Hack Your Hearthttps://h4cker.zip/tags/rsa/临渊羡鱼，不如退而结网Hugo 0.168.0enSun, 17 Aug 2025 16:17:12 +08002025宁波市第八届网络安全大赛初赛Writeuphttps://h4cker.zip/322f5b/Sun, 17 Aug 2025 16:17:12 +0800https://h4cker.zip/322f5b/今天上午初赛结束了，感谢两个队友带，dr0n ↗ 和Rot-will ↗

槽点有点多，比如几个题AI秒出，我觉得以后出题都要反AI了，可能就留一个签到题AI能做，如果反AI，可能我真啥都不会了

AI这东西做逆向和密码非常厉害，逆向读伪代码和分析加解密，而这正是LLMs擅长的理解能力。做Web和Misc不太行，因为这两个用到的工具和环境太多。

两个WEB都是平平无奇登陆框，都做不出来，给的hint也做不出来，但是闲鱼一把梭。。。

两个杂项也古怪，队友做麻了，什么二维码小点点、博多密码，rgb转换之类

还有，我们做完4题已经第三名了，其他实在做不出来，就算其他队伍跟我们做一样的题，我们有前三血，那也应该排前面

但是最后三十分钟，我们连掉三十多名，我刷新一下掉一名，他们全都爆发了，不过应该能进决赛

部分附件在：https://github.com/TajangSec/Ningbo-8th-CSC-Pre

PWN#

entity_cache#

存在uaf

REVERSE#

SEA_1#

我不太会逆向，AI告诉我这是AES_ECB

其他选手说，贴在claude desktop或者vscode 的mcp里面，AI秒出，自己根本不用动

伪代码：

AES加密，密钥是58453eec4d16ae234a10b597dfe1f6a6，密文是unk_458014，dump出来是29708f1980cce40f46abac148d488ca83716fe1d397202797b1999166265623e8f761285cb28e256b381167761e41094

cyberchef解开：DASCTF{75aab2560274ae21aa4554b993e658d1}

flower_world#

CRYPTO#

Three-prime RSA#

https://chatgpt.com/share/68a14d43-ab10-800b-bbee-7e4511dfaaea ↗

AI太猛了

从 r_cubed 恢复 r

r = int(round(r_cubed ** (1/3)))

从 D 中恢复 (p+q+r) 和 random_num 题目中 random_num 是一个 28 位质数，p+q+r 大约 512~514 位。因为 (p+q+r)*random_num < n，所以 D 没有取模的影响：

1random_num = 254766409  # 可通过简单分解D得到
2s = D // random_num     # 得到 s = p + q + r
python

求p+q

p_plus_q = s - r

解一元二次方程得到 p 和 q 方程：x^2 - (p+q)x + p*q = 0

1from sympy import symbols, solve
2x = symbols('x')
3sols = solve(x**2 - p_plus_q*x + pq, x)
4p, q = int(sols[0]), int(sols[1])
py

计算 d 并解密

1from Crypto.Util.number import long_to_bytes, inverse
2e = 65537
3d = inverse(e, (p-1)*(q-1)*(r-1))
4m = pow(c, d, n)
5flag = long_to_bytes(m)
6print(flag)
py

DASCTF{5521a971-9bed-11ef-bfda-14ac6024b6a8}

]]>UAFAESRSACRYPTOHACK学习https://h4cker.zip/9c5133/Wed, 04 Jun 2025 19:52:45 +0800https://h4cker.zip/9c5133/练习用，有空再更新。Introduction to CryptoHack#

Finding Flags#

crypto{y0ur_f1rst_fl4g}

Great Snakes#

crypto{z3n_0f_pyth0n}

ASCII#

1python
2Python 3.10.7 (tags/v3.10.7:6cc6b13, Sep  5 2022, 14:08:36) [MSC v.1933 64 bit (AMD64)] on win32
3Type "help", "copyright", "credits" or "license" for more information.
4>>> xl=[99, 114, 121, 112, 116, 111, 123, 65, 83, 67, 73, 73, 95, 112, 114, 49, 110, 116, 52, 98, 108, 51, 125]
5>>> print("".join([chr(i) for i in xl]))
6crypto{ASCII_pr1nt4bl3}
7>>> print("".join(map(chr,xl)))
8crypto{ASCII_pr1nt4bl3}
cmd

