[Tutorial] An interesting counting problem related to square product

Правка en55, от SPyofcode, 2021-11-01 08:53:10

The statement:

Given three integers $$$n, k, p$$$, $$$(1 \leq k \leq n < p)$$$.

Count the number of array $$$a[]$$$ of size $$$k$$$ that satisfied

  • $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$
  • $$$a_i \times a_j$$$ is perfect square $$$\forall 1 \leq i < j \leq n$$$

Since the number can be big, output it under modulo $$$p$$$.

For convenient, you can assume $$$p$$$ is a large constant prime $$$10^9 + 7$$$

Yet you can submit the problem for $$$k = 3$$$ here.

Notice that in this blog, we will solve in $$$O(n)$$$ for general $$$k$$$ from the original task

For harder variants you can see in this blog [Variants] An interesting counting problem related to square product

Solution for k = 1

The answer just simply be $$$n$$$

Solution for k = 2

Algorithm

We need to count the number of pair $$$(a, b)$$$ that $$$1 \leq a < b \leq n$$$ and $$$a \times b$$$ is perfect square.

Every positive integer $$$x$$$ can be represent uniquely as $$$x = u \times p^2$$$ for some positive integer $$$u, p$$$ and $$$u$$$ as small as possible ($$$u$$$ is squarefree number).

Let represent $$$x = u \times p^2$$$ and $$$y = v \times q^2$$$ (still, minimum $$$u$$$, $$$v$$$ ofcourse).

We can easily proove that $$$x \times y$$$ is a perfect square if and if only $$$u = v$$$.

So for a fixed squarefree number $$$u$$$. You just need to count the number of ways to choose $$$p^2$$$.

The answer will be the sum of such ways for each fixed $$$u$$$.

Implementation

Implementation using factorization
Implementation 1
Implementation 2
Implementation related to Möbius function

Complexity

So about the complexity....

For the implementation using factorization, it is $$$O(n \log n)$$$.

Hint 1
Hint 2
Proof
Bonus

For the 2 implementations below, the complexity is linear.

Hint 1
Hint 2
Hint 3
Hint 4
Proof

For the last implementation, the complexity is Linear

Hint 1
Hint 2
Proof

Solution for general k

Using the same logic above, we can easily solve the problem.

Now you face up with familliar binomial coefficient problem

This implementation here is using the assumption of $$$p$$$ prime and $$$p > max(n, k)$$$

You can still solve the problem for squarefree $$$p$$$ using lucas and CRT

Yet just let things simple as we only focus on the counting problem, we will assume $$$p$$$ is a large constant prime.

O(n) solution

Extra Tasks

These are harder variants, and generalization from the original problem. You can see more detail here

A: Can we also use phi function or something similar to solve for $$$k = 2$$$ in $$$O(\sqrt{n})$$$ or faster ?

B: Can we also use phi function or something similar to solve for general $$$k$$$ in $$$O(\sqrt{n})$$$ or faster ?

C: Can we also solve the problem where there can be duplicate: $$$a_i \leq a_j\ (\forall\ i < j)$$$ and no longer $$$a_i < a_j (\forall\ i < j)$$$ ?

D: Can we solve the problem where there is no restriction between $$$k, n, p$$$ ?

E: Can we solve for negative integers, whereas $$$-n \leq a_1 < a_2 < \dots < a_k \leq n$$$ ?

F: Can we solve for a specific range, whereas $$$L \leq a_1 < a_2 < \dots < a_k \leq R$$$ ?

G: Can we solve for cube product $$$a_i \times a_j \times a_k$$$ effectively ?

H: Can we solve if it is given $$$n$$$ and queries for $$$k$$$ ?

I: Can we solve if it is given $$$k$$$ and queries for $$$n$$$ ?

J: Can we also solve the problem where there are no order: Just simply $$$1 \leq a_i \leq n$$$ ?

K: Can we also solve the problem where there are no order: Just simply $$$0 \leq a_i \leq n$$$ ?

M: Can we solve for $$$q$$$-product $$$a_{i_1} \times a_{i_2} \times \dots \times a_{i_q} = x^q$$$ (for given constant $$$q$$$) ?

N: Given $$$0 \leq \delta \leq n$$$, can we also solve the problem when $$$1 \leq a_1 \leq a_1 + \delta + \leq a_2 \leq a_2 + \delta \leq \dots \leq a_k \leq n$$$ ?

O: What if the condition is just two nearby elements and not all pairs. Or you can say $$$a_i \times a_{i+1} \forall 1 \leq i < n$$$ is a perfect square ?

Теги combinatorics

История

 
 
 
 
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