⚠️ 自己写：去掉界限会涨到多少

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本节的积压离散模型（纯算术，不靠真线程）：

backlog(prod, cons, ticks)   无界队列：每 tick 进 prod 个、出 cons 个 → 每 tick 结束时队列长度（prod > cons 就一直涨）
bounded(n, cap, cons_every)  有界队列 cap：放 n 个，生产者每 tick 想放 1 个（满就等），消费者每 cons_every 个 tick 取 1 个 → (每 tick 队列长, 被挡下几次, 排空用了几 tick)
drops(n, cap, cons_every)    满了就丢（put_nowait）时，丢了几个

同样进 3 出 1 跑 6 个 tick，比一比「有界 cap=2」和「无界」队列各有多长。补全 compare() 的返回：有界看 bounded 的 max(trace)、无界看 backlog 的最后一项：

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← 补全：有界满了就丢怎么保证一个不被处理两次 →

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import threading

class Box:
    """有界队列：满了 put 等、空了 get 等；等的时候把锁交出去（wait_for 放锁，醒来再拿回）。"""
    def __init__(self, cap):
        self.cap = cap
        self.items = []
        self.cond = threading.Condition()

def put(self, x):
        with self.cond:
            self.cond.wait_for(lambda: len(self.items) < self.cap)
            self.items.append(x)
            self.cond.notify_all()

def get(self):
        with self.cond:
            self.cond.wait_for(lambda: len(self.items) > 0)
            x = self.items.pop(0)
            self.cond.notify_all()
            return x

def run_pc(cap, n):
    """一个生产者放 0..n-1，一个消费者取 n 个；交回消费者拿到的顺序（FIFO，确定）。"""
    box = Box(cap)
    got = []

def producer():
        for i in range(n):
            box.put(i)

def consumer():
        for _ in range(n):
            got.append(box.get())
    p = threading.Thread(target=producer, daemon=True)
    c = threading.Thread(target=consumer, daemon=True)
    p.start(); c.start()
    p.join(timeout=2); c.join(timeout=2)
    return got

def run_mpmc(cap, nprod, ncons, per):
    """nprod 个生产者各放 per 个（编号 生产者号*100 + 序号），ncons 个消费者一起取，直到取满；
    交回 (消费到的总数, 消费集合 == 生产集合)。只问聚合量——谁先谁后不确定，总数和集合确定。"""
    box = Box(cap)
    produced = set()
    got = []
    got_lock = threading.Lock()
    total = nprod * per

def producer(pid):
        for j in range(per):
            item = pid * 100 + j
            produced.add(item)
            box.put(item)

def consumer():
        while True:
            with got_lock:
                if len(got) >= total:
                    return
            x = box.get()
            with got_lock:
                got.append(x)

ps = [threading.Thread(target=producer, args=(k,), daemon=True) for k in range(nprod)]
    cs = [threading.Thread(target=consumer, daemon=True) for _ in range(ncons)]
    for t in ps + cs:
        t.start()
    for t in ps:
        t.join(timeout=2)
    # 生产完了，给消费者补足唤醒（放哨兵占位让还在等的消费者醒来看到 len>=total）
    for _ in range(ncons):
        box.put(-1)
    for t in cs:
        t.join(timeout=2)
    real = [x for x in got if x != -1]
    return len(real), set(real) == produced

def run_poison(cap, n, ncons):
    """一个生产者放 0..n-1，放完再放 ncons 颗毒丸（None）；ncons 个消费者见到毒丸就退出。
    交回 (消费到的真数据个数, 是不是所有消费者都退出了)。毒丸让「优雅退出」变确定。"""
    box = Box(cap)
    got = []
    got_lock = threading.Lock()
    stopped = []

def producer():
        for i in range(n):
            box.put(i)
        for _ in range(ncons):
            box.put(None)

def consumer():
        while True:
            x = box.get()
            if x is None:
                stopped.append(1)
                return
            with got_lock:
                got.append(x)

p = threading.Thread(target=producer, daemon=True)
    cs = [threading.Thread(target=consumer, daemon=True) for _ in range(ncons)]
    p.start()
    for t in cs:
        t.start()
    p.join(timeout=2)
    for t in cs:
        t.join(timeout=2)
    return len(got), len(stopped) == ncons

def run_poison_few(cap, n, ncons, npoison):
    """毒丸放少了：生产 0..n-1，只放 npoison 颗毒丸；ncons 个消费者。交回 (消费到的真数据数, 是否都退出了)。
    npoison < ncons 时会有消费者永远等——join(timeout) 兜住，返回 False。"""
    box = Box(cap)
    got = []
    got_lock = threading.Lock()
    stopped = []

def producer():
        for i in range(n):
            box.put(i)
        for _ in range(npoison):
            box.put(None)

def consumer():
        while True:
            x = box.get()
            if x is None:
                stopped.append(1)
                return
            with got_lock:
                got.append(x)

p = threading.Thread(target=producer, daemon=True)
    cs = [threading.Thread(target=consumer, daemon=True) for _ in range(ncons)]
    p.start()
    for t in cs:
        t.start()
    p.join(timeout=2)
    for t in cs:
        t.join(timeout=1)
    return len(got), len(stopped) == ncons

def backlog(prod, cons, ticks):
    """无界队列：每个 tick 进 prod 个、出 cons 个（出不能超过在队里的）；交回每个 tick 结束时队列的长度。
    生产快于消费（prod > cons）时会一直涨——积压。"""
    q = 0
    out = []
    for _ in range(ticks):
        q += prod
        q -= min(cons, q)
        out.append(q)
    return out

def bounded(n, cap, cons_every):
    """有界队列 cap：要放 n 个，生产者每 tick 想放 1 个（满了就等，不算这一次），消费者每 cons_every 个 tick 取 1 个。
    交回 (每 tick 结束时的队列长度, 生产者被挡下的次数, 全部排空用了多少 tick)。"""
    q = 0
    made = 0
    blocked = 0
    trace = []
    t = 0
    while made < n or q > 0:
        t += 1
        if (t % cons_every) == 0 and q > 0:          # 消费者到点，取一个
            q -= 1
        if made < n:                                 # 生产者想放一个
            if q < cap:
                q += 1; made += 1
            else:
                blocked += 1                          # 满了，被背压挡下
        trace.append(q)
        if t > 100000:
            break
    return trace, blocked, t

def drops(n, cap, cons_every):
    """满了就丢（put_nowait）：要放 n 个，满时这一个直接丢掉；消费者每 cons_every 个 tick 取一个。交回丢了几个。"""
    q = 0
    dropped = 0
    for t in range(1, n * cons_every + cons_every + 1):
        if (t % cons_every) == 0 and q > 0:
            q -= 1
        if n > 0:
            if q < cap:
                q += 1
            else:
                dropped += 1
            n -= 1
    return dropped

def compare():
    tr, _, _ = bounded(6, 2, 2)      # 有界 cap=2，消费者每 2 个 tick 取一个
    unb = backlog(3, 1, 6)           # 无界，进 3 出 1
    # TODO: 交回 (有界峰值, 无界终值)
    return None, None

b, u = compare()
print(str(b) + "/" + str(u))

本次输入：

输出：

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