Hello!
I was planning to solve some output-only tasks in preparation for IOI. As there aren't many output-only tasks, it's difficult to face them by just chance (And I didn't find much by googling). Does anyone have a list for output-only tasks?
№ | Пользователь | Рейтинг |
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1 | tourist | 4009 |
2 | jiangly | 3823 |
3 | Benq | 3738 |
4 | Radewoosh | 3633 |
5 | jqdai0815 | 3620 |
6 | orzdevinwang | 3529 |
7 | ecnerwala | 3446 |
8 | Um_nik | 3396 |
9 | ksun48 | 3390 |
10 | gamegame | 3386 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
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1 | cry | 164 |
1 | maomao90 | 164 |
3 | Um_nik | 163 |
4 | atcoder_official | 160 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 157 |
8 | TheScrasse | 154 |
8 | Dominater069 | 154 |
8 | nor | 154 |
Hello!
I was planning to solve some output-only tasks in preparation for IOI. As there aren't many output-only tasks, it's difficult to face them by just chance (And I didn't find much by googling). Does anyone have a list for output-only tasks?
In both of the rounds 664 and 663, I have placed in the same rank 783. The fact is one of them was Div1 and another one was Div2. Strangely enough, I got almost the same delta change from almost the same initial rating! So this made me think of placing behind 782 reds/orange/purple is the same as placing behind 782 cyan/blue/purple.
UPD: Um_nik gave a really nice explanation for this case (Also thanks to Radewoosh, though I couldn't understand the fact fully from him). The basic thing is you will get an expected rank based on the probability of beating other users. Now the confusion is how the rank is calculated, here is the quote from Um_nik: "Your expected place is not the number of participants with higher rating + 1. The probability of you placing higher than someone calculated as $$$\frac{1}{1+c^{r2-r1}}$$$ where $$$r1$$$ is your rating, $$$r2$$$ is their rating and $$$c$$$ is some constant. Now we sum this (1 — this, actually) over all participants and get the expected number of people who will place higher than you."
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