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命运如同手中的掌纹,无论多曲折,终掌握在自己手中。

RSA练习

0x00 RSA1 1 2 3 4 5 p = 8637633767257008567099653486541091171320491509433615447539162437911244175885667806398411790524083553445158113502227745206205327690939504032994699902053229 q = 12640674973996472769176047937170883420927050821480010581593137135372473880595613737337630629752577346147039284030082593490776630572584959954205336880228469 dp = 6500795702216834621109042351193261530650043841056252930930949663358625016881832840728066026150264693076109354874099841380454881716097778307268116910582929 dq = 783472263673553449019532580386470672380574033551303889137911760438881683674556098098256795673512201963002175438762767516968043599582527539160811120550041 c = 24722305403887382073567316467649080662631552905960229399079107995602154418176056335800638887527614164073530437657085079676157350205351945222989351316076486573599576041978339872265925062764318536089007310270278526159678937431903862892400747915525118983959970607934142974736675784325993445942031372107342103852 这道题泄露了dp和dq,分析参考自https://blog.csdn.net/xiao_han_

SPbCTF 2021 WriteUp

0x00 前言 国外的比赛题型很新颖,换句话说,也挺花里胡哨的,不过太菜了,只解出了easy的题目,555,摸鱼打一打。 0x01 CatStep 题目 1 2 3 Greeting human! We want to play a game

打靶-CloudAV

0x00 描述 云反病毒扫描器!是一种基于云的防病毒扫描服务。 目前,它处于测试模式。您被要求测试设置并查找漏洞并升级权限。 难度:容易 涉及的任务: 端口扫

打靶-SocialNetwork

0x00 准备工作 中等难度的靶机,Description如下: 1 Leave a message is a new anonymous social networking site where users can post messages for each other. They've assigned you to test their set up. They do utilize docker containers. You can conduct attacks against those too. Try to see if

打靶-MoriartyCorp

0x00 准备工作 查看靶机的描述信息,这里说明了flag的格式等。 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Description Hello Agent. You're here on a special mission. A mission to take down one of the biggest weapons suppliers which is Moriarty Corp.

近期一些CTF的writeup

0x00 签到 from Crypto.Util.number import * import random flag=b'flag{******************}' n = 2 ** 256 flaglong=bytes_to_long(flag) m = random.randint(2, n-1) | 1 c = pow(m, flaglong, n) print('m = ' + str(m)) print('c = ' + str(c)) # m = 73964803637492582853353338913523546944627084372081477892312545091623069227301 # c = 21572244511100216966799370397791432119463715616349800194229377843045443048821 这是一个求解离散对数的问题——经过查询在sage下

长城杯writeup

0x00 签到 题目给出了一串16进制的字符串 5a6d78685a3374585a57786a6232316c5833527658324e6f5957356e5932686c626d64695a544639 转ascii码,得到ZmxhZ3tXZWxjb21lX3RvX2NoYW5nY2hlbmdiZTF9,