基于中国剩余定理的秘密共享方案

from random import randint
import sympy
import sys
with open('C:\\Users\\吴文之\\Desktop\\验收数据\\plaintext4.txt') as file_object:
    lines = file_object.readlines()
m = int(lines[0])
def fast(a,e,m): #快速计算模幂
    a = a % m
    r = 1
    while e != 0:
        if e&1:
            r= (r*a)%m
        e >>= 1 #右移一位
        a = (a * a)%m
    return r
def primitive(p, q):  #调用快速取模幂求原根
    while True:
        g = randint(2,p-2)
        if fast(g,2,p) != 1 and fast(g,q,p) != 1:
            return g
def e_gcd(a,b):
    if b == 0:
        return a,1,0
    g,x,y = e_gcd(b,a%b)
    return g,y,x-a//b*y
while True: #生成一个大素数p
    q = sympy.randprime(pow(10,149),pow(10,150)/2-1)  #生成范围内的随机数
    if sympy.isprime(q): #判断q是否是素数
        p = 2 * q + 1 #生成强素数
        if sympy.isprime(p) and len(str(p)) == 150:
            break
print("q = " + str(q))
print("p = " + str(p))
g = primitive(p, q) #生成模𝑝的一个原根g
print("g = "+str(g))
a = randint(1,11) #随机选取整a
print("公钥对： (p,g,g^a) = "+"\n"+str(p)+"\n"+str(g)+"\n"+str(pow(g,a)%p))
y = fast(g,a,p)
#加密过程
k = randint(2, p-2)
c1 = fast(g, k, p)
c2 = (m*fast(y,k,p))%p
print("c1 = " + str(c1))
print("c2 = " + str(c2))
#解密过程
v = fast(c1,a,p)
v_ = e_gcd(v,p)[1]
m_ = c2*v_%p
if(m == m_):
    print("相同")
else:
    print("不同")