Given an array, and there are Q queries of two types.
Type-1 : given l , r , val
Update a[i] = a[i] xor val where i = [l , r]
Type-2 : given l , r
Return sum of a[i] where i = [l , r]
Can it be solved by segment tree with lazy propagation?
ThankYou :)
| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3843 |
| 2 | jiangly | 3705 |
| 3 | Benq | 3628 |
| 4 | orzdevinwang | 3571 |
| 5 | Geothermal | 3569 |
| 5 | cnnfls_csy | 3569 |
| 7 | jqdai0815 | 3530 |
| 8 | ecnerwala | 3499 |
| 9 | gyh20 | 3447 |
| 10 | Rebelz | 3409 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | maomao90 | 171 |
| 2 | awoo | 164 |
| 3 | adamant | 163 |
| 4 | TheScrasse | 159 |
| 5 | maroonrk | 154 |
| 5 | nor | 154 |
| 7 | -is-this-fft- | 152 |
| 8 | Petr | 147 |
| 9 | orz | 145 |
| 10 | pajenegod | 144 |
Yea, it can be solved using segment tree + lazy propagation, the idea is to keep the count of bits for all the values in a range.
Xoring a range basically means swapping the count of 1 bits with the count 0 bits for each subrange in the segment tree if val has 1 bit at that position.
The query is solved by adding 2^(bit_position)*cnt[bit_position] for each bit position and for every subrange.
ex problem from a past CF contest: http://codeforces.com/problemset/problem/242/E c++ solution: https://github.com/Jaskamalkainth/Codeforces/blob/master/xor_on_segment.cpp
Thanks a lot! :)
IMPOSIBLE