| № | 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 | 155 |
| 8 | TheScrasse | 154 |
| 9 | Dominater069 | 153 |
| 10 | djm03178 | 152 |
My hash has 3 fields
f1 = size of interval
f2 = sum of array elements over interval(a[i] +a[i+1]...a[j])
f2 = sum of squares of array elements over interval(a[i]^2 + a[i+1]^2 .. a[j]^2).
if all 3 matches, then elements are indeed same. Can this be broken?
P.S. While in attempt to solve D from previous contest(https://codeforces.me/contest/1284/problem/D). I was looking for this question.
Auto comment: topic has been updated by gagannagpal68 (previous revision, new revision, compare).
(5, 12, 10) and (6, 8, 13) have the same hash.
I think adding a cube power may make this hash a perfect hash.
Sorry to disappoint you, but probably not. It can be shown by a simple Dirichlet. There are a lot more arrays of size 1000 than there are possible outputs for your hash if you bound the numbers by let's say 1000.
On a contest, lets say we choose 3 random powers.
I guess that would be optimistically a good hash?
It will be perfect only if you consider at least $$$n$$$ distinct powers.
Adding powers up to k can easily be hacked with a test with n=2^(k+2).