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.
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.
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3408 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 135 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
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:
ksmallest elements, such that their sum is less thanl, but if you tookk+1it 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)kelements for one of theklargest 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
kis still smaller thanl, then there is no solution, because the next subset in order of sum is thek+1smallest 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.