符号约定
\(\left\lfloor x\right\rfloor\)：\(x\)下取整。
\(\left\{x\right\}\)：\(x\)的小数部分。
显然，对于\(\forall x\ge 0,x\in\mathbb{Q}\),有\(x=\left\lfloor x \right\rfloor+\left\{x\right\}\)
\(\left|V\right|\)：集合\(V\)的元素个数，称为集合的阶。
引理证明
引理\(1\)：
\(\forall a,b,c\in\mathbb{Z},\left\lfloor\frac{a}{bc}\right\rfloor=\left\lfloor\frac{\left\lfloor\frac{a}{b}\right\rfloor}{c}\right\rfloor\)
证明：
\(\because\frac{a}{b}=\left\lfloor\frac{a}{b}\right\rfloor+\left\{\frac{a}{b}\right\}\)
\(\therefore \left\lfloor\frac{a}{bc}\right\rfloor=\left\lfloor\frac{a}{b}\cdot \frac{1}{c}\right\rfloor=\left\lfloor\left(\left\lfloor\frac{a}{b}\right\rfloor+\left\{\frac{a}{b}\right\}\right)\cdot \frac{1}{c}\right\rfloor=\left\lfloor\frac{\left\lfloor\frac{a}{b}\right\rfloor}{c}+\frac{\left\{\frac{a}{b}\right\}}{c}\right\rfloor\)
\(\because 0\le \left\{\frac{a}{b}\right\}<1,0\le\left\{\frac{\left\lfloor\frac{a}{b}\right\rfloor}{c}\right\}<1\)
\(\therefore \left\lfloor\frac{a}{bc}\right\rfloor=\left\lfloor\frac{\left\lfloor\frac{a}{b}\right\rfloor}{c}\right\rfloor\)
引理\(2\)：
\(\forall n\in\mathbb{N_{+}},\left|\left\{\left\lfloor\frac{n}{d}\right\rfloor\mid d\in\mathbb{N_{+}},d\le n\right\}\right|\le\left\lfloor2\sqrt{n}\right\rfloor\)
标签：lfloor,mathbb,right,frac,分块,rfloor,整除,left From： https://www.cnblogs.com/jd122/p/16991956.html