密码学课设4 RSA实现(python实现)

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