Hex#

1>>> print(bytes.fromhex("63727970746f7b596f755f77696c6c5f62655f776f726b696e675f776974685f6865785f737472696e67735f615f6c6f747d"))
2b'crypto{You_will_be_working_with_hex_strings_a_lot}'
bash

Base64#

1Python 3.10.7 (tags/v3.10.7:6cc6b13, Sep  5 2022, 14:08:36) [MSC v.1933 64 bit (AMD64)] on win32
2Type "help", "copyright", "credits" or "license" for more information.
3>>> import base64
4>>> bt=bytes.fromhex("72bca9b68fc16ac7beeb8f849dca1d8a783e8acf9679bf9269f7bf")
5>>> print(bt)
6b'r\xbc\xa9\xb6\x8f\xc1j\xc7\xbe\xeb\x8f\x84\x9d\xca\x1d\x8ax>\x8a\xcf\x96y\xbf\x92i\xf7\xbf'
7>>> print(base64.b64encode(bt))
8b'crypto/Base+64+Encoding+is+Web+Safe/'
bash

Bytes and Big Integers#

1>>> from Crypto.Util.number import *
2>>> print(long_to_bytes(11515195063862318899931685488813747395775516287289682636499965282714637259206269))
3b'crypto{3nc0d1n6_4ll_7h3_w4y_d0wn}'
bash

XOR Starter#

 1python
 2Python 3.10.7 (tags/v3.10.7:6cc6b13, Sep  5 2022, 14:08:36) [MSC v.1933 64 bit (AMD64)] on win32
 3Type "help", "copyright", "credits" or "license" for more information.
 4>>> st = "label"
 5>>> print("".join(chr(ord(i)^13) for i in st))
 6aloha
 7>>> def xor(a,b):
 8...     return a^b
 9...
10>>> print("".join(chr(xor(ord(i),13)) for i in st))
11aloha
bash

XOR Properties#

1k1_ord = [o for o in bytes.fromhex("a6c8b6733c9b22de7bc0253266a3867df55acde8635e19c73313")]
2k23_ord = [o for o in bytes.fromhex("c1545756687e7573db23aa1c3452a098b71a7fbf0fddddde5fc1")]
3flag_123_ord = [o for o in bytes.fromhex("04ee9855208a2cd59091d04767ae47963170d1660df7f56f5faf")]
4
5k123_ord=[a^b for a,b in zip(k1_ord,k23_ord)]
6flag_ord=[a^b for a,b in zip(flag_123_ord,k123_ord)]
7
8print(bytes(flag_ord))
python

Favourite byte#

1st=bytes.fromhex('73626960647f6b206821204f21254f7d694f7624662065622127234f726927756d')
2for num in range(256):
3    string="".join((chr(num ^ i) for i in st))
4    if string.startswith("crypto"):
5        print(string)
python

You either know, XOR you don’t#

flag格式形如crypto{FLAG}

与密钥异或后得到0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104

那么这个字符串的前七位字节和最后一个字节与crypto{、}异或就能得到密钥的前七位。最后一位可能不准，因为密钥短的话就循环异或，不一定最后一位就和最后一个字节异或

先计算出来：

1enc_flag = '0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104'
2partial_flag = 'crypto{'
3
4encflag_ord=[o for o in bytes.fromhex(enc_flag)]
5par_ord=[o for o in partial_flag.encode()]
6key_start="".join([chr(a^b) for (a,b) in zip(encflag_ord,par_ord)])
7key_end=chr(encflag_ord[-1] ^ ord('}'))
8print(key_start)
9print(key_end)
bash

