数论分块

数论分块学习

用途

快速计算含有\(\lfloor{\frac{n}{i}}\rfloor\)的和式（\(i\)为变量）

引理

引理1

\[\forall a,b,c\in \mathbb{N_+},\quad \Big\lfloor \frac{a}{bc}\Big\rfloor=\bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor \]

证明1

\[\text{let}\quad \{x\}=x-\lfloor x\rfloor\quad\therefore \{x\}\in [0,1)\\ \Big\lfloor \frac{a}{bc}\Big\rfloor=\Bigg\lfloor \bigg(\Big\lfloor\frac{a}{b}\Big\rfloor+\Big\{\frac{a}{b}\Big\}\bigg)\frac{1}{c}\Bigg\rfloor\ge\bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor\\ \because \Big\{\frac{a}{b}\Big\}<1\le c-\bigg(\Big\lfloor\frac{a}{b}\Big\rfloor\mod c\bigg)\\ \therefore \Big\lfloor \frac{a}{bc}\Big\rfloor=\Bigg\lfloor \bigg(\Big\lfloor\frac{a}{b}\Big\rfloor+\Big\{\frac{a}{b}\Big\}\bigg)\frac{1}{c}\Bigg\rfloor <\Bigg\lfloor \bigg(\Big\lfloor\frac{a}{b}\Big\rfloor+c-\Big(\Big\lfloor\frac{a}{b}\Big\rfloor\mod c\Big)\bigg)\frac{1}{c}\Bigg\rfloor=\bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor+1\\ \therefore \bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor\le \Big\lfloor \frac{a}{bc}\Big\rfloor <\bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor+1\\ \therefore \Big\lfloor \frac{a}{bc}\Big\rfloor=\bigg\lfloor \frac{\lfloor \frac{a}{b}\rfloor}{c}\bigg\rfloor \]

注：该引理并不会被用到，这里只是把oiwiki上的证明丰富一下

引理2

\[\forall d\le n\in \mathbb{N_+},\quad|\{\lfloor \frac{n}{d}\rfloor\}|\le 2\lfloor\sqrt{n}\rfloor \]

证明2

\[\text{when}\quad d\le \lfloor\sqrt{n}\rfloor ,\quad|\{\lfloor \frac{n}{d}\rfloor\}|\le|\{d\}|=\lfloor\sqrt{n}\rfloor\\ \text{when}\quad d>\lfloor\sqrt{n}\rfloor,\because \lfloor \frac{n}{d}\rfloor\le\sqrt{n}\therefore |\{\lfloor \frac{n}{d}\rfloor\}|\le|\{i\le\lfloor\sqrt{n}\rfloor\in\mathbb{N_+} \}|=\lfloor\sqrt{n}\rfloor \]

注：该引理只用于证明复杂度

数论分块结论

\[\forall i\le n\in \mathbb{N_+}\quad,\max\{j\in\mathbb{N_+}|\lfloor\frac{n}{j}\rfloor=\lfloor\frac{n}i{}\rfloor\}=\bigg\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\bigg\rfloor \]

证明

\[\text{let}\quad\max\{j\in\mathbb{N_+}|\lfloor\frac{n}{j}\rfloor=\lfloor\frac{n}i{}\rfloor\}=r\\ \therefore \frac{n}{r}\ge\lfloor\frac{n}{r}\rfloor=\lfloor\frac{n}{i}\rfloor\\ \therefore r\le \frac{n}{\lfloor\frac{n}{i}\rfloor}\therefore r\le\bigg\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\bigg\rfloor\\ \text{let}\quad t=\bigg\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\bigg\rfloor,\quad \text{then prove}\quad \lfloor\frac{n}{t}\rfloor=\lfloor\frac{n}{i}\rfloor\\ \because t\ge \bigg\lfloor\frac{n}{(\frac{n}{i})}\bigg\rfloor=\lfloor i\rfloor=i\quad\therefore \lfloor\frac{n}{t}\rfloor\le\lfloor\frac{n}{i}\rfloor\\ \text{and}\because\lfloor\frac{n}{t}\rfloor=\Bigg\lfloor\frac{n}{\Big\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\Big\rfloor}\Bigg\rfloor\ge\Bigg\lfloor\frac{n}{\big(\frac{n}{\lfloor\frac{n}{i}\rfloor}\big)}\Bigg\rfloor=\lfloor\frac{n}{i}\rfloor\\ \therefore \lfloor\frac{n}{t}\rfloor=\lfloor\frac{n}{i}\rfloor\\ \therefore r=t=\bigg\lfloor\frac{n}{\lfloor\frac{n}{i}\rfloor}\bigg\rfloor \]