感觉题解证的有些冗长
对于 ∀i\forall i∀i 满足 ⌊ni⌋=⌊nL⌋\lfloor \frac{n} {i}\rfloor=\lfloor\frac{n}{L}\rfloor ⌊in⌋=⌊Ln⌋ ，R=⌊n⌊nL⌋⌋=⌊n⌊ni⌋⌋≥⌊nni⌋=iR=\lfloor\frac{n}{\lfloor \frac{n} {L}\rfloor}\rfloor=\lfloor\frac{n}{\lfloor \frac{n} {i}\rfloor} \rfloor\geq \lfloor\frac{n}{ \frac{n} {i}}\rfloor=iR=⌊⌊Ln⌋n⌋=⌊⌊in⌋n⌋≥⌊inn⌋=i
对于 ∀i′\forall i'∀i′ 满足 ⌊ni′⌋=⌊nL⌋−1\lfloor \frac{n} {i'}\rfloor=\lfloor\frac{n}{L}\rfloor-1 ⌊i′n⌋=⌊Ln⌋−1 ，R=⌊n⌊nL⌋⌋=⌊n⌊ni′⌋+1⌋<⌊nni′⌋=i′R=\lfloor\frac{n}{\lfloor \frac{n} {L}\rfloor}\rfloor=\lfloor\frac{n}{\lfloor \frac{n} {i'}\rfloor+1} \rfloor < \lfloor\frac{n}{ \frac{n} {i'}}\rfloor=i'R=⌊⌊Ln⌋n⌋=⌊⌊i′n⌋+1n⌋<⌊i′nn⌋=i′
即证 R=⌊n⌊nL⌋⌋R=\lfloor\frac{n}{\lfloor \frac{n} {L}\rfloor}\rfloorR=⌊⌊Ln⌋n⌋ 是这个块的右端点
