银行有n个柜员负责为顾客服务，顾客进入银行先取一个号码，然后等着叫号。当某个柜员空闲下来，就叫下一个号。 编程实现该问题，用P、V操作实现柜员和顾客的同步。

好的，这是一个经典的同步问题，可以使用信号量机制来实现。我们可以定义一个信号量数组，其中每个元素代表一个柜员，初值为1表示该柜员空闲，为0表示该柜员正在为顾客服务。同时还需要定义一个顾客号码的计数器，初值为0，表示还没有顾客。具体实现如下所示：

# 导入信号量模块
import threading

# 定义柜员数量
n = 5

# 定义信号量数组，初值为1
semaphores = [threading.Semaphore(1) for i in range(n)]

# 定义顾客号码计数器
counter = 0

# 定义顾客类
class Customer(threading.Thread):
    def __init__(self, num):
        threading.Thread.__init__(self)
        self.num = num
    
    def run(self):
        global counter
        global semaphores
        # 取一个号码
        my_num = 0
        semaphores[my_num].acquire()
        counter += 1
        my_num = counter
        semaphores[my_num % n].release()
        # 等待叫号
        semaphores[my_num % n].acquire()
        print("Customer", self.num, "is being served by Clerk", my_num % n)
        semaphores[my_num % n].release()

# 创建100个顾客线程
for i in range(100):
    c = Customer(i+1)
    c.start()

# 等待所有顾客线程结束
for c in threading.enumerate():
    if c != threading.current_thread():
        c.join()

在上面的代码中，我们定义了一个`Customer`类，用于表示顾客。在`run`方法中，每个顾客首先取一个号码，然后等待叫号。当某个柜员空闲下来时，顾客就被叫号，然后开始接受服务。在输出语句中，我们使用了`num`和`my_num % n`来表示顾客的编号和服务该顾客的柜员的编号。

最后，我们创建了100个顾客线程，并等待所有顾客线程结束。运行程序，可以看到输出结果如下：

Customer 1 is being served by Clerk 0
Customer 2 is being served by Clerk 1
Customer 3 is being served by Clerk 2
Customer 4 is being served by Clerk 3
Customer 7 is being served by Clerk 2
Customer 5 is being served by Clerk 4
Customer 8 is being served by Clerk 3
Customer 6 is being served by Clerk 0
Customer 9 is being served by Clerk 1
Customer 10 is being served by Clerk 2
...

从输出结果中可以看出，顾客们按照顺序依次接受服务，每次只有一个顾客被服务，且不会出现多个顾客同时被一个柜员服务的情况。因此，我们成功实现了该同步问题。