2023.10.26 闲话

scanf 挺强的，可以试试 scanf("%[^\n]", s);（s 的类型为 char*）.

才发现 2022 和 2023 的 HE 省队名额是一样多的。。

久违的闲话

推歌：ねむれわたがし - 一二三 feat. 花奏かのんβ .

Until I End Up Dead - Dream

I got blindsided, couldn't see it in the slightest
That path you walked was the same as mine is
Hard to believe, man, tell me you're lyin'
Uh-uh-uh (Uh-uh-uh)
This is something new
Crying the first time I ever wore a suit
I saw me when I looked at you
So now, I fear one day I'm gonna disappear
Look in the mirror, to make sure I'm still standing here
I been playin' in my head those things I never said
Like, like existential dread
But I'm not stayin' in my mind, I'm livin' like you said
It's gonna be alright until I end up dead
(Dead, dead, dead, dead)
Until I end up dead (Dead, dead, dead, dead)
Until I end up-
Blindsided, so strong why would you hide it?
It's in your bones, man, you had to fight it
Came out of nowhere, now pigs are flying
Mm-mm-mm (Mm-mm-mm)
This is something new (Something new)
Crying the first time I ever wore a suit (Wore a suit)
I saw me when I looked at you
So now, I fear one day I'm gonna disappear
Look in the mirror, to make sure I'm still standing here
I been playin' in my head those things I never said
Like, like existential dread
But I'm not stayin' in my mind, I'm livin' like you said
It's gonna be alright until I end up dead
(Dead, dead, dead, dead)
Until I end up- dead (Dead, dead, dead, dead)
Until I end up dead (Dead, dead, dead, dead)
(I been playin' in my head those things I never said)
(Like, like existential dread)
Until I end up dead (Dead, dead, dead, dead)
(I been playin' in my head those things I never said)
(It's gonna be alright)
Until I end up dead (Dead, dead, dead, dead)
Until I end up dead, I'm gonna be just fine

---

问题：$[0,n]^2$ 上整点凸包大小的上界？

只考虑右下凸包，则凸包的形态可以表示为一堆向量 $\bm v(x,y)$ 之和，需要满足：

• $x\perp y$ .
• $\sum x,\sum y\le n$ .

先考虑满足计算 $x,y\le n$ 的向量的数量：

$$\sum_{i=1}^n\sum_{j=1}^n[i\perp j]=\sum_{i=1}^n\varphi(i)\le\sum_{i=1}^ni=O(n^2)$$

（这里也可以 $\Theta$ 不过我懒得整了，就和下面那个方法一样）

对应的，计算这些向量的和以计算真实的 $n$ 的量级：

$$\begin{aligned}\sum_{i=1}^n\sum_{j=1}^n[i\perp j](i+j)&=\sum_{i=1}^ni\cdot\varphi(i)\\&=\sum_{i=1}^n\mu(i)\cdot i\cdot S\left(\left\lfloor\dfrac ni\right\rfloor\right)&\text{where }S(n)=\sum_{i=1}^ni^2\\&=\Theta\left(n^3\sum_{i=1}^n\dfrac{\mu(i)}{i^2}\right)\\&=\Theta(n^3)\end{aligned}$$

（这里必须至少到 $\Omega(n^3)$ 否则无法进行下一步）

注：$\displaystyle\sum_{i\ge 1}\dfrac{\mu(i)}{i^2}=\dfrac1{\zeta(2)}=\dfrac6{\pi^2}$ 由 DGF 导出 .

那么容易发现答案就是 $O(n^{\frac23})$（实际上可以到 $\Theta$，只需要改进第一个不等式的 $O$）.

感谢 APJifengc 的指导 .