func.py:
import sys sys.setrecursionlimit(100000) # 实现幂模函数 def power(a, b, c): a = a % c ans = 1 while b != 0: if b & 1: ans = (ans*a) % c b >>= 1 a = (a*a) % c return ans # 实现求最大公因数 def gcd(n, m): if n < m: n, m = m, n t1, t2, t3 = n, m, n % m while t3 != 0: t1 = t2 t2 = t3 t3 = t1 % t2 return t2 # 实现求逆元函数 def findModInverse(a, m): if gcd(a, m) != 1: return None u1, u2, u3 = 1, 0, a v1, v2, v3 = 0, 1, m while v3 != 0: q = u3 // v3 v1, v2, v3, u1, u2, u3 = (u1 - q * v1), (u2 - q * v2), (u3 - q * v3), v1, v2, v3 return u1 % m
rabin.py:
import random # 检测大整数是否是素数,如果是素数,就返回True,否则返回False # rabin算法的意思大家自己百度哈 def rabin_miller(num): s = num - 1 t = 0 while s % 2 == 0: s = s // 2 t += 1 for trials in range(5): a = random.randrange(2, num - 1) v = pow(a, s, num) if v != 1: i = 0 while v != (num - 1): if i == t - 1: return False else: i = i + 1 v = (v ** 2) % num return True def is_prime(num): # 排除0,1和负数 if num < 2: return False # 创建小素数的列表,可以大幅加快速度 # 如果是小素数,那么直接返回true small_primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997] if num in small_primes: return True # 如果大数是这些小素数的倍数,那么就是合数,返回false for prime in small_primes: if num % prime == 0: return False # 如果这样没有分辨出来,就一定是大整数,那么就调用rabin算法 return rabin_miller(num) # 实现取大素数函数 def get_prime(key_size=1024): while True: num = random.randrange(2**(key_size-1), 2**key_size) if is_prime(num): return num
RSA.py
import random from rabin import * from func import * # 实现自动求d,e函数 def get_de(): while True: d = random.randint(1, N) if gcd(d, N) == 1: e = findModInverse(d, N) return d, e # RSA算法 print("正在生成大素数...") p, q = get_prime(), get_prime() print("正在生成大小N...") n, N = p * q, (p-1)*(q-1) while True: print("正在生成d,e") d1, e1 = get_de() if d1 > 0 and e1 > 0: break # 这里输入明文,暂时只支持数字哈 m = input("请输入要加密的信息(数字)>>>") c = power(int(m), e1, n) print("密文为: %s" % c) with open('key.txt', 'w') as f: f.write("c: " + str(c) + "\nd1:" + str(d1) + "\nn:" + str(n)) print("密文和d1,n已经保存在key.txt中") m = power(c, d1, n) print("明文为: %s" % m) 运行rsa.py即可