I asked someone to explain my div1B solution to me in this comment

And you actually did that, thanks :D.

SirShokoladina please fix div2E. There is parsing error.

A lot of things to learn...

C shows up as "unable to parse markup" for me.

Can you please provide code to the solutions, especially Div 2D/1B... really hard to understand the solution for me. Thanks a lot

Also, Div 2E/1C problem is missing it's solution

I don't understand div2 C part where it says that answer can be no more than n-x. Can somebody explain me little bit better?

Great Contest! With mind blowing problems. SirShokoladina Sir, you should conduct contests more often.

LOL I did some kind of precalc on div1B. 40296044. My solution has complexity . But I can't implement it without map which contains vectors so it is very slow. Then i noticed that it can be precalculated and instead of 4³ * log with bad constant it gives clear 4³.

I think for the second solution to C, you mean to say that X and Y are the minimal possible values satisfying that condition (but the solution works).

Overall, I thought the logistics of the contest could have been managed better. Some of the problem statements could have been worded better, there were a lot of clarification announcements, etc.

That being said, I thought the problem quality was good, especially Div 1 C. If Div 1 BC were shifted down 1 problem (can't speak towards D or E), and another Div 1 B made, I think the problem scaling would have been perfect.

Can't understand the second solution for Div1.B. The solution says:

"In such way every satisfying this criteria parallelepiped (not considering the orientation) with three different side lengths was counted 6 times."

Let's consider the large parallelepiped to be 4x6x10, the small parallelepiped 2x3x5 will only be counted once. Why the solution says that it is counted 6 times?

(Actually, can't understand the first solution either... What does "mask" mean in the solution?)

so my understanding of div2D was correct but problem statement and number of submissions prevented me,huh,sad

can anyone help fix my mistake for div2 B?

Submission Link

?detaR tI sI

Why did my code of div1C which used "exit(0);" get "Idleness Limit Exceeded" in pretest 12?

Can Someone Please tell me a more intuitive way to understand the solution of div2 C (div1 A) or different easier approach if possible???

Can someone please elaborate on the Div1B Second solution? (The Inclusion exclusion one) ...?!

It is not very clear to me how the "ans2" part is calculated by the author's solution...

If you had a different combinatorial solution please share that as well ..

It seems that SirShokoladina forgot to link the editorial to the problems in Div2 (contest 1008). In any of the five problems we can't see the link of editorial.

Can anyone explain div2d in understandable way?

For div2C does cycle decomposition work if the permutation contains repeated elements? I mean for the proof it is assumed that there is a cycle decomposition?

Haha, I have a passing O(number of divisors * constant) solution of div1B. It's only barely passing though.

We can build bitmasks for each divisor describing which input numbers it divides. The key is that for each triple of bitmasks corresponding to the smallest, middle and largest number, we can pick an arbitrary assignment of dimensions to these three bitmasks (or ordering of three chosen divisors). Then, for each query and each ordering of divisors (6 in total), we can process the divisor list corresponding to the middle number, use 2 pointers to keep the number of divisors from the divisor list corresponding to the smallest number (similarly for the largest) that are smaller than the current divisor from that middle list, then check all pairs of bitmasks (smallest, largest number) that are valid for that current divisor's bitmask and the current ordering; since we have the number of ways to choose the remaining two divisors, computing the answer is straightforward.

The advantage of this approach is that there's no need to think about what to add, what to subtract, which orderings to use etc. It's all just iterating over initially arbitrarily chosen things. The obvious disadvantage is that it's slow, I needed some optimisations in the last test (namely "if all pairs of bitmasks give 0, skip innermost loop").

anybody got any implementation for Div1C second solution?!

Hello, I have a question about my algorithm on div 2D/ div 1B.

First, I made some observations (not sure if they are correct):

1. In order to divide the large parallelepiped into smaller (and equal) ones, we need each dimension of the small to be divisible by the corresponding dimension of the large.

2. By adhering to the above observation, we enumerate a lot of duplicate parallelepipeds. For example, the parallelepiped (2, 2, 2) can be split into the following ones:

(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)

but, in reality, we only need the (1, 1, 1), (1, 1, 2), (1, 2, 2) and (2, 2, 2).

So, I have thought that we can just create the dimensions in increasing order. That means if we have a tuple (a, b, c), we have the constraint: a <= b <= c.

Now, on the programming part:

For each query,

1. I find the divisors of a, b and c (the dimensions of the large parallelepiped).

2. I find the number of all the possible combinations of the divisors I found above such as if I have the (a, b, c) tuple, I accept only the ones where a <= b <= c. On this step, I do some optimizations in order not to trigger a TLE.

Here is my code for reference:

http://codeforces.net/contest/1008/submission/40527094

(it is a bit unorganized, but, on a top level I implement the above steps)

So, my question is what am I doing wrong?? It is difficult for me to make sense of the editorial and compare my solution to the ones provided in order to find my mistake!

Thanks in advance! Any help is highly appreciated!

Hi, can anyone please help me with my div2C submission, it is getting error on test case 8

#include<bits/stdc++.h>
using namespace std;

int main(){
	int n;
	cin>>n;
	std::vector<int> v(n);
	for(int i=0; i<n; i++)
		cin>>v[i];
	sort(v.begin(),v.end());
	int k=0;
	for(k; k<n; k++)
		if(v[k]>v[0])
			break;
	std::vector<int> num(v.begin()+k,v.end());
	int count=0;
	for (int i = 0; i < num.size(); ++i)
	{
		if(num[i]>v[i])
			count++;
		//cout<<num[i]<<" ";
	}
	cout<<count;
}

1007A — Reorder the Array — this problem can be solved by using two pointers . Here is my solution Link