ANUGCD - Editorial

march14
medium
prime-factor
segment-tree

#1

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 * … * pi^ki * … * pn^kn. We can call all pi as X’s factors (of course, pis are all different primes). For example, 18 = 2 * 3^2. So we can say that 18, has factors 2 and 3. Because pi >= 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[pi], which records the positions of the numbers who has the factor pi. From the analysis above, we know that the sum of position[pi].size() is O(NlogN)
  2. Build several segment trees, the i-th one corresponds to the position[pi], 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.


#2

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


#3

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


#4

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


#5

@anudeep2011: Thank You for such a nice problem :slight_smile:

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 :slight_smile: 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.


#6

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


#7

@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


#8

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


#9

@Anudeep, Sir can you please tell where my code fails http://www.codechef.com/viewsolution/3622900


#10

@anudeep2011
@xellos0
@rodrigozhou
I did this question using the segment tree and prime factorisation and I am interested in the other methods for doing this problem mentioned by you above. So, I request you to explain your solution solution a bit more so that i and others can have benefit of that because it is quite difficult to understand the code directly.


#11

I used Segment Tree and sparse tree in separate solution. My program was running in under 2s for the given constraints, but i kept getting TLE. Can you plz look at my solution, http://www.codechef.com/viewsolution/3557742 http://www.codechef.com/viewsolution/3557556


#12

@anudeep2011 @rodrigozhou

Hi, I was going through the Binary Indexed Tree solution linked in the editorial. I could not quite understand how a BIT was used to perform range maximum query, specifically the following two blocks of update and query code:

void update(vector &bit, int x, int val) { while (x < bit.size()) { bit[x] = max(bit[x], val); x += x & -x; } }

int query(const vector &v, const vector &bit, int l, int r) {
int ret = 0;
l–;
while (r > l) {
int n = r - (r & -r);
if (n >= l) {
ret = max(ret, bit[r]);
r = n;
}
else {
ret = max(ret, v[r–]);
}
}
return ret;
}

I haven’t come across and do not know of a method to get max/min value in a range using BIT. The above method looks buggy to me too. Can you guys please explain the above approach?


#13

@sultanofswing

Just remember the usual query in BIT. It would be like this:

int query(const vector<int> &bit, int x) { int ret = 0; while (x > 0) { ret += bit[x]; x -= x & -x; } }

In the usual BIT query, bit[x] represents the accumulated sum in the range [x-(x&-x)+1, x]. That’s why you do “x -= x & -x”: you’re jumping to the next range not summed yet.

So, the idea to do a range query in a BIT is to check if the whole interval that bit[x] represents is contained in the range that I want.

In the header of the range query function, the parameters means: v is the array of values, bit is the BIT of v and the range query is for [l, r].

In line, “int n = r - (r & -r)” means that bit[r] has the accumulated sum of the range [n+1, r]. As I have already decreased the value of l, if n >= l, then [n+1, r] is fully contained in (l, r] and I can sum bit[r] to my answer. Otherwise, I cannot do this. So I just get the value set in the v[r] and advance r in just one index.

I used the sum function in the explanation instead of max, and I hope it is not a problem.


#14

@anudeep2011
Can’t We do solve this question using Mo’s Algorithm. I have tried it but getting TLE.
Here is my code:https://www.codechef.com/viewsolution/8398234


#15

Here is my solution if anyone wants to refer https://www.codechef.com/viewsolution/22490044 as I guess it’s pretty readable. Feel free to comment if anything isn’t clear. (PS: implemented after referring editorial :slight_smile:


#16

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


#17

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


#18

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


#19

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


#20

@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