Hello everyone,

here is a very simple idea that can be useful for (cp) number theory problems, especially those concerning multiples, divisors, $$$\text{GCD}$$$ and $$$\text{LCM}$$$.

## Idea

Let's start from a simple problem.

*You are given $$$n$$$ pairs of positive integers $$$(a_i, b_i)$$$. Let $$$m$$$ be the maximum $$$a_i$$$. For each $$$i$$$, let $$$f(i)$$$ be the sum of the $$$b_j$$$ such that $$$i \mid a_j$$$. Output all pairs $$$(i, f(i))$$$ such that $$$f(i) > 0$$$.*

An obvious preprocessing is to calculate, for each $$$i$$$, the sum of the $$$b_j$$$ such that $$$a_j = i$$$ (let's denote it as $$$g(i)$$$). Then, there are at least $$$3$$$ solutions to the problem.

#### Solution 1: $$$O(m\log m)$$$

For each $$$i$$$, $$$f(i) = \sum_{j=1}^{\lfloor m/i \rfloor} g(ij)$$$. The complexity is $$$O(m(\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{m})) = O(m\log m)$$$.

#### Solution 2: $$$O(n\sqrt m)$$$

There are at most $$$n$$$ nonzero values of $$$g(i)$$$. For each one of them, find the divisors of $$$i$$$ in $$$O(\sqrt i)$$$ and, for each divisor $$$j$$$, let $$$f(j) := f(j) + g(i)$$$.

If $$$m$$$ is large, you may need to use a map to store the values of $$$f(i)$$$ but, as there are $$$O(n\sqrt[3] m)$$$ nonzero values of $$$f(j)$$$, the updates have a complexity of $$$O(n\sqrt[3] m \log(nm)) < O(n\sqrt m)$$$.

#### Solution 3: $$$O(m + n\sqrt[3] m)$$$

Build a linear prime sieve in $$$[1, m]$$$. For each nonzero value of $$$g(i)$$$, find the prime factors of $$$i$$$ using the sieve, then generate the divisors using a recursive function that finds the Cartesian product of the prime factors. Then, calculate the values of $$$f(i)$$$ like in solution 2.

Depending on the values of $$$n$$$ and $$$m$$$, one of these solutions can be more efficient than the others.

Even if the provided problem seems very specific, the ideas required to solve that task can be generalized to solve a lot of other problems.

## 1154G - Minimum Possible LCM

**Hint 1**

**Hint 2**

**Hint 3**

**Solution**