补全：找出最坏的一次交错

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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。）

贯穿全条的模拟器：

一条线程 = 一串「步」    inc_steps() = [("read",), ("add", 1), ("write",)]   读共享的 n 进寄存器、寄存器加一、写回
run_schedule(threads, schedule)   schedule 是线程编号的序列，每个编号出现一次就让那条线程走一步；交回 (最终 n, 记录)
all_schedules(a, b)               两条线程（a 步、b 步）的全部交错
outcomes(threads)                 各种最终值各出现几次
locked(steps)                     把一串步包成一个原子步——模拟器遇到它一口气做完

补全 worst(threads, expect)：所有交错里最终值离 expect 最远（最小）的那个值，和得到它的一张调度表（第一张）。两条各加两次。

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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 worst(threads, expect):
    # TODO
    return expect, []

def thread_of(times):
    steps = []
    for _ in range(times):
        steps += inc_steps()
    return steps

v, s = worst([thread_of(2), thread_of(2)], 4)
print(str(v) + "/" + "".join(str(x) for x in s))

本次输入：

输出：

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