摘要:
很巧妙的一道题 题目源码: import random from secret import flag ror = lambda x, l, b: (x >> l) | ((x & ((1<<l)-1)) << (b-l)) N = 1 for base in [2, 3, 7]: N *= pow 阅读全文
摘要:
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_l 阅读全文
摘要:
题目给的CBC: #!/usr/bin/python2.7 # -*- coding: utf-8 -*- from Crypto.Cipher import AES from Crypto.Random import random from Crypto.Util.number import lo 阅读全文