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EPIC Institute of Technology Round Summer 2024 (Div. 1 + Div. 2) — Video Tutorial
By Shayan, history, 6 months ago, In English

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.

1987A — Upload More RAM

1987B — K-Sort

1987C — Basil's Garden

1987D — World is Mine

1987E — Wonderful Tree!

1987F2 — Interesting Problem (Hard Version)

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6 months ago, # |
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C was the weirdest problem

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    6 months ago, # ^ |
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    Agreed. I still am unable to wrap my head around it completely.

    Can someone help me get a proof or a solution for it.......

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      6 months ago, # ^ |
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      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].

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6 months ago, # |
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Did you link this blog before even the official editorial, and named the official editorial Tutorial 2?? or was it the authors themselves

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6 months ago, # |
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Good contest

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6 months ago, # |
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Great one

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6 months ago, # |
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B although trivial is pretty elegant i must say

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6 months ago, # |
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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.

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