输出myXORke和y，拼在一起有意义：myXORkey

所以认为它是密钥

下面就计算他和异或后的字符串就行

长度不一致，一般是循环使用短序列

这里默认使用pwn库的xor函数

1import pwn
2key="myXORkey"
3enc="0e0b213f26041e480b26217f27342e175d0e070a3c5b103e2526217f27342e175d0e077e263451150104"
4flag=pwn.xor(key.encode(),bytes.fromhex(enc)).decode()
5print(flag)
python

crypto{1f_y0u_Kn0w_En0uGH_y0u_Kn0w_1t_4ll}

Modular Arithmetic#

Greatest Common Divisor#

我自己写的gcd脚本：

 1def gcd(a,b):
 2    x=max(a,b)
 3    y=min(a,b)
 4    while x%y!=0:
 5        a=y
 6        y=(x%y)
 7        x=a
 8    print(y)
 9
10gcd(52920,66528)
python

DeepSeek优化的：

1def gcd(a, b):
2    while b != 0:
3        a, b = b, a % b
4    return a
5
6print(gcd(52920, 66528))  # 输出结果：1512
python

它好聪明

Extended GCD#

原理知道，但是写不出来脚本

GingerCarrot ↗ 的脚本，不是很好理解这个脚本，可能我不明白递归，代码能力退化了

 1
 2def egcd(a, b):
 3    if a == 0:
 4        return b, 0, 1
 5    else:
 6        gcd, x, y = egcd(b % a, a)
 7        return gcd, y - (b // a) * x, x
 8
 9
10print(egcd(26513, 32321))
python

Modular Square Root#

1a = 8479994658316772151941616510097127087554541274812435112009425778595495359700244470400642403747058566807127814165396640215844192327900454116257979487432016769329970767046735091249898678088061634796559556704959846424131820416048436501387617211770124292793308079214153179977624440438616958575058361193975686620046439877308339989295604537867493683872778843921771307305602776398786978353866231661453376056771972069776398999013769588936194859344941268223184197231368887060609212875507518936172060702209557124430477137421847130682601666968691651447236917018634902407704797328509461854842432015009878011354022108661461024768
2p = 30531851861994333252675935111487950694414332763909083514133769861350960895076504687261369815735742549428789138300843082086550059082835141454526618160634109969195486322015775943030060449557090064811940139431735209185996454739163555910726493597222646855506445602953689527405362207926990442391705014604777038685880527537489845359101552442292804398472642356609304810680731556542002301547846635101455995732584071355903010856718680732337369128498655255277003643669031694516851390505923416710601212618443109844041514942401969629158975457079026906304328749039997262960301209158175920051890620947063936347307238412281568760161
3Mod(a,p).sqrt()
python

Chinese Remainder Theorem#

我是手算的

算出了6482，但解是

kM+∑i=1naitiMi ; k∈Z kM + \sum_{i=1}^{n}a_{i}t_{i}M_{i}\ ;\ k\in \mathbb{Z}

一般都是要求最小的解，所以我这边模M即可，得到872

wp有代码版本，推荐几个：

1from Crypto.Util.number import *
2"""
3x ≡ 2 mod 5
4x ≡ 3 mod 11
5x ≡ 5 mod 17
6"""
7print((2*11*17*inverse(11*17,5) + 3*5*17*inverse(5*17,11) + 5*5*11*inverse(5*11,17)) % 935)
python

这个不好理解的话，看下面这个

 1from Crypto.Util.number import inverse
 2
 3M = 5 * 11 * 17
 4M1 = M // 5
 5M2 = M // 11
 6M3 = M // 17
 7
 8N1 = inverse(M1, 5)
 9N2 = inverse(M2, 11)
10N3 = inverse(M3, 17)
11
12ans = pow(2 * M1 * N1 + 3 * M2 * N2 + 5 * M3 * N3, 1, M)
13print(ans)
python
1from sympy.ntheory.modular import crt
2 
3r = [2, 3, 5]
4m = [5, 11, 17]
5 
6print(crt(m, r)[0])
python

Adrien’s Signs#

正经CTF题目了

代码注释如下：

]]>密码学RSA