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simple xor
importbase64encrypted_str="XUBdTFdScw5XCVRGTglJXEpMSFpOQE5AVVxJBRpLT10aYBpIVwlbCVZATl1WTBpaTkBOQFVcSQdH:)"data=base64.b64decode(encrypted_str) xor_key=[58,41]#':'(58),')'(41)decrypted=b''foriinra...
- Crypto
- 8天前
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HackINI -2023-The Mistery of Paillier 1题解
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1 金币
- Crypto
- 17天前
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HackINI -2023-PCryptor题解
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1 金币
- Crypto
- 19天前
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lz77
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3 金币
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- 25天前
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lz77writeup
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3 金币
- Crypto
- 25天前
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HackINI -2023-OTPx题解
首先向目标服务发送命令2p=remote("49.232.142.230",10320)p.sendlineafter(b">>",b"2")p.recvuntil(b"encryptedflagis:")enc_flag_hex=p.recvline().decode().strip()print(enc_flag_hex)获取对应的密文b'a\xf2\xbd\r\xe2\x96m\...
- Crypto
- 27天前
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【Write Up】Whispers_of_the_Ancients
只有大小写字母会按照某个key进行移位操作。直接把原文丢进去逐个位置试一下即可。#xprotfetwq{t1e3L3p3_c4$w_ApdBc0}#shellmates{v1g3N3r3_e4$y_CryTp0}---这一行是试出来的flaga=ord('c')-(ord('n')-ord('a'))ifa<ord('a'):a+=26print(chr(a))
- Crypto
- 1月前
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【Write Up】crtrsa
fromCrypto.Util.numberimport*fromgmpy2import*e=295354426800286670387207655193095372257231712277786129929340705339180819922065528923598308898637263014182104911801575201741264214893411372317485523614288...
- Crypto
- 1月前
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【Write Up】RSA5
fromgmpy2import*fromCrypto.Util.numberimport*n=158307578375429142391814474806884486236362186916188452580137711655290101749246194796158132723192108831610021920979976831387798531310286521988621973910776...
- Crypto
- 1月前
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【Write Up】Easy RSA
fromCrypto.Util.numberimport*fromgmpy2import*e=65537defget_p():x=111244400217481271590920768614054548149815751447445088571785765729293214350029429985314209857710901672622568778059021353041122716410744...
- Crypto
- 1月前
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【Write Up】院校组-easyRSA
factordb得到p,q，直接套RSA板子。fromgmpy2import*fromCrypto.Util.numberimport*p=8307103755174226983699771812499382664784661030503034013965679561410051699975573257899430944515587916063550418050690024796566861042...
- Crypto
- 1月前
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【Write Up】大帝的征程#2
观察密码表table='a~z'+'0~9'，能看出来这个密码是一个简单的替代密码。观察前面mpgfxk和moectf的关系，发现m=m，p、o是相邻的，g、e是隔着一个的，自然可以得到根据位置决定变换的变种凯撒密码。moectf{c0nquer_th3_un1v3rs3}
- Crypto
- 1月前
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【Write Up】Asymmetric
importgmpy2importrandomimportbase64importsysfromCryptodome.Util.numberimport*defexgcd(a,b):#ax+by=gcd(a,b)ifb==0:return[1,0]c=exgcd(b,a%b)return[c[1],c[0]-a//b*c[1]]definv(a,b):#ax+bk=1return(exgcd(a,...
- Crypto
- 1月前
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NCTF_2022_coloratura
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1 金币
- Crypto
- 1月前
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RSA3-BYUCTF-2023
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1 金币
- Crypto
- 1月前
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RSA1-BYUCTF-2023
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1 金币
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- 1月前
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仿射加密
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1 金币
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- 1月前
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laying with xor
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3 金币
- Crypto
- 1月前
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MapleCTF2022
给定了p^4-q^3=hint,p*q=n直接解方程就可以了#sagefromCrypto.Util.numberimport*N=e=enc=hint=#p=var('p')#eq =p^4-hint-N^3/p^3==0#solve=solve(eq,p)#print(solve)p=11917573148183173444338385104784582231114229409447...
- Crypto
- 1月前
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Binarandom
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4 金币
- Crypto
- 4月前
