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.
Why there is no leaderboard? Where can I ask clarification?
Can anyone just clarify the problem "Minimum Cost"? The statement is not doing much justice. Just need clarification nothing else.
You can go from i to p[i] with cost 0.
Sorry, refresh the page.
Constraint on N is showing the Constraint of T..
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.
Sorry, please refresh the page.
I refreshed the page and all my submissions disappeared. Should I try to solve problems or give up? :)
I don't know about the back-end. I'll ask them.
Fixed.
Please, add limitations for N in problem 4.
Fixed.
Not easy contest. Please next time only give name June contest
In contrast with other contests, it's an easy one.
Strongly agree :)
Why there are no explanations in the editorials?
They'll be posted today.
Can someone explain the use of XOR in Mosaics and holes. Or any other method to solve.
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