UNCTF2020 Crypto Writeup

队伍：打CTF不靠实力靠运气

作者：wjhwjhn

easy_rsa

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from Crypto.Util import number
import gmpy2
from Crypto.Util.number import bytes_to_long

p = number.getPrime(1024)
q = number.getPrime(1024)
if p > q:
    a = p + q
    b = p - q
    print(a,b)

n = p * q
e = 65537
phi = (p-1)*(q-1)
d = gmpy2.invert(e,phi)
m = bytes_to_long(b'msg')
c = pow(m,e,n)
print(c)

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#320398687477638913975700270017132483556404036982302018853617987417039612400517057680951629863477438570118640104253432645524830693378758322853028869260935243017328300431595830632269573784699659244044435107219440036761727692796855905230231825712343296737928172132556195116760954509270255049816362648350162111168
#9554090001619033187321857749048244231377711861081522054479773151962371959336936136696051589639469653074758469644089407114039221055688732553830385923962675507737607608026140516898146670548916033772462331195442816239006651495200436855982426532874304542570230333184081122225359441162386921519665128773491795370
#22886015855857570934458119207589468036427819233100165358753348672429768179802313173980683835839060302192974676103009829680448391991795003347995943925826913190907148491842575401236879172753322166199945839038316446615621136778270903537132526524507377773094660056144412196579940619996180527179824934152320202452981537526759225006396924528945160807152512753988038894126566572241510883486584129614281936540861801302684550521904620303946721322791533756703992307396221043157633995229923356308284045440648542300161500649145193884889980827640680145641832152753769606803521928095124230843021310132841509181297101645567863161780

a + b = 2p
p = (a + b) / 2
q = p - a

用a和b，得到p和q，然后利用long_to_bytes(pow(c,d,n))解密。

鞍山大法官开庭之缺的营养这一块怎么补

某日，鞍山大法官在点外卖时点了2个韭菜盒子，商家只送了1个，大法官给了该商家一个差评次日，该大法官又在该商家点了1个韭菜盒子，希望商家能补上上次的韭菜盒子，而商家又只发了一个韭菜盒子这名大法官一天正常要吃2个韭菜盒子，而该商家每天只给他1个韭菜盒子，请问该名大法官缺的营养这一块怎么补

ottttootoootooooottoootooottotootttootooottotttooootttototoottooootoooottotoottottooooooooottotootto

看到这么多ot联想到培根密码，把o换成a，把t换成b

得到

abbbbaabaaabaaaaabbaaabaaabbabaabbbaabaaabbabbbaaaabbbababaabbaaaabaaaabbabaabbabbaaaaaaaaabbabaabba

得到flag

简单的RSA

正解：使用RSA-Wiener-Attack(维纳攻击)进行攻击。这种类型常见的情况就是e是一个较大或者较小的数字，一般是较大。已经有先辈帮我们实现好了: [github repo=“pablocelayes/rsa-wiener-attack” /] 我们就直接使用他提供的脚本进行计算。瞬间解出了结果。