Right now IOI 2016 is in progress. I'm sure a lot of you guys will be watching, so this post was created to let you guys talk about the scores and the scoreboard in the comments :) The link to the scoreboard is here.
# | User | Rating |
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1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
# | User | Contrib. |
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1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 156 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | nor | 152 |
Right now IOI 2016 is in progress. I'm sure a lot of you guys will be watching, so this post was created to let you guys talk about the scores and the scoreboard in the comments :) The link to the scoreboard is here.
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Bolivia is in the top Scoreboard for the first time in history! I have great expectations for this year! GO BOLIVIA!
Day 1 is over, and a lot of contestants have fully solved Task 1. What are your thoughts on the problems (and maybe your approaches)?
I think this solves 1:
k
smallest elements, such that their sum is less thanl
, but if you tookk+1
it would be larger thanu
. (If you can't do that, the problem is either easy: a sum lands inside the interval; or impossible: the total element sum is smaller thanl
)k
elements for one of thek
largest elements.When you change an element, the sum can never increase by more than
w_max - w_min <= u - l
, so there is no risk of 'jumping' over the[l, u]
interval.If the sum of the largest
k
is still smaller thanl
, then there is no solution, because the next subset in order of sum is thek+1
smallest elements, which we know is too large.This solution should take time equal to sorting, which should be fine for n = 200,000.
I thought it was a fun problem :-)
your answer in good but there is a better one that is very similar instead of sorting, just take the first k elements that the sum of the first k is smaller then l and the sum of the first k+1 is higher then u. then we can use the same trick that you did for k and k+1 (if there is a solution then thre is a solution that uses k elements or k+1 elements)
You could also easily use two pointer on tge sorted array?
Yes, I think that's a nice way to accomplish the third step.
3 from China ,2 form Russia in top 5 . but how??
When are day 2 problems coming out?
When day 2 comes.
Oh, I thought today was day 2. Anyways, just a little question: how difficult are the IOI problems usually? From my point of view the first problem from day 1 would be around div1. A/B. What do you guys think?
problem 2 and 3 are much harder than Div1 A and B. a lot of those participants are red or orange and they all solve Div1 A and B easily in CF rounds of normal difficulty. but only 4 solved 2nd problem and just 1 solved the third (solved == 100 points)
Its hard and possibly impossible to compare IOI problems with CF problems due to the completely different style and format. The first problem was easy, but the other 2 were very difficult.