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)。
A bundle of kanten, from the Encylopedia of Food (1923).
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it only serves to prove my point. It’s an inevitability and while game DRMs arguably serve a different purpose compared to two-bit JS based DRMs on a fucking NSFW ASMR site, the point is, yet again, the same.
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