you will be given an array. and q query. q and n all less than 100000.
In every query you will be given a k. how many subarray xor less than k?
** I think it can be solved by persistent trie. but i need another solution.
# | User | Rating |
---|---|---|
1 | ecnerwala | 3649 |
2 | Benq | 3581 |
3 | orzdevinwang | 3570 |
4 | Geothermal | 3569 |
4 | cnnfls_csy | 3569 |
6 | tourist | 3565 |
7 | maroonrk | 3531 |
8 | Radewoosh | 3521 |
9 | Um_nik | 3482 |
10 | jiangly | 3468 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 162 |
4 | TheScrasse | 159 |
5 | nor | 158 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 151 |
8 | SecondThread | 147 |
9 | orz | 146 |
10 | pajenegod | 145 |
you will be given an array. and q query. q and n all less than 100000.
In every query you will be given a k. how many subarray xor less than k?
** I think it can be solved by persistent trie. but i need another solution.
Name |
---|
Lets rephrase the problem a little bit. Denote by P[i] the XOR of numbers a[1], a[2], ..., a[i]. Now the XOR of subarray [L, R] becomes P[R] XOR P[L — 1] and the problem boils down to finding how many pairs of numbers from a list, when XOR-ed give a number less than K. This problem can be solved in O(NlogN) with the help of the Walsh-Hadamard transform and you can get some insight for the implementation from here: https://csacademy.com/blog/fast-fourier-transform-and-variations-of-it
i only know fast walsh hadamard gives us... all pairs xor/and/or sum. i don't know about this variation. Thanks.