【公钥密码】RSA 算法加解密 公钥密码体制 Python

【公钥密码】RSA 算法加解密 公钥密码体制 Python

目录

1.依赖的基本数学算法、编码流 *** 作

2.RSA 算法加解密、例题实现

---

1.依赖的基本数学算法、编码流 *** 作

# 本代码文件名: __util.py

import math

def gcd(x, y):
	""" 欧几里德算法 """
	while x > 0:
		x, y = y % x, x
	return y

def extendGcd(a, b):
	""" 扩展欧几里德算法 """
	if b == 0:
		return 1, 0
	else:
		x, y = extendGcd(b, a % b)
		x, y = y, x - (a//b) * y
		return x, y

def modInvElem(a:int, m):
	""" 求整数a关于1模m的乘法逆元 """
	if (gcd(a, m) !=1):
		return -1
	inva, _ = extendGcd(a, m)
	inva %= m
	return inva

def fastModExponent(a, m, n):
	""" 快速模指数算法
		@return: (a^m) mod n
	"""
	ans = 1
	a = a % n # 若a比n大则取模简化
	while m != 0:
		if m & 1: ans = (ans * a) % n
		m = m >> 1
		a = (a * a) % n
	return ans


def enCode(text):
	""" 53编码 by Herk
		@字符串:"abc" -> 编码流:"010203"
	"""
	pool = " abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
	code = [(2-len(str(pool.index(i))))*'0'+str(pool.index(i)) for i in text]

	return ''.join(code)

def deCode(code):
	""" 53解码 by Herk
		@编码流:"010203" -> 字符串:"abc"
	"""
	pool = " abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
	text = [pool[int(code[i:i+2])%53] for i in range(0, len(code), 2)]

	return ''.join(text)

def divideGroup(code, glen=10):
	""" 对编码进行分组补全
		@code: 待分组的编码流
		@glen: 分组长度, 最后一组不足则右端补0
		@return: 编码流的分组列表
	"""
	group = [code[i:i+glen] for i in range(0, len(code), glen)]
	group[-1] = group[-1] + (glen - len(group[-1])) * '0'
	return group

2.RSA 算法加解密、例题实现

# 本代码文件名: rsa.py

from Crypto.Util import number
from __util import *

class RSA():
    """ RSA 加解密 """
    
    def __init__(self, p=None, q=None, e=None):
        """ 初始化公私钥
            @获取p,q,e生成公私钥
            @若为空则随机生成公私钥
        """
        key = self.genParam(p, q, e)
        print("公钥为:", key[0])
        print("私钥为:", key[1])
    
    def genParam(self, p, q, e):
        """ 参数生成函数
            @生成公钥, 生成私钥
        """
        if (p==None or q==None):
            __p = number.getPrime(10)
            __q = number.getPrime(10)
            phi = (__p-1) * (__q-1)
            self.e = phi - 1  # e比phi小1, 两者必定互素
        else:
            self.e = e
            __p, __q = p, q
            phi = (__p-1) * (__q-1)
        self.n = __p * __q
        self.d = modInvElem(self.e, phi)
        return [self.e, self.n], [self.d, self.n]

    def cryptRSA(self, codes, flag):
        """ RSA算法
            @编码流A->编码流B
        """
        codeB = '' # 待返回的编码流B
        codelistA = divideGroup(codes, 10)  # 分组长度要小于 n
        if flag == 1: exponent = self.e     # RSA正向算法
        else: exponent = self.d             # RSA逆向算法
        for code in codelistA:
            leng = len(code)   # 传入编码流长度, return时补全不足
            code = int(code)
            code = fastModExponent(code, exponent, self.n)
            codeB += str(code).zfill(leng) #zfill:编码流左端补0
        return codeB

    def encrypt(self, messagcode):
        """ RSA加密算法
            @mscode: 明文编码流
            @return: 密文编码流
        """
        return self.cryptRSA(messagcode, 1)

    def decrypt(self, ciphercode):
        """ RSA解密算法
            @cipher: 密文编码流
            @return: 明文编码流
        """
        return self.cryptRSA(ciphercode, 0)


if __name__ == '__main__':

    mscode = enCode("please wait for me")
    cicode = "0763222127199153452800748825533895624854"

    crypto = RSA(71593, 77041, 1757316971)
    cilist = crypto.encrypt(mscode)
    mslist = crypto.decrypt(cicode)
    messag = deCode(mslist)

    print("密文编码流为:", cilist)
    print("明文编码流为:", mslist)
    print("解密后的明文:", messag)