SUBGROUPX Repo: https://github.com/n-WN/SubgroupX Griffin 这题的关键是 Phase1 恢复的 $x$ 存在仿射不确定性，导致 Phase2 求根得到的是 $x = a m + b$ 而不是 $m$，需要对 $(a,b)$ 做 affine 枚举。 / 结论 flag: alictf{**} 关键坑点: Phase1 恢复的 $x$ 存在仿射不确定性，导致 Phase2 求根得到的是 $x = a m + b$ 而非 $m$，必须做 affine 枚举，最终命中 $(a,b)=(-1,256)$ Full solve code (gist) 复现参数 以下数值来自一次成功复现的检查点和附件内容:

SUBGROUPX Repo: https://github.com/n-WN/SubgroupX 解了几道分数较高的题目，部署后继续修复公式渲染（theme/模板问题）。 / AstralCat 服务端每轮随机 99 字节 message，用 Luo Shu 做 CBC。关键点是同一个对象连续调用两次 Encrypt()，第二次加密的输入其实是第一次的密文。

SUBGROUPX Repo: https://github.com/n-WN/SubgroupX 整理版汇总：按仓库内文档与脚本串主线，每题只保留关键思路和 exp 核心片段（截图已脱敏）。 / 整理说明 本文按仓库内文档与脚本整理，每题只保留解题主线与 exp 核心片段。 文中用 HOST:PORT 代替真实地址，flag 统一写为 LilacCTF{...}，不包含外链 URL 与真实 IP。

8abyRSA 题面 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 from Crypto.Util.number import getPrime, bytes_to_long from LCGRandom import lcg flag = b"DASCTF{xxxx}" m = bytes_to_long(flag) q = getPrime(512) p = getPrime(512) Q = pow(q, 2, p) # Q = q^2 mod p n = p*q e = getPrime(512) c = pow(m, e, n) print('c = ' + str(c)) mul = p n = e c = Q lcg1 = lcg(2023, mul, n, c) print(lcg1.generate(6)) # Data # c = 14968774972802568447907734980942381885170753466134942777553237930032293557412276601708303806296932877792965437305452274767013479132983066980557420348521340748653518789837988349171582558831336243036976332252071289409948879158346086243451785505819420501498240344839999096449458990633841326838003742773196414924 # [1829291767649461103161355195365566822501625317566021355553611375584303624765389047237072963403651896280059309840080549881138083110457632256619677785411889, 8614784647131959905489143385414473366220060118109917221659356152996027731619256154868253763867569947876938642677951734943013867186530789820165520197525548, 7845112879564311528346861967994968539141359532994486000551038947867807602664345244223622318955917969589790931596150225393028516278795565957075910272882168, 464696663556289552651401659156226806419638205935729756668755007052945250273596319671267391086858010496056118762904740103810918505838370732992982394379753, 6368769952056267497431070536699202267447921981615807369090285698513090854334005214066202915932322373348204466447234330868540529876844912401907159350901200, 4401214958628797991805307100256403706658940216552816808894942142670769563264366171079551655390635613048161714797548328158857275108940238519111063046341755] c = 14968774972802568447907734980942381885170753466134942777553237930032293557412276601708303806296932877792965437305452274767013479132983066980557420348521340748653518789837988349171582558831336243036976332252071289409948879158346086243451785505819420501498240344839999096449458990633841326838003742773196414924 lcg_gen_6 = [1829291767649461103161355195365566822501625317566021355553611375584303624765389047237072963403651896280059309840080549881138083110457632256619677785411889, 8614784647131959905489143385414473366220060118109917221659356152996027731619256154868253763867569947876938642677951734943013867186530789820165520197525548, 7845112879564311528346861967994968539141359532994486000551038947867807602664345244223622318955917969589790931596150225393028516278795565957075910272882168, 464696663556289552651401659156226806419638205935729756668755007052945250273596319671267391086858010496056118762904740103810918505838370732992982394379753, 6368769952056267497431070536699202267447921981615807369090285698513090854334005214066202915932322373348204466447234330868540529876844912401907159350901200, 4401214958628797991805307100256403706658940216552816808894942142670769563264366171079551655390635613048161714797548328158857275108940238519111063046341755] seed = 2023 Exploit SageMath’s Code

