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.
Extra Tasks
Solved A: Can we also use phi function or something similar to solve for $$$k = 3$$$ in $$$O(\sqrt{n})$$$ ?
Solved B: Can we also use phi function or something similar to solve for general $$$k$$$ in $$$O(\sqrt{n})$$$ ?
Solved C: Can we also solve the problem where there can be duplicate: $$$a_i \leq a_j\ (\forall\ i < j)$$$ and no longer $$$a_i < a_j (\forall\ i < j)$$$ ?
Solved D: Can we solve the problem where there is no restriction between $$$k, n, p$$$ ?
Solved 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 solve for $$$q$$$-product $$$a_{i_1} \times a_{i_2} \times \dots \times a_{i_q} = x^q$$$ (for given constant $$$q$$$) ?
M: Given $$$0 \leq \delta \leq n$$$, can we also solve the problem when $$$1 \leq a_1 \leq a_1 + \delta + \leq a_2 \leq a_2 + \delta \leq \dots \leq a_k \leq n$$$ ?
*Marked as solved only if tested with atleast $$$10^6$$$ queries
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
Complexity
So about the complexity....
For the implementation using factorization, it is $$$O(n \log n)$$$.
For the 2 implementations below, the complexity is linear.
For the last implementation, the complexity is Linear
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.
A better solution for k = 2
Idea
In the above approach, we fix $$$u$$$ as a squarefree and count $$$p^2$$$.
But what if I fix $$$p^2$$$ to count $$$u$$$ instead ?
Yet you can see that the first loop now is $$$O(\sqrt{n})$$$, but it will still $$$O(n)$$$ total because of the second loop
Approach
Let $$$f(n)$$$ is the number of pair $$$(a, b)$$$ that $$$1 \leq a < b \leq n$$$ and $$$(a, b, n)$$$ is a three-term geometric progression.
Let $$$g(n)$$$ is the number of pair $$$(a, b)$$$ that $$$1 \leq a \leq b \leq n$$$ and $$$(a, b, n)$$$ is a three-term geometric progression.
Let $$$F(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f(p)$$$.
So it is no hard to prove that $$$g(n) = f(n) + 1$$$.
This interesting sequence $$$g(n)$$$ is A000188, having many properties, such as
- Number of solutions to $$$x^2 \equiv 0 \pmod n$$$.
- Square root of largest square dividing $$$n$$$.
- Max $$$gcd \left(d, \frac{n}{d}\right)$$$ for all divisor $$$d$$$.
Well, to make the problem whole easier, I gonna skip all the proofs to use this property (still, you can use the link in the sequence for references).
$$$g(n) = \underset{d^2 | n}{\Large \Sigma} \phi(d)$$$.
From this property, we can solve the problem in $$$O(\sqrt{n})$$$.
Yet this paper also takes you to something similar.
Implementation
A better solution for general k
Extra task A, B
Algorithm
Let $$$f_k(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 < a_2 < \dots < a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression.
Let $$$g_k(n)$$$ is the number of set $$$(a_1, a_2, \dots, a_k, n)$$$ that $$$1 \leq a_1 \leq a_2 \leq \dots \leq a_k \leq n$$$ and $$$(a_1, a_2, \dots, a_k, n)$$$ is a $$$(k+1)$$$-term geometric progression.
Let $$$F_k(n) = \overset{n}{\underset{p=1}{\Large \Sigma}} f_k(p)$$$.
Let $$$s_k(n)$$$ is the number of way to choose $$$p^2$$$ among those $$$k$$$ numbers when you fix squarefree $$$u$$$ (though we are doing in reverse).
Implementation
Complexity
The complexity of the first implementation is $$$O(\sqrt{n} \log \sqrt{n})$$$
The complexity of the second implementation is $$$O(\sqrt{n} \log \log \sqrt{n})$$$
Solution for duplicates elements in array
Extra task C
Idea
It is no hard to proove that we can use the same algorithm as described in task A, B or in original task.
Using the same algorithm, the core of calculating is to find out the number of non-decreasing integer sequence of size $$$k$$$ where numbers are in $$$[1, n]$$$.
Can you proove it ?
Now it is done, just that it
The idea is the same as what clyring described here but represented in the other way
Implementation
Complexity
In the first implementation it is obviously linear.
The second and third implementation is also easy to show its complexity
Sadly, since $$$k \leq n$$$. We also conclude that the complexity is $$$O(n)$$$, and even worse it also contains large constant factor compared to that in the first implementation.
