HOW TO SOLVE THE INEQUALITY i*(i + 1) <= n such that i is maximum possible in O(1) time complexity.is their exist any method.
→ Обратите внимание
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| Страны | Города | Организации | Всё → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 1 | Qingyu | 162 |
| 2 | adamant | 148 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 143 |
| 5 | errorgorn | 141 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 133 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2026 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 18.05.2026 16:41:52 (g1).
Десктопная версия, переключиться на мобильную.
При поддержке
This is the one I can think of. Solve $$$i^{2}+i-n=0$$$
So just take floor(sol) from the previous equation.
The problem is that solution involves computing square root which is not O(1).
yeah that is the problem, i can find it in sqrt complexity but i guess constant complexity is not possible
Computing the square root of a number is actually very close close to O(1) !
I don't know the complexity exactly but it is described here : https://en.wikipedia.org/wiki/Computational_complexity_of_mathematical_operations