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# PRMDIV - Editorial

Practice

Contest

Author: Ivan Fekete

Tester: Misha Chorniy

Editorialist: Bhuvnesh Jain

EASY

# Prerequisites

Sieving Techniques

# Problem

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 $S(A[j])$.

# Explanation

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 = 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]:
good[i].append(j)
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 AND TESTER'S SOLUTIONS:

Author's solution can be found here.

Tester's solution can be found here.

Editorialist's solution can be found here.

This question is marked "community wiki".

asked 28 Jul '18, 17:31 6★likecs
3.7k2481
accept rate: 9% 2.7k1618

1

Author's solution link is wrong. It should be for the tester. Correct one is : https://ideone.com/b1dJXm

(29 Jul '18, 01:16) 6★

I was so close to a full AC in this one. I didn't realize we could precompute the good pairs and store them to speed up each test case. TLE on the absolute last test case :-(

(29 Jul '18, 01:31) 4★

 1 # 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 answered 29 Jul '18, 02:02 31●3 accept rate: 0% Correct. Check again. (29 Jul '18, 02:03) likecs6★
 1 @likecs could you please clear the linear sieve and an additional O(n) array approach? answered 29 Jul '18, 12:33 3★titin_ 13●3 accept rate: 0% Hint 2: Linear sieve stores smallest prime factor for every number. In the second loop, we will use a DP solution. Say we have computed answer till some point, how can handle transition for "x", say using the answer from "x / lp[x]" where lp[x] denotes lowest prime factor of "x" (29 Jul '18, 13:05) likecs6★
 0 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. answered 29 Jul '18, 01:46 2.7k●1●6●18 accept rate: 10% @aryanc403, think you. Hint: The answer is yes and instead of using only the linear sieve, we can use an additional $O(N)$ loop again. Do you get some hint or more is required? (29 Jul '18, 02:05) likecs6★
 0 I am sure I did the same, but still wasn't able to pass subtask 6. Could someone provide some pointers: https://www.codechef.com/viewsolution/19375192 Thanks. answered 29 Jul '18, 02:01 86●4 accept rate: 12% 2 Overflow error. Your soln https://www.codechef.com/viewsolution/19384172 . Changes made - pairs += a[i] * ((long long)a[k]); and pairs += a[i] * (a[i] - 1L); E.g. - Test case T=1; n=1000000; a[i]=3; ans ~ 1e12 > INT_MAX (29 Jul '18, 02:34) Congrats man you again proved that we should not use \% until and unless it is absolutely necessary. Your soln with removed if(a[k]) results in TLE. https://www.codechef.com/viewsolution/19384237 With this correct my soln (which I used in contest) similar to your now passes. (29 Jul '18, 02:40) @vijju123 I think there is some problem with counting upvotes on this answer. I upvoted but it is showing 0 upvotes. And his profile does not show any downvote. And If I un upvote this answer then it is showing -1 votes. (29 Jul '18, 02:43) Its always stuff like this! Thanks @aryanc403, I hope this will make me remember about overflow during int arithmatic. (29 Jul '18, 19:14)
 0 Does it EASY? :) answered 29 Jul '18, 02:20 126●4 accept rate: 5%

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 ;)
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 $O(NlogN)$ .

answered 29 Jul '18, 04:54 2.7k1618
accept rate: 10%

 0 @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]] ? answered 29 Jul '18, 11:40 4★radon12 36●1 accept rate: 50% Correct. Please check again. (29 Jul '18, 13:02) likecs6★
 0 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:- https://ideone.com/XUhDs1 answered 29 Jul '18, 15:14 2★rtrajat 1●1 accept rate: 0% Reason is already answered. Check this page again. (29 Jul '18, 16:47)
 0 Can someone point out what was wrong with my code (I was targeting for subtask 1) https://www.codechef.com/viewsolution/19376505 answered 30 Jul '18, 11:11 11●2 accept rate: 0% 1 for(int i=2;i<=sqrt(MAX);i++) => Mistake is here, this is enough to compute if a number is prime is or not but not enough to calculate S[x]. Think about counter examples. (30 Jul '18, 23:22) tamiliit3★
 0 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. answered 30 Jul '18, 21:19 78●6 accept rate: 7% ans += freq[i]*(freq[i] - 1) ; ans += freq[i]*freq[good[i][j]] ; Either use ans += 1LL*freq[i]*(freq[i] - 1) ; etc or use long long. (30 Jul '18, 21:29) okay got it...looks like someone already asked this question before thanks :)) (31 Jul '18, 18:52)
 0 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 answered 31 Jul '18, 00:36 36●4 accept rate: 6% https://www.codechef.com/submit/complete/19404615 Your AC code. A judicious and appropriate use of data structures is a must. Your map gave a heavy, extra $logN$ factor. Also, use arrays wherever possible. (31 Jul '18, 01:00) Okay, got it, Thanks :) (31 Jul '18, 14:18)
 0 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 answered 31 Jul '18, 02:15 533●1●12 accept rate: 3% waiting for reply... plz someOne help (31 Jul '18, 21:31) Because at any instance, only ${10}^{4}$ of the ${10}^{6}$ elements will be present in an array. This reduces time complexity. Further, if operations are basic,such as addition etc. then its easy to get ${10}^{8}-{10}^{9}$ operations done in a second. (01 Aug '18, 12:29)
 0 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. answered 31 Jul '18, 08:54 350●10 accept rate: 13% Its typo. Corrected it. (31 Jul '18, 18:56)
 0 https://ideone.com/LskyCZ I have been trying to solve for subtask 1 but not getting an AC. Please someone help. answered 01 Aug '18, 05:39 1 accept rate: 0%
 0 @likecs It will have 10^9 (approx) computations.How will it be completed in 1 sec? answered 02 Aug '18, 01:02 3★utchinu 0●1 accept rate: 0%

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]" ?

answered 02 Aug '18, 10:18 213
accept rate: 0%

 0 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. answered 02 Aug '18, 19:19 73●7 accept rate: 4%
 0 Can someone help me out ? I can't find reason why it still TLE Here is my code https://www.codechef.com/viewsolution/19452525 Thank you!! answered 04 Aug '18, 09:46 1 accept rate: 0%
 0 I tried implementing the solution in the editorial in Java. Its giving me TLE. Can someone please tell me why? Solution Link: https://www.codechef.com/viewsolution/19435752 answered 13 Aug '18, 10:10 2★svr8 1 accept rate: 0%
 0 In java you can implement editorial's method like this: https://www.codechef.com/viewsolution/19737889 (I was getting TLE if I pre-processed using Lists of Java) answered 16 Aug '18, 00:42 5★vjvjain0 91●2●10 accept rate: 6%
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question asked: 28 Jul '18, 17:31

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last updated: 16 Aug '18, 01:04