python中Miller Rabin检验中的素数计数问题

import random

def is_prime(n, k=5):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False

    # 进行Miller-Rabin检验
    def miller_rabin(n, d):
        a = random.randint(2, n - 2)
        x = pow(a, d, n)
        if x == 1 or x == n - 1:
            return True
        while d != n - 1:
            x = pow(x, 2, n)
            d *= 2
            if x == 1:
                return False
            if x == n - 1:
                return True
        return False

    # 将n-1表示为(2^r)*d的形式
    d = n - 1
    while d % 2 == 0:
        d //= 2

    # 进行k次Miller-Rabin检验
    for _ in range(k):
        if not miller_rabin(n, d):
            return False
    return True

def prime_count(start, end):
    count = 0
    for num in range(start, end + 1):
        if is_prime(num):
            count += 1
    return count

start = 1
end = 100
count = prime_count(start, end)
print(f"There are {count} prime numbers between {start} and {end}.")