文章分类 - CyberSpace Security / CTF / Crypto
摘要：密码题杂记 Shamir门限 跟着tcxjk师傅的博客学习了一下 题目: 公司使用Shamir门限密钥设计了一个秘密保存方案，将fag保存了起来，最终的设计效果如下密钥总共有9份，拿到任意5个密钥即可解出保存的flag.现在我们知道公共的密钥: p=0x3b9f64aeadae9545d899102
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摘要：目录Week1VigenèrebabyencodingSmall dAffinebabyaesWeek2滴啤不止一个pihalfcandecodeRotate Xorbroadcastpartial decryptWeek3Rabin's RSA小明的密码题knapsackbabyrandomez_
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摘要：R_r paillier同态加密 大致介绍: https://zhuanlan.zhihu.com/p/557034854 例题: from Crypto.Util.number import * import random from gmpy2 import * from secret impor
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posted @ 2023-11-21 16:14
N0zoM1z0
摘要：Danger RSA from Crypto.Util.number import * m = bytes_to_long(flag) def get_key(a, nbit): assert a >= 2 while True: X = getRandomInteger(nbit // a) s
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摘要：MiniMiniPack from gmpy2 import * from Crypto.Util.number import * import random from FLAG import flag def gen_key(size): s = 1000 key = [] for _ in ra
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摘要：Crpyto ezRSA from Crypto.Util.number import * from secret import secret, flag def encrypt(m): return pow(m, e, n) assert flag == b"dasctf{" + secret +
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摘要：chall.py import gmpy2 from Crypto.Util.number import * from flag import flag assert flag.startswith(b"flag{") assert flag.endswith(b"}") message=bytes
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摘要：跟着blog系统地学习一遍 ECB模式 ECB模式首先将明文分块 然后用相同的加密方式和密钥进行加密 因此相同的明文在不同次的加密后是一样的 注意: plaintext和key都必须是16的倍数 SWPU2020 ECB chall.py from Crypto.Cipher import AES
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摘要：Simple 综合考察了 维纳攻击得phi 利用私钥d相关攻击分解n 和扩展维纳攻击 然而你会惊喜的发现 chall.py from Crypto.Util.number import * from Crypto.Cipher import DES import gmpy2 from secret
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摘要：chall.py #Subset Sum Problem #https://imgs.xkcd.com/comics/np_complete.png import random choices = list(set([random.randint(100000,1000000000) for _ i
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posted @ 2023-11-10 16:00
N0zoM1z0
摘要：目录XOReasymath撤退3!!!easyrsa XOR n = 2081029853064313977972537933555768796028190509610710141158522091867265332387523434454034280165112366755381286645879
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摘要：1. chall.py from Crypto.Util.number import * from flag import flag import gmpy2 assert(len(flag)==38) flag = bytes_to_long(flag) p = getPrime(512) q =
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摘要：密码部分的题出的挺好的 很多题目都可以起到很好的锻炼学习作用 比赛时很多没做出来 赛后尽可能都复现一下 Week1 Take My Bag chall.py from Crypto.Util.number import * from secret import flag def encrypt(m)
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摘要：chall.py #python3 import sys sys.path.append("..") from Crypto.Util.number import * from random import * from sage.all import * from secret import fla
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摘要：chall.py from Crypto.Util.number import getPrime,long_to_bytes,bytes_to_long from Crypto.Cipher import AES import hashlib from random import randint d
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摘要：chall.py from secret import flag from Crypto.Util.number import * p = getPrime(1024) q = getPrime(1024) N = p*p*q d= inverse(N, (p-1)*(q-1)//GCD(p-1,
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posted @ 2023-10-29 12:16
N0zoM1z0
摘要：先记录着 有时间好好学学再理解 from sage.all import * p = 75206427479775622966537995406541077245842499523456803092204668034148875719001 a = 4039928064153768526323636
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posted @ 2023-10-29 12:14
N0zoM1z0
摘要：学习sage用法 学习p高位泄露 chall.py from Crypto.Util.number import getPrime, bytes_to_long from secret import flag p = getPrime(1024) q = getPrime(1024) n = p *
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posted @ 2023-10-22 17:24
N0zoM1z0
摘要：解方程组 sympy: from z3 import * from sympy import * from libnum import * from primefac import * n = 22307913740463468357754335486410675936913694858280706
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posted @ 2023-10-22 16:29
N0zoM1z0
摘要：chall.py from Crypto.Util.number import * from secret import flag p = getPrime(1024) q = getPrime(16) n = p*q m = bytes_to_long(flag) for i in range(1
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posted @ 2023-10-22 16:08
N0zoM1z0
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