2023.11.14 闲话

晚上吃晚饭回来，看到眼前光影节。

等到天气暑热郁郁枯寂，都仿照媒拾成功，我说不出，仿照怜悯的行为，也同逍遥快乐。

早就不带初心。这就是“初心”是什么？

当我在水上社，我的朋友说我变了，然而，人每天都在变化，应该有所准备，拿着这足以找到的。

以初心不一定容易。改变他的心为“初心”。

然而现在的魏天老师不改，这有当初的志向，我先为大家后面了。

然而，虽然有变化，一定要回头光辉，重感受所遇。

也许是化不可脱离，每相转折，尚能回忆。

不转，在坡吗？

Owo Song

{intro}
This song is for everybody who
Loves somebody (my mom doesn't love me)
Yeah, that works (or my father)

{chorus}
Owo what is this?
I think i found love within
Deep inside your heart and soul
Owo (ay, yuh) (what does owo mean? again, i have no idea)
Owo what is this?
I think i found love within3333
Deep inside your heart and soul
Owo, yeah, yeah (owo, baby) (seriously! what does owo mean?!)
Owo what is this?
I think i found love within
Deep inside your heart and soul
Owo, oh (owo, baby) (my mom just grounded me from soccer practice please help)

{verse 1}
Tell me, do you want me, baby? (yeah, yeah)
Tell me, do i drive you crazy? (yeah, yeah)
Tell me, do i make you go 'owo'? (owo, yeah, girl) (somebody please help, my mom won't give me back my x-box)
Tell me, do you want me, baby? (owo, owo, owo, yeah, yeah)
Tell me, do i drive you crazy? (owo, owo, owo)
Tell me, do i make you go 'owo'? (yeah) (yeah, yeah)

{chorus}
(my mom has yet to give me back my)
Owo what is this? (x-box360, i am going to)
I think i found love within (call the cops on her or-)
Deep inside your heart and soul (*laughs*)
Owo (yeah, ay)
Owo what is this?
I think i found love within
Deep inside your heart and soul
Owo (uwo, girl) (i am abused at home, oh)
Owo what is this?
I think i found love within
Deep inside your heart and soul
Owo (uwo, darling)

{outro}
(owo, o-owo!
I have burn marks on me from my fathers cigarettes *giggles*
Owo! *laughs*)

---

问题：

给一个长度为 \(n\) 的整数序列，\(q\) 次操作每次区间 checkmax，最后输出整个序列 .

\(n\le 10^5\)，\(q\le2\times 10^7\)，值域 \([0,2^{32}]\) . 数据随机 .

模拟赛的题，做法比较逆天，记录一下：

Solution 1   二区间合并，\(O(n\log n+q)\) .

Solution 2   每次把区间拆成两个长为 \(2^k\) 区间的并打标记最后 pushdown，\(\Theta(n\log n+q)\) .

Solution 3   离线下来排序就变成区间覆盖，时间倒流并查集维护即可，时间复杂度 \(\Theta(\operatorname{sort}(q)+q\alpha(n))\) .

Solution 4   分块，\(B\) 是块长 . 对于整块的记 \(t_{l,r}\) 表示块 \(l\dots r\) 的标记，最后 pushdown . 对于散块前后缀直接打标记，否则暴力扫 . 因为数据随机所以只有 \(\frac Bn\) 的概率两端点在同一块需要暴力扫，所以这部分的期望时间复杂度是 \(O(B^2)\) 的 . 取 \(B=\sqrt n\) 问题被期望 \(O(n+q)\) 解决 .

仔细思考一下其实只要运算有交换律结合律 Solution 4 就都能跑，哈哈 .

upd. 由乃救爷爷 .

感谢 APJifengc 和一些人 .

一个有趣的事实是，原题的数据生成器质量不高所以直接把 \(q\) 和 \(2\times10^5\) 取 min 即可通过 .