发送一条消息，让多个Genserver同时执行操作，并知道它们都执行了操作？

defmodule MySupervisor do
  use Supervisor

  def start_link do
    Supervisor.start_link(__MODULE__, :ok)
  end

  def init(:ok) do
    children = [
      worker(MyGenServer, []),
      worker(MyGenServer, []),
      worker(MyGenServer, [])
    ]

    supervise(children, strategy: :one_for_one)
  end
end

defmodule MyGenServer do
  use GenServer

  def start_link do
    GenServer.start_link(__MODULE__, :ok)
  end

  def init(:ok) do
    {:ok, []}
  end

  def handle_info(:execute_operation, state) do
    # 执行操作的逻辑代码
    # ...

    {:noreply, [self() | state]}
  end
end

{:ok, supervisor_pid} = MySupervisor.start_link()

# 向监督树中的每个进程发送消息
Enum.each(supervisor_pid.children, fn child ->
  GenServer.cast(child, :execute_operation)
end)

# 等待所有进程执行完操作
:timer.sleep(1000) # 等待足够的时间

# 获取每个进程的状态列表
states = Enum.map(supervisor_pid.children, fn child ->
  GenServer.call(child, :get_state)
end)

# 判断是否所有进程都执行了操作
if Enum.all?(states, fn state -> state == [:pid1, :pid2, :pid3] end) do
  IO.puts("所有进程都执行了操作")
else
  IO.puts("有进程未执行操作")
end