Arpa's blog

By Arpa, history, 6 months ago, In English,

Hello!

Welcome to June Easy '19, from our easy and educational contest series. It's a 3-hour competition with six algorithmic tasks. We are going to hold it on Sunday at 16:00 GMT. Check contest page for more details.

I helped Danylo Danylo99 Mocherniuk, Mohammad-Mahdi Kerpoo Taheri, and, Dishant Dishant_18 Trivedi setting this round. AmirHossein amsen Pashaee and I are testers of the contest. As usual, here are the prizes for the top three contestants:

  • $75 Amazon gift card
  • $50 Amazon gift card
  • $25 Amazon gift card

Note that prizes and T-shirts are given to top participants with ratings < 1600 (beginners).

GL & HF.

Let's discuss the problems after the contest!

P. S. Sorry for problems occurred. Now everything fixed.

 
 
 
 
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6 months ago, # |
  Vote: I like it +27 Vote: I do not like it

Why there is no leaderboard? Where can I ask clarification?

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6 months ago, # |
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Can anyone just clarify the problem "Minimum Cost"? The statement is not doing much justice. Just need clarification nothing else.

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6 months ago, # |
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In prob 5, why does sample contain 8 rows while in the input section described 6 rows? Also, it states to output answer with exactly 5 digits while sample is with 6 digits after dot.

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    6 months ago, # ^ |
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    Sorry, please refresh the page.

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      6 months ago, # ^ |
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      I refreshed the page and all my submissions disappeared. Should I try to solve problems or give up? :)

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        6 months ago, # ^ |
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        I don't know about the back-end. I'll ask them.

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        6 months ago, # ^ |
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        Fixed.

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6 months ago, # |
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Please, add limitations for N in problem 4.

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6 months ago, # |
  Vote: I like it -18 Vote: I do not like it

Not easy contest. Please next time only give name June contest

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Why there are no explanations in the editorials?

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6 months ago, # |
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Can someone explain the use of XOR in Mosaics and holes. Or any other method to solve.

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    6 months ago, # ^ |
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    Take update array of size M and vector <pair<int,int>> v[N]. By traversing for each index starting from top left, take current position as i,j.

    case 1. if value at this position is 0 and no. of total updates is even at this position so, it means value is still zero here. To change it,we will flip K*K matrix having top left at this position. Now, I have taken update array I will update this position with +1 and upd[j+k] with -1. And v[i+k].push_back({j,j+k}). This is because when I will be at (i+k)th row, I no longer need these updates.So at that time do this upd[j]--; and upd[j+k]++;

    case 2. if value at (i,j) is 1 and no. of updates at this position is odd, It means now value is zero. To change it, we will have to flip k*k matrix with top left at this position and do same as in step 1.

    In both cases, If it is unable to flip K*K matrix with top left at that position. I mean if((j+k-1)>=m || (i+k-1)>=n) then answer is -1.

    My code