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 | jiangly | 3846 |
2 | tourist | 3799 |
3 | orzdevinwang | 3706 |
4 | jqdai0815 | 3682 |
5 | ksun48 | 3590 |
6 | Ormlis | 3533 |
7 | Benq | 3468 |
8 | Radewoosh | 3463 |
9 | ecnerwala | 3451 |
9 | Um_nik | 3451 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 165 |
2 | -is-this-fft- | 161 |
3 | Qingyu | 160 |
4 | Dominater069 | 158 |
5 | atcoder_official | 157 |
6 | adamant | 154 |
7 | Um_nik | 151 |
8 | djm03178 | 150 |
9 | luogu_official | 149 |
10 | awoo | 147 |
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 |
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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?