→ Обратите внимание
До соревнования
WorldQuant Round 1 (Codeforces Round 1128, Div. 1 + Div. 2)
3 дня
Зарегистрироваться »
WorldQuant Round 1 (Codeforces Round 1128, Div. 1 + Div. 2)
3 дня
Зарегистрироваться »
*есть доп. регистрация
→ Лидеры (рейтинг)
| № | 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 → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | AmShZ | 144 |
| 5 | Um_nik | 141 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | BledDest | 134 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2026 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 14.10.2026 21:54:10 (g2).
Десктопная версия, переключиться на мобильную.
При поддержке
there will be many more case according to your solution
https://codeforces.me/contest/478/submission/54276055 the logic is this plain and any other logic will have multiple sub-cases
Can you explain about your solution.
Observe that balloon of smallest and 2nd smallest type can always be used so either the 3 would be in comparable amounts such that they can be grouped in 3 in triples of (1,1,1) or (1,2,0) and similar. the extra ballon which cannot be mixed is left out. the only case where its not possible is when the balloon of largest type is extra. the worst case is when you take one ballon of smallest and mix it with two of largest. similarly you take 1 of middle size and 2 of largest. this will lead to ans as (smaller + 2nd smaller balloons). now the answer would be the minimum of two according to the above conditions