2er0!=O的博客

[watevrCTF 2019]Swedish RSA 附件： polynomial_rsa.sage： flag = bytearray(raw_input()) flag = list(flag) length = len(flag) bits = 16 ## Prime for Finite Field. p = random_prime(2^bits-1, False, 2^(bits-1)) file_out = open("downloads/polynomial_rsa.txt", "w

[NPUCTF2020]共 模 攻 击 附件： hint.py： from gmpy2 import * from Crypto.Util.number import * from secret import hint m = bytes_to_long(hint) p = getPrime(256) c = pow(m, 256, p) print(p) p, q = getPrime(256), getPrime(256) n = p * q e1, e2 = getPrime(32), getP

[NPUCTF2020]认清形势，建立信心 附件task.py： from Crypto.Util.number import * from gmpy2 import * from secret import flag p = getPrime(25) e = # Hidden q = getPrime(25) n = p * q m = bytes_to_long(flag.strip(b"npuctf{").strip(b"}")) c = pow(m, e, n) print(c) print(p

[QCTF2018]Xman-RSA 附件： ciphertext： 1240198b148089290e375b999569f0d53c32d356b2e95f5acee070f016b3bef243d0b5e46d9ad7aa7dfe2f21bda920d0ac7ce7b1e48f22b2de410c6f391ce7c4347c65ffc9704ecb3068005e9f35cbbb7b27e0f7a18f4f42ae572d77aaa3ee189418d6a07bab7d93beaa365c9834

[INSHack2019]Yet Another RSA Challenge - Part 1 附件： description.md Buy an encrypted flag, get a (almost intact) prime factor for free ! You can find a harder version of this challenge in the Programming category. output.txt 71957974565330311902587309.

[INSHack2017]rsa16m 附件： description.md Challenge description: When you need really secure communications, you use RSA with a 4096 bit key. <br> I want really really really secure communications to transmit the nuclear launch codes (yeah IoT is ever

[De1CTF2019]babyrsa 附件 import binascii from data import e1,e2,p,q1p,q1q,hint,flag n = [2012961535249176549934011294318831718054876159786130084730582714151046561967053684463455824643923037165883692810306343287024570718035590719428486151090607126535240957

[MRCTF2020]Easy_RSA 附件： import sympy from gmpy2 import gcd, invert from random import randint from Crypto.Util.number import getPrime, isPrime, getRandomNBitInteger, bytes_to_long, long_to_bytes import base64 from zlib import * flag = b"MRCTF{XXXX}" base

[NPUCTF2020]EzRSA 附件： from gmpy2 import lcm , powmod , invert , gcd , mpz from Crypto.Util.number import getPrime from sympy import nextprime from random import randint p = getPrime(1024) q = getPrime(1024) n = p * q gift = lcm(p - 1 , q - 1) e = 54722 fla

[GWCTF 2019]BabyRSA 查看题目 secret N=636585149594574746909030160182690866222909256464847291783000651837227921337237899651287943597773270944384034858925295744880727101606841413640006527614873110651410155893776548737823152943797884729130149758279127430044739254