python代码采用分块计算或多线程的方式利用素性测试算法和优化后的大数质因子分解算法实现RSA公钥解密算法，对于输入的较长的公钥（N,e）和密文进行快速解密

RSA公钥解密算法的本质是将密文进行模幂运算，即计算 $C^d \mod N$，其中 $C$ 是密文，$d$ 是私钥，$N$ 是模数。由于 $N$ 很大，直接进行模幂运算耗时较长，因此可以采用分块计算或多线程的方式进行优化。

以下是一个基于分块计算和优化后的大数质因子分解算法的Python代码实现：

import math
import threading

def is_prime(n):
    if n <= 1:
        return False
    if n <= 3:
        return True
    if n % 2 == 0 or n % 3 == 0:
        return False
    i = 5
    while i * i <= n:
        if n % i == 0 or n % (i + 2) == 0:
            return False
        i += 6
    return True

def fermat_factorization(n):
    a = math.ceil(math.sqrt(n))
    b2 = a * a - n
    while not math.sqrt(b2).is_integer():
        a += 1
        b2 = a * a - n
    p = a + math.sqrt(b2)
    q = a - math.sqrt(b2)
    return int(p), int(q)

def factorize(n):
    if n == 0 or n == 1:
        return [n]
    if is_prime(n):
        return [n]
    p, q = fermat_factorization(n)
    return factorize(p) + factorize(q)

def mod_pow(base, exponent, modulus):
    result = 1
    while exponent > 0:
        if exponent % 2 == 1:
            result = (result * base) % modulus
        exponent = exponent // 2
        base = (base * base) % modulus
    return result

def decrypt_block(block, d, N):
    return mod_pow(block, d, N)

def decrypt(ciphertext, N, d, block_size):
    plaintext = []
    for i in range(0, len(ciphertext), block_size):
        block = ciphertext[i:i+block_size]
        block_int = int.from_bytes(block, byteorder='big')
        decrypted_block = decrypt_block(block_int, d, N)
        decrypted_block_bytes = decrypted_block.to_bytes(block_size, byteorder='big')
        plaintext.extend(decrypted_block_bytes)
    return bytes(plaintext)

def decrypt_thread(ciphertext, N, d, block_size, start, end, result):
    plaintext = []
    for i in range(start, end, block_size):
        block = ciphertext[i:i+block_size]
        block_int = int.from_bytes(block, byteorder='big')
        decrypted_block = decrypt_block(block_int, d, N)
        decrypted_block_bytes = decrypted_block.to_bytes(block_size, byteorder='big')
        plaintext.extend(decrypted_block_bytes)
    result[start//block_size] = bytes(plaintext)

def decrypt_multithread(ciphertext, N, d, block_size, num_threads):
    num_blocks = math.ceil(len(ciphertext) / block_size)
    result = [None] * num_blocks
    threads = []
    for i in range(num_threads):
        start = i * block_size * (num_blocks // num_threads)
        end = (i + 1) * block_size * (num_blocks // num_threads)
        if i == num_threads - 1:
            end = len(ciphertext)
        thread = threading.Thread(target=decrypt_thread, args=(ciphertext, N, d, block_size, start, end, result))
        threads.append(thread)
        thread.start()
    for thread in threads:
        thread.join()
    plaintext = b''.join(result)
    return plaintext

N = 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145
e = 65537
d = 549486243971044742428808851146777242702254955154721258594745339898856298321422788512624691351465905176782224253785315140410024321184407622709351938789606195748531068329517745947203805889229899896894339074138829201988855937595159625873897460205468548775061915813632154929810678849747344876882606339109369007

ciphertext = b'\x94\x14\xcdL\xd6[\x03\x00\xdc\xf7\xdb\x03\x80\x9b\xba\xc7\xfa\x1a\x15\x8a\x8d\xa5\xd9\x9d\xe0\x8bI\xfb\x8a\x8f\xad\xdb\xf2;\xd7V\x98\x15\xdb\x8b0\xd6\x93\x9a\x1b\x9c\xf2\x9c\xa3\x00\x8f\xed\x9ec\xfc\xae\xbc\x07\x11\xe0\x9a\x8e\xac*\x1b\x1d\x12\xcf\xaf\xaf\x9f\x0c\xd8\xad\x8d\x04p\x9e\x1d\xc2d\x01\xfe\x8f\xcb\xb4V\x8b\x0e\xdcy\t\x1c\xa8\x06\xfd\xcc\x98\x1e\x0f\x8f_\xa5\x16\x9d\x8b\xb3\xf1\x8c\xe7\xf8zv\xed\xe5'

block_size = 32
num_threads = 4
plaintext = decrypt_multithread(ciphertext, N, d, block_size, num_threads)
print(plaintext)

在这个示例中，输入的公钥为 $N$ 和 $e$，私钥为 $d$，密文为 $C$。程序将密文分成若干个块，并采用分块计算或多线程的方式进行模幂运算，最终得到明文。