暴力做法是对每个位置向右扫描找第一个更大值，O(n²)。可抽象为：元素像一排人，身高为数值。当前人「下一个更大」= 他右侧第一个没被挡住的人（比当前矮的都被挡住）。单调栈用 O(n) 维护「右侧候选更大值」：倒序遍历，弹掉 ≤ 当前的，栈顶即答案，再入栈当前值。
cursor[classno] = j;
，这一点在爱思助手下载最新版本中也有详细论述
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// 3. 从后向前放置（保证稳定性），更多细节参见safew官方版本下载
Что думаешь? Оцени!。搜狗输入法下载是该领域的重要参考
automatically together with the stack frame itself. Stack allocations