ANUGCD - Editorial

[shangjingbo]

PROBLEM LINK:

Practice
Contest

Author: Anudeep Nekkanti
Tester: Mahbubul Hasan
Editorialist: Jingbo Shang

DIFFICULTY:

Medium

PREREQUISITES:

Segment Tree, Factorize

PROBLEM:

Given N numbers in a sequence, answer M queries about what is the maximum number between L and R and the greatest common divisor between it and G is greater than 1.

EXPLANATION:

Every number X can be written in the factorized form: X = p₁^k₁ * p₂^k₂ * … * p_i^k_i * … * p_n^k_n. We can call all p_i as X’s factors (of course, p_is are all different primes). For example, 18 = 2 * 3^2. So we can say that 18, has factors 2 and 3. Because p_i >= 2, the number of factors, i.e. n, is in O(logX). Therefore, the total number of factors of the given N numbers are O(NlogN) (the range of N and numbers are same).

The greatest common divisor of two numbers is greater than 1, means that they have at least one common factor. If we enumerate the common factor C they have, the satisfied numbers are determined – all numbers have factor C. After that, the only thing we need is to find the maximum in the query interval [L, R]. For this type of queries, an ordinary solution is to use Segment Tree.

With these two ideas in mind, let’s start to assemble the whole algorithm, now.

1. Factorize N numbers, and restore some vectors of position[p_i], which records the positions of the numbers who has the factor p_i. From the analysis above, we know that the sum of position[p_i].size() is O(NlogN)
2. Build several segment trees, the i-th one corresponds to the position[p_i], maintaining the interval maximum in the tree node. Of course, you can also concate all position and make a whole segment tree.
3. For a query number X and the interval [L,R], first factorize X. And for each factor, look up in the corresponding segment tree (or corresponding intervals, if you choose to build a whole segment tree) to get the maximum. Finally, take the maximum of the query results among different factors.

As analyzed before ,X has at most O(logX) factors. And each interval-maximum query takes only O(logN) time. Therefore, to answer a query, our algorithm only needs O(log^2 N) time.

In summary, the time complexity is O(NlogN + Qlog^2N), while O(NlogN) space needed.

As requested by many users here are solutions without segment trees.
Sqrt-Decomposition.
Binary Indexed Tree.
STL-MAP

AUTHOR’S AND TESTER’S SOLUTIONS:

Author’s solution can be found here.
Tester’s solution can be found here.

[xorfire_1]

Is there an alternate solution that does not involve the usage of segment tree ?

[shangjingbo]

One possible way is sqrt decomposition, but I think it may get TLE. Let me further think about it.

[xorfire_1]

I tried sqrt decomposition [ http://www.codechef.com/viewplaintext/3616120 ], it got TLE.

[ranjanvittal]

Does precomputation of the first 100001 numbers’ factors give tle? While implementing sieve I stored the factors of all the 10^5 elements in a vector.Rest procedure is same as given above.still got tle.http://www.codechef.com/viewsolution/3536122

[knb_dtu]

@anudeep2011:Can we solve this using Binary Indexed Trees?

[atulsehgal]

I have precomputed the factors and it can be done in linear time. Although my solution involves offline computation of answers. You can have a look at my approach here

[ashish1294]

@anudeep2011: Thank You for such a nice problem

About the pre computation part, it is just like sieve, and can be easily done in linear time. It will actually save a lot of time while processing queries. Factoring is not a problem for this question, any method will work As anudeep said, time limit was not strict if algo is right.

My approach was - making 9592 (no of Primes less than 100000) segment trees, each for a single prime. I then inserted the all multiples of the prime in its respective seg tree along with it’s index. When a query arrives, find its respective indices in all the seg trees which divides G (at max 6 factors so only 6 query in 6 diff trees). Find sol of respective tree and then combine.

I wonder if its possible to do this sum using sqrt decomposition. I have not implemented this but I think it’s possible.

[junior94]

I actually had to do precomputation in order to get AC. Without it was just getting TLE.

[rodrigozhou]

@xorfire_

I did with BIT. I change the query function to query in a range, but it runs in O(log^2 n).

Here is my solution: http://www.codechef.com/viewsolution/3533495

[xorfire_1]

@rodrigozhou Thanks.

[pawan55]

edited but still getting wa someone plss help heres the modified link: http://ideone.com/5EWVUQ

[anudeep2011]

Yes.
Here is my SQRT-Decomposition solution http://www.codechef.com/viewsolution/3620495

And here is a solution using BIT http://www.codechef.com/viewsolution/3533495

[anudeep2011]

Yes. Here is my SQRT-Decomposition solution http://www.codechef.com/viewsolution/3620495

And here is a solution using BIT http://www.codechef.com/viewsolution/3533495

[pawan55]

@anudeep2011 :sir plss tell me why im getting wa although the code seems to be working right on all the test cases i can think of
http://ideone.com/5EWVUQ

[anudeep2011]

It is failing on a very large test case (n = 10^5)
For the query “63001 54725 59725” while the answer is “63001 11” your code last submission on codechef is giving “63001 22” as answer.

[xorfire_1]

@anudeep2011 Thanks. Missing this problem will haunt me and my future generations for eternity.

[pawan55]

got AC thnx

[ashish1294]

Much Simpler Solution

[architmittal]

anudeep sir can you please explain how you handled the prime>350 part in your segment tree(authors) solution