PRMDIV - Editorial

Problem Link



Author: Ivan Fekete

Tester: Misha Chorniy

Editorialist: Bhuvnesh Jain




Sieving Techniques


Let S(x) denote the sum of distinct prime divisors of x. Find the number of pairs (i, j) in array A such that if A[i] divides A[j] then S(A[i]) also divides S(A[j]).


Let us first calculate the function S for all integers efficiently. This is a simple modification of the prime sieve where we add the contribution each prime divisor to the numbers as well.

	S[1] = 0
	for i in [2, 1000000]:
		if S[i] == 0:
			# number is prime
			j = i
			while j <= 1000000:
				S[j] += i
				j += i

The time complexity of the above approach will be O(\sum_{p = prime} X/p) = O(X \log{\log{X}}), where X = 1000000.

Let us also calculate the good pairs of numbers. Below is a pseudo-code for it:

	# good[i] stores the "j" such that
	# "i" and "S[i]" divide "j" and "S[j]" respectively.
	for i in [2, 1000000]:
		j = i
		while j <= 1000000:
			# Ensured that "i" divides "j". See the loop conditions.
			if S[j] % S[i]:
			j += i

The time complexity of the above approach is O(\sum_{i=2}^{i=N} X/i) = O(X \log{X}), where X = 1000000. If you have any doubts regarding the above 2 precomputations, I suggest you to learn and read Sieve of Eratosthenes and try to understand how the code is modified here.

Now, coming back to the problem. We need to find the number of pair of (i, j) in array A such that if A[i] divides A[j] then S(A[i]) also divides S(A[j]). For this, if we do it naively for every pair using the help of above pre-computation to check if A[i] and A[j] is good, then it will inefficient and only pass subtask 1.

The important thing to note that even if we iterate over all “good” arrays, the total size of all arrays is bounded by O(X \log{X}), considering all the numbers are appended. Actually, this bound is lower, but it doesn’t matter.

So, let say if we have the frequency of all elements in the array A. When iterating over the “good” array, if 2 numbers have frequency x and y then they contribute (x * y) to the answer. The only caveat is that pair (i, i) i.e. same pair of indices is also counted in it. So, we need to subtract N from the final answer as well.

Thus, the below psuedo-code can easily solve our complete problem:

	for i in [1, n]:
		freq[a[i]] += 1
	ans = 0
	for i in [2, 1000000]:
		if freq[i] > 0:
			# Element is present
			for j in good[i]:
				ans += freq[i] * freq[j]
	# Account for i == j
	ans -= n
	print ans
	# Clear freq array for next test case.
	for i in [1, n]:
		freq[a[i]] -= 1

Once, you are clear with the above idea, you can see the editorialist implementation below for help.

Feel free to share your approach, if it was somewhat different.

Some Thoughts

There exist a linear sieve for prime numbers as well. Can you modify the first algorithm for pre-computation of S to work in linear time? If yes, how? If no, what is your counter argument?

Time Complexity

O(X \log{X} + N) per test case.

Space Complexity

O(X \log{X})


Author’s solution can be found here.

Tester’s solution can be found here.

Editorialist’s solution can be found here.


We cannot use linear sieve for pre-computation of S. Because linear sieve makes sure that each n is visited by it’s lower prime no. And is only visited once. But For computation of we have to visit each distinct prime factor of no. And there are atmost 6 distinct factors for nos upto 1e6.

I am sure I did the same, but still wasn’t able to pass subtask 6. Could someone provide some pointers:


1 Like

Clear freq array for next test case.

for i in [1, n]:
    freq[a[i]] += 1

This should be freq[a[i]] -= 1. Correct it

1 Like

Does it EASY? :slight_smile:

Thanks @fr4nkesti3n ,

My AC approach here -
I was just trying luck along with double sieve. Hence I used a variable named lucky. xD. In case I get AC this is reason behind it.

Approach -

Sieve to find S[i]

Maintain Counter Array.

Used Map. Although Set is more appropriate then Map :wink:

Brute force for 1<=a[i]<=lucky. With lucky<=100 Atmost 100 iterations.

For a[i]>lucky. One loop for 1<=i<=lucky and rest sieve. Atmost 200 iterations.

But still in TL .

Unfortunately it failed on last test case. And I made 15 submissions with different values of lucky with hope of getting AC.

Finally I got AC after seeing @fr4nkesti3n soln.

And bingo it resulted in A.C. Soln.

Never use % until and unless it is absolutely necessary.

I repeat Never use % until and unless it is absolutely necessary.

It can take you from 100 pts to 20 pts And you will never realise reason behind it.

He added condition if(cnt[j]) then only compute s[j]\%s[i]

Editorials soln can be optimised further by using set/map for outer loop. By in that case insertion complexity will be O(logn) And overall complexity will be again O(NlogN) .


@likecs in the pseudocode to calculate the final answer

for i in [2, 1000000]:
        if freq[a[i]] > 0:
            # Element is present

Should’nt it be freq[i] instead of freq[a[i]] ?

@likecs could you please clear the linear sieve and an additional O(n) array approach?

1 Like

Hello everyone,I m getting TLE in last test cases and i m not able to figure out the reason…please help me out.
Here is the link to my solution:-

Can someone point out what was wrong with my code (I was targeting for subtask 1)

After trying to understand it for 10hrs and after doing nearly 15 submissions, I am still struggling to get AC.

please check my PA Solution here and tell me what I am missing.

I tried the way as mentioned in the editorial, but am getting TLE on the 1st, 2nd and last Test Cases. I find my and ediorialist’s codes almost similar but couldn’t figure out why I am getting TLE. Is the O(MAX^2) complexity in every precalculation is causing TLE.

Secondly, I tried to sieve only up to the maximum element in the input array. It decreased the time of execution but all tasks gave WA.(Some still gave TLE)

My code _ TLE One

My code _ WA One

Time Complexity:- O(XlogX+N) per test case, where X=1000000
On Computing time, it is similar to 2*10^7
and there can be 100 test cases at maximum, when i am assuming N=10^4
then total time will be over >10^9 still no TLE according to editorial… why? please SomeOne Explain where I am wrong

In the last section psuedo-code line no. 5 if freq[a[i]] > 0: should it not be changed to freq[i] > 0: here is the AC Code.

I have been trying to solve for subtask 1 but not getting an AC. Please someone help.

@likecs It will have 10^9 (approx) computations.How will it be completed in 1 sec?

Can someone please explain, why the author is

ensuring that “i” divides “j” ?

when he is calculating good[i] array,

does he mean if “i” doesn’t divide “j” => “s[i]” doesn’t divide “s[j]” ?

Need Help !!! Since I am a beginner I don’t understand How Counting is performed to clear subtask 2. I try to do the dry run in my rough copy but was not able to solve due to the huge size of the array. Can u plz tell me where I am lacking and which concept should I study to understand this. You can also share some link.

Can someone help me out ? I can’t find reason why it still TLE
Here is my code
Thank you!!

I tried implementing the solution in the editorial in Java. Its giving me TLE. Can someone please tell me why?
Solution Link: