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Easy Inverse 1234567891011121314151617from Crypto.Util.number import * p = getPrime(1024)bits = 100from secret import flagm = bytes_to_long(flag)hints = [pow(m , -1 , p) , pow(m+1 , -2 , p)]hints_leak = [(i>>bits)<<bits for i in hints]print(f"hints_leak = {hints_leak}")print(f"p = {p}") 12hints_leak =...

Full Weak Engineer CTF 2025 baby-crypto 1sjrpgs{ebg13rq_zrffntr!} 直接凯撒解密即可 1fwectf{rot13ed_message!} base🚀 1234567891011121314151617181920212223242526272829303132333435\#!/usr/bin/env python🚀with open('emoji.txt', 'r', encoding='utf-8') as f: emoji = list(f.read().strip())table = {i: ch for i, ch in enumerate(emoji)}def encode(data): bits = ''.join(f'{b:08b}' for b in data) pad = (-len(bits))...

one hint 12345678910111213141516171819202122232425262728293031\#!/usr/bin/env python3def vigenere(pt,key): res = "" for i in range(len(pt)): if pt[i] in "LITCTF{_}": res += pt[i] continue res += chr((ord(pt[i])+ord(key[i%len(key)])-2*ord('a'))%26+ord('a')) return resdef power(pt,n): a = pt for i in range(n): a = vigenere(a, vigenere(a, a)) return aflag = "LITCTF{redacted}"print("hint 1:...

EzRSA 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859import sysimport randomfrom sympy import nextprimeimport binasciiFLAG = b"flag{xxxxxxx}"BIT_SIZE = 512DIFF = 1000def generate_close_primes(bit_size, diff): start = random.getrandbits(bit_size) | 1 p = nextprime(start) q = nextprime(p + diff) return p, qdef flag_to_int(flag_bytes): return int.from_bytes(flag_bytes, byteorder='big')def main(): ...

Nepnep2025 Lattice Bros 1234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253#已知α的极小多项式为三次多项式f(x),即f(α)=0,且α≈54236.606188881754809671280151541781895183337725393#上述极小多项式的常数项为a0from secret import a0,alphaimport gmpy2from Crypto.Util.number import long_to_bytesimport randomfrom math import sqrt,log2d=981020902672546902438782010902608140583199504862558032616415p = d - a0k=sqrt(log2(p))+log2(log2(p))B = 2**30assert B < p/2**km = 30assert m >...

这里就对mix这题进行记录 mix 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475from Crypto.Util.number import *from secret import flagimport randomp = getStrongPrime(2048)q = getStrongPrime(2048)s = getStrongPrime(2048)r = int(裴波那契数列[2022])mask = b'******'enc1 = b'******'flag1 = b'CMCTF{' +enc1 + mask + b'}'e = 0o10001m = bytes_to_long(enc1)n = p * q * rc =...