FT Digital Edition: our digitised print edition
// 题目要求找「右侧第一个 ≤ cur」的元素 → 弹出所有 cur 的，栈顶即为折扣
，更多细节参见同城约会
for (int i = 0; i < n; i++) {
玩法四：8:1 超长图，无限拓展的画布
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