ZCO 2015 Discussion

Solution to rectangles. Disclaimer: I gave the morning session so I am not 100% sure
1.Sort all coordinates by y- coordinate
2.Pick the one with the lowest y coordinate.
3.The area covered by this rectangle is y*width, width = 10000 for 1st case.
4. Add its x coordinate to a fresh list.
5.Now we have two subproblems, One right to the marked x-coordinate and One left to the marked coordinate
6.Repeat the procedure from 2 till here, keeping in mind that the limits have changed for the width. Also if there are two coordinates with same y add both of their x coordinates to the maintained list.
Has anybody got 100% in this question ? I want to see their code badly. This was a very good question I wish it was in the morning. Our session was boring. I was doing it half minded.

Well there is a catch here… The vertices can lie on the edge of the rectangle. Moreover if you add point one by one in a order, you may not count a newly created area that you assume you have. So DP seems out of question
(I initially thought that we redistribute the area disturbed by new points (added by ascending y coordinate) but failed))

It seems I am doing something wrong here. I meant the area which you get from bottom top approach

The morning session was way easier.
One was a DP and the other was really also easy.

Afternoon the covering was a greedy.

But the rectangles one
… Ugh it was so bad.
I unfortunately gave the afternoon session and got only 1.

Wish I’d given the morning one.
I really I hope I get through :confused:

I had the option to choose between the morning and the afternoon session. I chose the afternoon session. And now I absolutely regret it. T_T I spent the whole three hours sending numerous solutions modifying minute things and tweaking but nothing worked. :frowning: And the morning session problems do seem a lot easier. At least that’s what I feel after reading the problem statements.

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I did the cover question very easy algorithm . got it in just last minutes . was doing a very bad bad error.12 attempts to reach a 100
Shit man seriosly should have given the morning question . Altough i was happy afternoon session had an adhoc problem .

Dont you think that the problems that appeared in the second session were many times more difficult than that of the first half. I think there ought to be more fairness in the evaluation procedure. Infact I think we ought to file a petetion against this. Please mail all your grievances to ico@iarcs.org.in also, we should get together and do something about it. All those who are like minded please email me: soumadeep.saha97@gmail.com
Thank you
A victim of ZCO afternoon session

@deepcoder18: You are right!

The cutoffs won’t be different for afternoon and morning sessions right?

I gave the afternoon session … @Organic-Shilling For Rectangles, I tried exactly the same method that you posted but it gave me correct in only two sub-test-cases so my total score was zero. I was pretty sure I got covering right, but here, my solution must have been far too long as I got TLE in all except one, which was the very first one which gave correct.

Covering can be solved with a greedy approach where for every interval you check whether there is any number that is already picked in some earlier chosen interval such that it lies between the current interval and if not you always add the rightmost number of the interval to the set of picked numbers.
Complexity: O(n^2)

P.S Sort the intervals in increasing order of its rightmost number.

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Does anyone know when the results are going to be out?

This is my code for variation . It fetched me 100 pts in the grader but i want to know whether its

full efficient or not .( PLEASE COMMENT )

#include

#include

using namespace std;

int main()

{

ios::sync_with_stdio(0);

int n,k;

cin>>n>>k;

int a[n];

for(int i=0;i<n;++i)

cin>>a[i];

int c=0;

sort(a,a+n);

for(int i=n-1;i>=1;i–)
{

for(int j=i-1;j>=0;j–)

{

if(a[i]-a[j]>=k)

{

c=c+j+1;

break;

}

}

}

cout<<c;

return 0;

}

guys this was my code for covering .i got 100 points (till now) dont now future .I can explain my code if theres a problem .
please tell me if any case is coming wrong

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I think the cutoff for both morning and afternoon would be 100/200 .

I took the morning session and got 100 in Variation.
I ran out of time for Breakup.
Let’s see whether 100 would be enough to qualify.

Yay! I got an email saying I had qualified! I got 100.

@potatio Can you please confirm if you took part in the morning session or the afternoon one? I had a full score in the morning session, but haven’t received anything yet.

Took part in the morning session. You will definitely receive one then.

Its definitely very hard. I wouldn’t have been able to complete it. Although Rectangles is pretty easy. Would have done that in 30 minutes. Better Luck next time unless you get selected via ZIO.

@Organic-Shilling How did you go about rectangles? I couldn’t think of a good algo :frowning: