除法分块

part 1

对于

[sumlimits_{i=1}^n left lfloor frac n i ight floor ]

最多只有 (2 * sqrt n) 个取值

证明

显然

part 2

当 $ l = i $ 时 ， (r = left lfloor frac n {left lfloor frac n l ight floor} ight floor) , 此时

[forall x in [l,r] ,forall y in [l,r] ]

[left lfloor frac n x ight floor = left lfloor frac n y ight floor ]

且 (r) 就是这个值 $ (left lfloor frac n l ight floor)$ 的右边界

证明 ( 反证法 )

首先 (left lfloor frac n l ight floor = left lfloor frac n r ight floor) 即 (left lfloor frac n l ight floor = left lfloor frac n {left lfloor frac n {left lfloor frac n l ight floor} ight floor} ight floor)

设 (t = left lfloor frac n l ight floor , n = t * l + r , l> r) , 那么原式 $ = left lfloor frac n {left lfloor frac {t * l + r} {t} ight floor} ight floor $ , 对 $ t < rand t >= r $，分类讨论即可

如果 (r) 不是右边界，则一定有 (t = left lfloor frac n {r+1} ight floor = left lfloor frac n r ight floor)

则 $ n geq t * (r+1) , left lfloor frac n t ight floor geq frac{(r + 1) * t}{t}, left lfloor frac n t ight floor geq r + 1 $

根据定义 (left lfloor frac n t ight floor = r , r geq r + 1) ，与事实不符