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 | 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 → |
| № | 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 |
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?