| № | 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 | 133 |
| 8 | maroonrk | 133 |
| 10 | qwexd | 130 |
This is the problem of cses coin piles. I wanted to know can this problem be generalized? like for removing p coins from left pile and q coins from right pile and vice versa.
I tried doing it but wasn't able to figure it out. And didn't even know how to check if the solution is correct or not.
Auto comment: topic has been updated by death_bringer (previous revision, new revision, compare).
Auto comment: topic has been updated by death_bringer (previous revision, new revision, compare).
You need to solve equation: $$$a = l \cdot p + r \cdot q$$$ and $$$b = l \cdot q + r \cdot p$$$ where $$$l, r \geq 0$$$