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.
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3880 |
2 | jiangly | 3669 |
3 | ecnerwala | 3654 |
4 | Benq | 3627 |
5 | orzdevinwang | 3612 |
6 | Geothermal | 3569 |
6 | cnnfls_csy | 3569 |
8 | jqdai0815 | 3532 |
9 | Radewoosh | 3522 |
10 | gyh20 | 3447 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | awoo | 161 |
2 | maomao90 | 160 |
3 | adamant | 156 |
4 | maroonrk | 153 |
5 | -is-this-fft- | 148 |
5 | atcoder_official | 148 |
5 | SecondThread | 148 |
8 | Petr | 147 |
9 | nor | 144 |
9 | TheScrasse | 144 |
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.
Название |
---|
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.