| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | jiangly | 3670 |
| 3 | Kevin114514 | 3657 |
| 4 | maroonrk | 3355 |
| 5 | Um_nik | 3353 |
| 6 | squareOf105 | 3346 |
| 7 | tourist | 3343 |
| 8 | strapple | 3324 |
| 9 | heuristica | 3322 |
| 10 | jiangbowen | 3291 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | AmShZ | 143 |
| 5 | Um_nik | 141 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | BledDest | 134 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
I solved this problem(https://www.codechef.com/problems/SORTSUB7EZ) by using the assumption that:- "all the remainders of ai(wrt any integer) belong to set {0, 1, 2, ....,(ai+1)/2-1}" but idk the proof, can anyone help?
PS: i tried a few values of ai and found out that assumption.
to get remainder $$$x \ (2 * x \lt a)$$$ you can do always do $$$a_i mod (a_i - x)$$$ as
$$$y = a_i - x$$$
$$$2y = 2a_i - 2x$$$
$$$a_i + (a_i - 2x) \gt a$$$
$$$2y \gt a$$$