Initially the array contain all 1s.
There are two type of operation:
1 A: update arr[A] = 0.
2 A: Find index of Ath 1 in the array.
Number of elements, 1<=N<=(1e6)
Number of queries, 1<=Q<=(1e6)
I tried tree statistic. However, it didn't pass.
→ Обратите внимание
→ Трансляции
→ Лидеры (рейтинг)
| № | Пользователь | Рейтинг |
|---|---|---|
| 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 → |
→ Лидеры (вклад)
| № | Пользователь | Вклад |
|---|---|---|
| 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 |
→ Найти пользователя
→ Прямой эфир
Codeforces (c) Copyright 2010-2024 Михаил Мирзаянов
Соревнования по программированию 2.0
Время на сервере: 14.11.2024 08:19:00 (k3).
Десктопная версия, переключиться на мобильную.
При поддержке
build a segment tree wich in each node you save the number of 1's in [l, r)
update is simple.
and for answering a query in the query function check if A is smaller or equal to the number of 1's in the left node then go left in the segment tree,
otherwise go to the right child and decrease A with the number of 1's in the left child;
o(nlogn)Can you provide a link for that problem?
Answer queries in the reverse order
Can be done in O(logN) per query using a BIT.
how is it O(logN) ? , considering you are doing binary search on bit .isn't it O(log^2N)
You can binary search on the bit itself. Check topcoder tutorial for BIT.