDJKX'+chr(k)+'ZM')
            des=m.hexdigest()
            if 'e9032' in des and 'da' in des and '911513' in des:
                print des

4个字符以内md5码补全

 13、Alice与Bob

14、大帝的密码武器

 15、rsarsa

Math is cool! Use the RSA algorithm to decode the secret message, c, p, q, and e are parameters for the RSA algorithm.

p =  9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483

q =  11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407

e =  65537
c =  83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034


Use RSA to find the secret message

import gmpy2
e = 65537
p = 9648423029010515676590551740010426534945737639235739800643989352039852507298491399561035009163427050370107570733633350911691280297777160200625281665378483
q = 11874843837980297032092405848653656852760910154543380907650040190704283358909208578251063047732443992230647903887510065547947313543299303261986053486569407
n = p*q
#密文
C = 83208298995174604174773590298203639360540024871256126892889661345742403314929861939100492666605647316646576486526217457006376842280869728581726746401583705899941768214138742259689334840735633553053887641847651173776251820293087212885670180367406807406765923638973161375817392737747832762751690104423869019034
 
d =gmpy2.invert(e,(p-1)*(q-1))
print(d)
#求明文
M = pow(C,d,n)    #快速求幂取模运算
print(M)

 16、Windows系统密码

 md5解码

17、 信息化时代的步伐

18、 

凯撒?替换?呵呵!

MTHJ{CUBCGXGUGXWREXIPOYAOEYFIGXWRXCHTKHFCOHCFDUCGTXZOHIXOEOWMEHZO} 注意:得到的 flag 请包上 flag{} 提交, flag{小写字母}

quipqiup - cryptoquip and cryptogram solverhttps://quipqiup.com/

19、old-fashion

Os drnuzearyuwn, y jtkjzoztzoes douwlr oj y ilzwex eq lsdexosa kn pwodw tsozj eq ufyoszlbz yrl rlufydlx pozw douwlrzlbz, ydderxosa ze y rlatfyr jnjzli; mjy gfbmw vla xy wbfnsy symmyew (mjy vrwm qrvvrf), hlbew rd symmyew, mebhsymw rd symmyew, vbomgeyw rd mjy lxrzy, lfk wr dremj. Mjy eyqybzye kyqbhjyew mjy myom xa hyedrevbfn lf bfzyewy wgxwmbmgmbrf. Wr mjy dsln bw f1_2jyf-k3_jg1-vb-vl_l

词频分析,替换密码

https://quipqiup.com/

20、世上无难事

https://quipqiup.com/

21、Unencode

89FQA9WMD<V1A<V1S83DY.#<W3$Q,2TM]

22、[AFCTF2018]Morse

-..../.----/-..../-..../-..../...--/--.../....-/-..../-..../--.../-.../...--/.----/--.../...--/..---/--.../--.../....-/...../..-./--.../...--/...--/-----/...../..-./...--/...--/...--/....-/...--/...../--.../----./--.../-..

 

23、还原大师

4个字符以内md5补全

我们得到了一串神秘字符串:TASC?O3RJMV?WDJKX?ZM,问号部分是未知大写字母,为了确定这个神秘字符串,我们通过了其他途径获得了这个字串的32位MD5码。但是我们获得它的32位MD5码也是残缺不全,E903???4DAB????08?????51?80??8A?,请猜出神秘字符串的原本模样,并且提交这个字串的32位MD5码作为答案。 注意:得到的 flag 请包上 flag{} 提交 

 

本文含有隐藏内容,请 开通VIP 后查看

网站公告

今日签到

点亮在社区的每一天
去签到