Given an array,how to solve queries (10^5) of type Q:Value,L,R where value is a number and we need to report the count of all the numbers in the range [L,R] of the array such that gcd(Value,A[i])>1 where L<=i<=R. Given ARRAY SIZE :- 10^5 Each number 1<= A[i] <=10^5 TL = 1 Sec

What are the constraints for the number of elements and the values (

a_{i})?Sorry for this ,the Value of each 1<= A[i] <=100000 & Array size <=10^5

I'd like to know more details (time limit, amount of numbers in array and constraints on the numbers), but anyway, I'll share the solution that came to mind.

^{9}has more than 1344 divisors, so we can find divisors of every element of array and store a list of positions that divide every number. This may or may not fit in memory or time, depending on the constraints. This step is , whereMis the maximum value an array element can take.Value. There are at most 9 such primes, because the product of the first 10 primes is $>10^9$. Now for every possible mask of primes, check how many positions of range [L,R] divide their product with a binary search and use inclusion-exclusion principle. This process isO(2^{9}*_{log}N) for each query, making the whole programO(Q* 2^{9}*_{log}N). This, again, may or may not fit in time depending on the constraints.TL = 1 sec Array size is 10^5 And Each number is 1<=A[i]<=100000

Then it should work. of the first step is OK for

M≤ 10^{5}is OK. Now the second step has less complexity because the number of prime factors is at most 6, so it'sO(Q* 2^{6}*_{log}N), which is also OK.Do you have an online judge where I can test the solution?

This was a subproblem of some other problem (tree based) .The link to the problem is https://www.codechef.com/problems/GTREE

I think codechef problem is much easier than the one mentioned by you. Solution is similar but we can solve it without binary search just doing dfs and storing for each p how many numbers in current subtree are divisible by p.

Can you explain this " store a list of positions that divide every number"?

Maybe after the CodeChef xD

Ok.No problem.

.

How would you perform binary search in [L,R] to count no of positions?

can u elaborate how you are doing the binary search to obtain count of elements in [L,R] which divisible a certain mask of primes?

Auto comment: topic has been updated by vidit_123 (previous revision, new revision, compare).this problem is from codechef march long challange..

it still running dude, don't cheat

look when the blog was published :)

I think this is a problemsetter's codeforce account...

Exactly.Thank you @Dalgerok.

At first generate a list of primes that divide a number, for all number till

M, the maximum number. This takes .Now, for each number of the array

a_{i}, lets say the primesp_{1},p_{2}, ...,p_{k}divide that number, you can get this quickly from step 1. Go through all subsets of these primes. If a subset isp_{i1},p_{i2}, ...,p_{ir}, then insertito listpos_{pi1·pi2... pir}. This step takesO(2^{6}n).Now when you receive a query

X, take the primesp_{1},p_{2}, ...,p_{k}, that dividesX, from the list generated at first. Now go through all of its subsets. Let the chosen subset bep_{i1},p_{i2}, ...,p_{ir}, then count how many times a number from [l,r] appears inpos_{pi1·pi2... pir}. It can be done with two binary searches. Ifris odd then add the count to answer, else subtract this count. It can be proven that this gives the answer, by inclusion-exclusion principal. Basically this counts number of numbers with at least one prime common with the queryX. This step takestalk is cheap, send code

Which part you didn't understand?

talk is cheap, send code

Code is costly, spend time

o, really?

Ok, you can generate the list of prime divisors this way —

As , this runs in .

talk is cheap, send full ACed code

Send problem link.

There you go: http://goo.gl/b8D7F

There you go: http://goo.gl/b8D7F

What about the update in array element in the query?

I don't see any mention of 'update' in this post. What are you talking about?

Its my personal query.

Lets say you change

a_{i}tox. First you need to reverse all changes thata_{i}did, that means, go through all subsets of it's prime divisors and deleteifrom the list of their product. Then changea_{i}tox, and again add it back.But as you need to keep the lists sorted to do binary search later, you can use balanced binary search trees, or just the policy based data structure in C++.

If you have updates, then both query and update become

For those who are wondering who is "CodeChefFuckYou": please upvote this comment. Unless it has less than 100 upvotes I won't say his username(only that he is very good on Codeforces and you'll be surprised ^__^). He has a unique style of posting comments.

For most interested: "&BYK&FM" — encrypted username, good luck!!

Why are you upvoting this post? It is cheating, and this post should be deleted immediately, I think because it violates CodeChef rules and principles of the competitive programming.

This blog post was created 6 months ago. However the current interest in this post can be attributed to CodeChef's problem.

This blog has nothing to do with CodeChef, the thing should be deleted is CodeChef problem. CodeChef should really care about problems of long challenge, because in 10 days even we can find code of problem from internet.

Sorry, I missed this fact. So it's good that later comments are downvoted and unclarified.

You also can do it with bitset. numbers less than 10^5 has less than 7 prime divisors. So for every primes less than 1e5 set the bit of positions of those numbers which are divided by.

GCD(x, a[i]) has no common prime divisors If their GCD is 1. you can travel through prime divisors and check how many numbers you can divide in the given range. then the ans might be (r-l+1- NumberofSetBit).

you can solve it easily than other technique.