领克回应「高速语音关大灯」：已完成优化方案
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思路：倒序单调栈。弹出所有 ≤ 当前身高的元素（这些人都能被看到），count 为弹出数量；若栈非空，还能看到栈顶（第一个更高的人），故 +1。能看到的人数 = count + (栈非空 ? 1 : 0)。，详情可参考下载安装汽水音乐
ParakeetNemotron
這是美國、以色列與伊朗之間一場新戰爭的第三天而已。