Help me please.
I have problem.
Given array.
You have Q queries.
You need to find most frequent number between l and r. n, Q <= 500 000.
№ | 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 → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
4 | atcoder_official | 161 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 156 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
9 | nor | 153 |
I have problem.
Given array.
You have Q queries.
You need to find most frequent number between l and r. n, Q <= 500 000.
Название |
---|
What about Time Limit?
Maybe, it can be solved by MO's algorithm with big constant(~5).
Let's maintain:
cntx -> the number of occurance of x,
cntBlockx -> the number of elements, that occurrence lie in this block(if K is size of blocks, occurrence must be between [K * Block: (K + 1) * Block)).
valsx -> the vector with elements, that occurrence is equal to x.
When adding/removing elements, we must update each of them. Deleting can be done in 1 operation. Swap with last, delete last(maintain extra array posx, the position of x in vector).
Finding answer will be ez. Find maximum block mx that, cntBlockmx > 0, find maximum number x in this block, that vals[x].size() > 0.
Sorry for my poor English.
Do you know smth about segment tree?