补全：漏掉一处会怎样

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（每道题开头都有同一段：上面的模拟器。判题机要确定的答案，所以本条用「调度表」代替真线程的运气。）

补全 leak_test(protect_fn)：两条线程各是 inc + inc（两个读-改-写）；用 protect_fn 保护后跑所有交错，交回错误交错的种数。给一个「只保护第一个 inc」的坏保护函数，看它漏了多少。

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代码编辑器

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def inc_steps(k=1):
    return [("read",), ("add", k), ("write",)]

def check_act_steps(need=1):
    return [("read",), ("check", need), ("add", -need), ("write",)]

def run_schedule(threads, schedule, start=0):
    """threads: 每条线程是「步」的列表；schedule: 线程编号的序列，每个编号出现一次就走一步。
    共享变量 n；每条线程有自己的寄存器 reg。交回 (最终 n, 记录列表)。"""
    n = start
    pc = [0] * len(threads)
    reg = [0] * len(threads)
    log = []
    for t in schedule:
        if pc[t] >= len(threads[t]):
            continue
        step = threads[t][pc[t]]
        pc[t] += 1
        n, reg[t], note, go_on = do_step(step, n, reg[t])
        log.append("T" + str(t) + ":" + note)
        if not go_on:
            pc[t] = len(threads[t])      # 检查没过：这条线程后面的步全部跳过
    return n, log

def do_step(step, n, r):
    if step[0] == "read":
        return n, n, "read " + str(n), True
    if step[0] == "add":
        return n, r + step[1], "add→" + str(r + step[1]), True
    if step[0] == "write":
        return r, r, "write " + str(r), True
    if step[0] == "check":
        ok = r >= step[1]
        return n, r, "check " + str(r) + (">=" if ok else "<") + str(step[1]), ok
    if step[0] == "atomic":
        notes = []
        ok = True
        for s in step[1]:
            n, r, note, ok = do_step(s, n, r)
            notes.append(note)
            if not ok:
                break
        return n, r, "atomic(" + ",".join(notes) + ")", ok
    raise ValueError(step[0])

def all_schedules(a, b):
    """两条线程（a 步和 b 步）的全部交错：每个交错是一串 0/1。"""
    if a == 0:
        return [[1] * b]
    if b == 0:
        return [[0] * a]
    return [[0] + s for s in all_schedules(a - 1, b)] + [[1] + s for s in all_schedules(a, b - 1)]

def outcomes(threads, start=0):
    counts = {}
    for s in all_schedules(len(threads[0]), len(threads[1])):
        n, _ = run_schedule(threads, s, start)
        counts[n] = counts.get(n, 0) + 1
    return counts

def locked(steps):
    return [("atomic", steps)]

def protect_all(thread):
    return [step for seg in thread for step in locked(seg)]

def protect_first(thread):
    # TODO: 只把第一段 locked，其余原样
    return []

def leak_test(protect_fn):
    # TODO: thread = [inc_steps(), inc_steps()]；两条线程；数最终值不等于 4 的交错
    return 0

print(str(leak_test(protect_all)) + "/" + str(leak_test(protect_first)))

本次输入：

输出：

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