对于任意一个块的左端点 \(i \quad (i\leq n)\) ，需要找到块的右端点 \(j \quad (i\leq j\leq n)\) 使得
\[
\left \lfloor\frac{n}{i}\right\rfloor = \left\lfloor\frac{n}{j} \right \rfloor
\]
若能求出 \(j\)，那么下一个块的 \(i = j + 1\)。
令 \(\large \left \lfloor \frac{n}{i} \right \rfloor = k\)，有 \(\large i \times k + p = n\)。其中 \(\large p = n \mod i\)。
将 \(\large \frac{n}{j} = k\) 表示为 $ \large \frac{n}{i + d_{max}} = k$。对于 $ \large \frac{n}{i + d} = k$，同上有 \((i+d) \times k + p'=n\)。有
\[
(i+d) \times k + p' = \large i \times k + p = n
\]
\[ p' = i \times k + p - i \times k - d \times k = p - d \times k \]
因为 \(p' \ge 0\) 且 \(d\) 为整数，所以有 \(p - d \times k \ge 0\)，进而推出 \(d_{max} = \left \lfloor \frac{p}{k} \right \rfloor\)
推式子
\[
j = i + d_{max} = i + \left \lfloor \frac{p}{k} \right \rfloor
\]
\[ = i+ \left \lfloor \frac{n \mod i}{\left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor = \left \lfloor i +\frac{n \mod i}{\left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor \]
\[ = \left \lfloor i+ \frac{n - \left \lfloor \frac{n}{i} \right \rfloor \times i}{ \left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor \]
\[ = \left \lfloor \frac{\left \lfloor \frac{n}{i}\right \rfloor \times i}{\left \lfloor \frac{n}{i} \right \rfloor }+ \frac{n - \left \lfloor \frac{n}{i} \right \rfloor \times i}{ \left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor \]
\[ = \left \lfloor \frac{n - \left \lfloor \frac{n}{i} \right \rfloor \times i + \left \lfloor \frac{n}{i} \right \rfloor \times i}{ \left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor \]
\[ = \left \lfloor \frac{n}{ \left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor \]
所以 \(j = \left \lfloor \frac{n}{ \left \lfloor \frac{n}{i} \right \rfloor} \right \rfloor\)。