Hill Cipher(不知道题目叫什么名字) 下载题目附件后打开看到 / 拖到 010 Editor 提示报错，注意到文件末尾存在字符串，且为三的倍数 / 同时结合图片名为 hill.png 以及图片中数字的排列，推导可能是 Hill Cipher

Crypto easyLCG 题目代码 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 from Crypto.Util.number import * from random import randint FLAG = ? ROUND = randint(200, 300) class LCG: lcg_a = getPrime(498) lcg_b = getPrime(498) lcg_n = getPrime(512) def __init__(self, lcg_seed): self.state = lcg_seed def next(self): self.state = (self.state * self.lcg_a + self.lcg_b) % self.lcg_n return self.state seed = bytes_to_long(FLAG) print(seed.bit_length()) lcg = LCG(seed) for _ in range(ROUND-6): lcg.next() print([lcg.next() for _ in range(6)]) # 351 # [2485483242304449696161151168576736302336140244327446722677621064961717587947642623655706309294371876714311165214262071924932913930440142186509325733360885, 6174672247406972581092780254648964828729162316422924118545162215261641830776919624579500836375440017861960302702691972683448546062254126073514097080361044, 4584872703321313263026316988830140564935997972340636369234263637913498924218112980629201945528119162637762726584003527994733552009906632734086040880127542, 4829175497283310360340169484343201154685159906303099960915060755507745973302279262836696523131741537828200567942990339422885197644614494869008027862313162, 6669041483112643196441450289748743294802963984583343809912658359929976814854869038886397237458253328867571805574485994275429182399171275201561798677581761, 2732859498560958306654933055106793656103744294844703560692860713169132266734427400301681246251133210447387861776419850678194913478737337289544516324708390] 思路 比较基础的参数恢复后逆推 seed

初赛密码题只有一道，还沾了点猜谜… / 绝密文件-代号P 题目代码 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 import cv2 import numpy as np from Crypto.Util.number import * from secret import p, q, c def arnold(img, shuffle_times, a, b): r, c, d = img.shape # d: dimension p = np.zeros(img.shape, np.uint8) # new image for s in range(shuffle_times): # shuffle times for i in range(r): # height for j in range(c): # width x = (i + b * j) % r # new position y = ((a * i) + (a * b + 1) * j) % c # new position p[x, y, :] = img[i, j, :] img = np.copy(p) return p Img_path = "flag_test.png" Img = cv2.imread(Img_path) assert isPrime(p) and isPrime(q) and p**2 + \ q**2 == c and c == 179093209181929149953346613617854206675976823277412565868079070299728290913658 Img_arnold = arnold(Img, c, p, q) cv2.imwrite("flag_enc.png", Img_arnold) 思路 分析得知 arnold()管理图片像素置换，而参数 $a,b$ 未知

部分位已知 通过分析题目附件中的代码发现：素数总是隐藏了 50%bit，反过来就是 50%bit 已知 / DFS + 枚举状态 + 通过 $N$ 的低部分位来 Check 枚举的 $p,q$ 低位是否准确即可解出 $p,q$ / 随后即为常规 RSA

可通过优化代码的方式打表，但这里只给出以古典方式分析题目 确认 c 与 secrets 的对应关系（使得可以通过 c 恢复 secrets） / 确认 Time_Machine() 的单调性（使得可以根据大小关系推断） / 词频分析（找出空格、e的对应值） / 特判（Case） / 枚举状态、NLP 处理小技巧、使用等宽字体、使用辅助函数恢复（磨刀不误砍柴工）、生成 ASCII 码表、生成多种排序方式、生成 Key Value Pair，且使用 Tuple 来实现更多展示信息，方便分析

RSA是什么 特征：给出 p、q，考察 RSA 解密原理 SageMath’s Code / 1 2 3 4 5 6 7 8 9 10 from Crypto.Util.number import long_to_bytes p = q = phi = (p - 1) * (q - 1) n = p * q e = d = inverse_mod(e, phi) m = power_mod(c, d, n) print(long_to_bytes(int(m))) # 强制类型转换 RSA $e=3$ 特征：$e$ (公钥指数) 很小 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 from gmpy2 import iroot import itertools n = e = 3 c = """ m = iroot(c, e) if m[1]: print(long_to_bytes(m[0])) """ for i in itertools.count(0): m = iroot(c + i * n, e) if m[1]: print(long_to_bytes(m[0])) break RSA 因数分解 特征：$N$ 的因子很小 这里为 ($80\text{bit} * 80\text{bit}$)