浅谈整除分块

模型:

\(\sum_{i=1}^{n}\left\lfloor\frac{n}{i}\right\rfloor\)
假设 \(n = 8\),那么可得:

\(i\) 1 2 3 4 5 6 7 8
\(8/i\) 8 4 2 2 1 1 1 1

概念:

表中同样的值会连续出现,而相同的值所划分的区间成为一个块。
整除的性质使得从 \(1\)\(n\) 的表可根据数值划分为不同的块,且分块数远远小于 \(n\)
利用这种性质,我们可以推导出每个分块具体的左右端点位置在哪,这样就可以快速求解出来了。

推导:

假设我们已知某一个分块的左端点 \(l\),要求解出该分块的右端点 \(r\)
设该分块的数值为\(k\),对于该分块中的每个数\(i\),有 \(k=\left\lfloor\frac{n}{i}\right\rfloor=\left\lfloor\frac{n}{l}\right]\),即 \(i\times k \le n\)
那么显然满足 \(i \times k \le n\) 成立的最大的 \(i\) 就是我们要的右端点 \(r\)
于是可得:\(\left\{\begin{array}{l}k=\left\lfloor\frac{n}{l}\right\rfloor \\ r=\max (i), i \times k \leq n\end{array}\right.\)
推导得:\(r=\left\lfloor\frac{n}{k}\right\rfloor=\left\lfloor \frac{n}{\left\lfloor\frac{n}{l}\right\rfloor}\right\rfloor\)

变形0:

已知 \(n,m\),求 \(\sum_{i=1}^{min(n,m)}\lfloor\frac{n}{ i}\rfloor\lfloor\frac{m}{i}\rfloor\)
\(\left\{\begin{array}{l}k1=\left\lfloor\frac{n}{l}\right\rfloor,k2= \left\lfloor\frac{m}{l}\right\rfloor\\ r=\max (i), i \times k1 \leq n,i\times k2 \le m\end{array}\right.\)
\(r = min(\lfloor\frac{n}{k1}\rfloor,\lfloor\frac{m}{k2}\rfloor)\)

变形1:

已知 \(n,a,b\),求 \(\sum_{i=1}^{n}\left\lfloor\frac{n}{a i+b}\right\rfloor\)
\(i' = a\times i + b\),先求出 \(\left\lfloor\frac{n}{i'}\right\rfloor\) 的整除分块
可得 \(k=\left\lfloor\frac{n}{a\times l+b}\right\rfloor\)\(r'=\left\lfloor\frac{n}{k}\right\rfloor=\left\lfloor\frac{n}{\left\lfloor\frac{n}{a \times l + b}\right\rfloor}\right\rfloor\)
\(\left\{\begin{array}{l}i'=a\times i +b \\ r'=\max (i')\end{array}\right.\)\(\begin{cases}a\times i=i'-b\\ i=\dfrac{i'-b}{a}\end{cases}\)\(r=max(i)= max(\left\lfloor\frac{i'-b}{a}\right\rfloor) = \lfloor\frac{r'-b}{a} \rfloor\)
$r =\left \lfloor \dfrac{\left \lfloor \dfrac{n}{\lfloor \dfrac{n}{a\times l+b}\rfloor }\right \rfloor -b}{a}\right \rfloor $

变形2:

做法同上。
已知 \(n\),求 \(\sum_{i=1}^{n}\left\lfloor\frac{n}{i^2}\right\rfloor\)
\(i' = i^2\),先求出 \(\left\lfloor\frac{n}{i'}\right\rfloor\) 的整除分块
可得 \(k=\left\lfloor\frac{n}{l^2}\right\rfloor\)\(r'=\left\lfloor\frac{n}{k}\right\rfloor=\left\lfloor\frac{n}{\left\lfloor\frac{n}{l^2}\right\rfloor}\right\rfloor\)
\(\left\{\begin{array}{l}i=\sqrt{i'} \\ r'=\max (i')\end{array}\right.\)\(r = max(i) = max(\lfloor\sqrt{i'} \rfloor) = \lfloor r' \rfloor = \lfloor\sqrt{\left\lfloor\frac{n}{\left\lfloor\frac{n}{l^2}\right\rfloor}\right\rfloor}\rfloor\)

变形3:

已知 \(n\),求 \(\sum_{i=1}^{n}\left\lceil\frac{n}{i}\right\rceil\)
\(n\) 整除 \(i\) 时,\(\left\lceil\frac{n}{i}\right\rceil = \left\lfloor\frac{n}{i}\right\rfloor\)
\(n\) 不整除 \(i\) 时,\(\left\lceil\frac{n}{i}\right\rceil = \left\lfloor\frac{n}{i}\right\rfloor + 1\)
于是我们可以用 $\lfloor \frac{n+i-1}{i}\rfloor $ 来代替 \(\lceil\frac{n}{i}\rceil\)
原式就转换为了 \(\sum_{i=1}^{n}\lfloor\frac{n+i -1}{i}\rfloor\)

那么有 \(\left\{\begin{array}{l}k=\left\lfloor\frac{n+l-1}{l}\right\rfloor \\ r=\max (i), i\times k \leq n+i-1 \end{array}\right.\)
\(\because i \times k \le n+i-1 \Rightarrow i\times(k-1) \le n-1\)
\(\therefore r=\left\lfloor\frac{n-1}{k-1}\right\rfloor=\left[\frac{n-1}{\left\lfloor\frac{n+l-1}{l} \mid-1\right.}\right\rfloor\)

posted @ 2021-05-10 15:48  GsjzTle  阅读(168)  评论(9编辑  收藏  举报