| № | Пользователь | Рейтинг |
|---|---|---|
| 1 | tourist | 3993 |
| 2 | jiangly | 3743 |
| 3 | orzdevinwang | 3707 |
| 4 | Radewoosh | 3627 |
| 5 | jqdai0815 | 3620 |
| 6 | Benq | 3564 |
| 7 | Kevin114514 | 3443 |
| 8 | ksun48 | 3434 |
| 9 | Rewinding | 3397 |
| 10 | Um_nik | 3396 |
| Страны | Города | Организации | Всё → |
| № | Пользователь | Вклад |
|---|---|---|
| 1 | cry | 167 |
| 2 | Um_nik | 163 |
| 3 | maomao90 | 162 |
| 3 | atcoder_official | 162 |
| 5 | adamant | 159 |
| 6 | -is-this-fft- | 158 |
| 7 | awoo | 157 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 9 | nor | 153 |
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 :)
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.net/problemset/problem/242/E c++ solution: https://github.com/Jaskamalkainth/Codeforces/blob/master/xor_on_segment.cpp
Thanks a lot! :)
IMPOSIBLE