But it is still effecient enough to solve problem where $$$k$$$ is small.
Solution when there are no restriction between k, n, p
Extra task D
Idea
So first of all, the result do depend on how you calculate binomial coefficient but they are calculated independently even if you can somehow manage to use the for loop of binomial coeffient go first.
Therefore even if there is no restriction between $$$k, n, p$$$, the counting part and the algorithm doesnt change.
You just need to change how you calculate binomial coefficient, and that is all for this task.
Let just ignore the fact that though this need more detail, but as the blog is not about nck problem I will just make it quick
For large prime $$$p > max(n, k)$$$
- Just using normal combinatorics related to factorial (since $$$p > max(n, k)$$$ nothing will affect the result)
- For taking divides under modulo you can just take modular inversion (as a prime always exist such number)
- Yet this is standard problem, just becareful of the overflow part
- You can also optimize by precalculating factorial, inversion number and inversion factorial in linear too
For general prime $$$p$$$
- We can just ignore factors $$$p$$$ in calculating $$$n!$$$.
- You also need to know how many times factor $$$p$$$ appears in $$$1 \dots n$$$
- Then combining it back when calculating for the answer.
- If we dont do this $$$n!$$$ become might divides some factors of $$$p$$$.
- By precalculation you can answer queries in $$$O(1)$$$
For squarefree $$$p$$$
- Factorize $$$p = p_1 \times p_2 \times p_q$$$ that all $$$p_i$$$ is prime.
- Ignore all factors $$$p_i$$$ when calculate $$$n!$$$.
- Remember to calculate how many times factors $$$p_i$$$ appear in $$$1 \dots n$$$.
- When query for the answer we just combine all those part back.
- Remember you can just take modulo upto $$$\phi(p)$$$ which you can also calculate while factorizing $$$p$$$.
- Remember that $$$n!$$$ must not divides any factor $$$p_i$$$ otherwise you will get wrong answer.
- By precalculation you can answer queries in $$$O(\log p)$$$
For general positive modulo $$$p$$$
- Factorize $$$p = p_1^{f_1} \times p_2^{f_2} \times p_q^{f_q}$$$ that all $$$p_i$$$ is unique prime.
- We calculate $$$C(n, k)$$$ modulo $$$p_i^{f_i}$$$ for each $$$i = 1 \dots q$$$.
- To do that, we need to calculate $$$n!$$$ modulo $$$p_i^{f_i}$$$ which is described here.
- To get the final answer we can use CRT.
- Yet this is kinda hard to code and debug also easy to make mistake so you must becareful
- I will let the implementation for you lovely readers.
- Yet depends on how you calculate stuffs that might increase your query complexity
- There are few (effective or atleast fully correct) papers about this but you can read the one written here
Implementation
Complexity
In the first implementation it is obviously linear.
And for the second implementation.
So you got $$$O(n \times \log p + \sqrt{p})$$$ in final.
Though you can still optimize this but by doing that why dont you just go straight up to solve for non squarefree $$$p$$$ too ?
Solution when numbers are also bounded by negative number
Extra task E
Idea
Yet this is the same as extra task C where only the counting part should be changed.
As we only care about integer therefore let not use complex math into this problem.
If there exist a negative number and a positive number, the product will be negative thus the sequence will not satisfied.
Becareful, there are the zeros too.
When the numbers are all unique, or $$$-n \leq a_1 < a_2 < \dots < a_k \leq n$$$
Thus give us the formula of $$$task_E(n, k) = 2 \times task_B(n, k) + 2 \times task_B(n, k - 1)$$$.
Remember that when $$$k = 0$$$ the answer is $$$0$$$ otherwise you might somewhat having wrong result for negative number in binomial coefficients formula
So what if I mix the problem with task C too ?
When the numbers can have duplicates, or $$$-n \leq a_1 \leq a_2 \leq \dots \leq a_k \leq n$$$
Yet once again you can simplified it with less cases for easier calculation.
Thus give us the formula of $$$task_E(n, k) = 1 + 2 \times \overset{k}{\underset{t = 1}{\Large \Sigma}} task_B(n, t)$$$.
But this give you a $$$O(k)$$$ solution.
You can do better with math
Implementation
Complexity
So.. you might be tired of calculating complexity again and again for those are too familar to you.
So I gonna skip this as the proof is as the same as what you can read above.