「配枪朱丽叶。」

RootのCTF学习笔记。

rsa

AFCTF2018/BUUCTF-Crypto:One Secret, Two encryption

题目下载 先分别求一下n和e。 SSL在线工具-公钥解析public1.pub: e = 16666266329603682390011594080477659912702500422062441574471711881956573029330195019321017779995100012357363388431077098717859067493930042576141298020610811558614337223801450…

BUUCTF/AFCTF2018-Crypto:可怜的RSA

题目下载 拿到公钥: -----BEGIN PUBLIC KEY----- MIIBJDANBgkqhkiG9w0BAQEFAAOCAREAMIIBDAKCAQMlsYv184kJfRcjeGa7Uc/4 3pIkU3SevEA7CZXJfA44bUbBYcrf93xphg2uR5HCFM+Eh6qqnybpIKl3g0kGA4rv tcMIJ9/PP8npdpVE+U4Hzf4IcgOaOmJiEWZ4smH7LWudMlOekqFTs2dWKbqzlC…

i春秋圣诞CTF答题夺旗赛(第四季)做题记录(WEB+MISC+CRYPTO)

WEB部分 【nani】 啥也没有,看源代码发现一句话/index.php?file=show.php。 访问show.php得到user.php,因为是file参数考虑是否有LFI。 ?file=php://filter/read=convert.base64-encode/resource=user.php得到user.php的源码: warn); } function __wakeup(…

NCTF2019/BUUCTF-childRSA

题目: from random import choice from Crypto.Util.number import isPrime, sieve_base as primes from flag import flag def getPrime(bits): while True: n = 2 while n.bit_length() < bits: n *= choice(primes) if isPrime(n + 1): return n + 1 e = …

BUUCTF/HDCTF2019 - bbbbbbrsa

题目下载 给了两个文件。 第一个enc: p = 177077389675257695042507998165006460849 n = 37421829509887796274897162249367329400988647145613325367337968063341372726061 c = ==gMzYDNzIjMxUTNyIzNzIjMyYTM4MDM0gTMwEjNzgTM2UTN4cjNwIjN2QzM5ADMwIDNyMTO4…

RSA-只有密文

怎么办,已经截获了密文和模数n!能够破解吗?请分解出RSA中的两个大素数q和p。提交格式是key{x}x为两个素数中较小的那个的MD5前8位。 题目下载 他让求的是p和q,200个密文里可能会有和n相同的质数q, 所以大概意思就是让密文分别和N求最大公约数 n = 135176…

2018百越杯Crypto-RSA

题目下载 第一步利用openssl把n和e求出来: 可以发现n不算太大,丢到大数分解网站里可以得到: p:184333227921154992916659782580114145999 q:336771668019607304680919844592337860739 到这里常规解就会报错了,参考了下大佬的wp: 同时看encrypt函数,可…

掘安杯-Crypto:RSA脚本一把梭 (模不互素)

N is 186743751083130949285851565811389413685700222221909454612844026732040180753540698271860858513098065923987216288453368405327795791973029849876615472454231807609580228985464965242492016795434211588421034964528619321831443433159251061543…

BUUCTF/HDCTF2019 together

题目下载 开局四个文件,分别来看一下嘛。 myflag1 R3Noy6r3WLItytAmb4FmHEygoilucEEZbO9ZYXx5JN03HNpBLDx7fXd2fl+UL5+11RCs/y0qlTGURWWDtG66eNLzGwNpAKiVj6I7RtUJl2Pcm3NvFeAFwI9UsVREyh7zIV6sI9ZP8l/2GVDorLAz5ULW+f0OINGhJmZm8FL/aDnlfTElhQ87LPicWpXYoMt…

BUUCTF/GUETCTF2019 BabyRSA

p+q : 0x1232fecb92adead91613e7d9ae5e36fe6bb765317d6ed38ad890b4073539a6231a6620584cea5730b5af83a3e80cf30141282c97be4400e33307573af6b25e2ea (p+1)(q+1) : 0x5248becef1d925d45705a7302700d6a0ffe5877fddf9451a9c1181c4d82365806085fd86fbaab08b6fc66a…

UNCTF复现赛-Crypto部分做题记录

【 不仅仅是RSA 】 通过这个网站得到C1和C2的值 C1:4314251881242803343641258350847424240197348270934376293792054938860756265727535163218661012756264314717591117355736219880127534927494986120542485721347351 C2:485162209351525800948941613977942…

SameMod

题目下载 n相同,很明显是共模攻击。 # -*- coding: utf-8 -*- from libnum import n2s,s2n from gmpy2 import invert # 欧几里得算法 def egcd(a, b): if a == 0: return (b, 0, 1) else: g, y, x = egcd(b % a, a) return (g, x - (b // a) * y, y) def ma…

RSAROLL

题目下载 给了n,e,其中n可以分解: p = 18443 q = 49891 依次进行解密即可。 import gmpy2 import libnum p = 18443 q = 49891 e = 19 s = (p- 1) * (q - 1) d = long(gmpy2.invert(e, s)) n = p *q b = '' a=[704796792,752211152,274704164,18414022,368…

2018网鼎杯第3场- hafuhafu (生成pem)

pk = (25572000680139535995611501720832880791477922165939342981900803052781801299380515116746468338767634903543966903733806796606602206278399959935132433794098659859300196212479681357625729637405673432324426686371817007872620401911782200407…

RSA-Rabin加密中的N可被分解

Rabin加密是RSA的衍生算法,e==2是Rabin加密典型 解密方法: 先通过其他方法获得p和q,然后上脚本例子hardRSA import gmpy2,libnum n=0xC2636AE5C3D8E43FFB97AB09028F1AAC6C0BF6CD3D70EBCA281BFFE97FBE30DD p=275127860351348928173285174381581152299 q=319…

RSA-低加密指数攻击&密文为base64编码的脚本

n = 2726203073819016290606853330921824831931203741685679481453245986613019667356183308473904817176947989380667149952264380341210827990722389551789796990625362602827028902864659689742964113891300156194755778484031101439997331209805689653990…

BUUCTF-Dangerous RSA (低加密指数攻击) WriteUp

题目如下: #n: 0x52d483c27cd806550fbe0e37a61af2e7cf5e0efb723dfc81174c918a27627779b21fa3c851e9e94188eaee3d5cd6f752406a43fbecb53e80836ff1e185d3ccd7782ea846c2e91a7b0808986666e0bdadbfb7bdd65670a589a4d2478e9adcafe97c6ee23614bcb2ecc23580f4d2e3cc…

bugku-rsa writeup

题目: N : 460657813884289609896372056585544172485318117026246263899744329237492701820627219556007788200590119136173895989001382151536006853823326382892363143604314518686388786002989248800814861248595075326277099645338694977097459168530898…

BUUCTF-RSA2

今天遇到了一道题,代码如下: N = 10199180977755325347027675139926474013115768232925267350179215450700615843443200914199536724196252570595004625340018888465826249653470643879151507188586089755273665689956691573129722581725063987364337631010…