Hi,
Here is the video editorial of Problems A-F2 of EPIC Institute of Technology Round Summer 2024 (Div. 1 + Div. 2). I hope it helps.
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Hi,
Here is the video editorial of Problems A-F2 of EPIC Institute of Technology Round Summer 2024 (Div. 1 + Div. 2). I hope it helps.
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C was the weirdest problem
Agreed. I still am unable to wrap my head around it completely.
Can someone help me get a proof or a solution for it.......
The solution should be computed from right to left.
For the last element, the time needed to become zero is just the beginning height. Once we know the timing for the last element, we can go from right to left and calculate the total time each flower needs.
For any flower h[i], we know that it will take time h[i] after the time at which h[i+1] == h[i] — 1. In other words, a flower of height 7 will take 7 seconds after the next flower is at height 6. We know when h[i+1] is at height 6 by adding 6 to the time at which it is at 0. So, flower i will take h[i] + timing[i+1] — (h[i] — 1) which simplifies to timing[i+1] + 1. If there is a particularly larger h[i], its value may be more than timing[j+1]+1 so we just set timing[i] to h[i].
Did you link this blog before even the official editorial, and named the official editorial Tutorial 2?? or was it the authors themselves
It happens even if we link this one later (and I think it was added later), so It's just not really obvious behavior of Codeforces.
no
Good contest
Great one
B although trivial is pretty elegant i must say
Problem B, i declared my answer variable as int and all those 10^9 added up to long. C's solution passed pretest 1 but gave clueless answers in pretest 2.