1 It’s also possible to use 4 candidates per pixel and compute the barycentric coordinates of the resulting tetrahedron, but using 3 candidates forming a triangle is more straightforward. ↑
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，更多细节参见Line官方版本下载
for (int i = 0; i < n; i++) {
// 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