| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | jiangly | 3670 |
| 3 | Kevin114514 | 3657 |
| 4 | maroonrk | 3355 |
| 5 | Um_nik | 3353 |
| 6 | squareOf105 | 3346 |
| 7 | tourist | 3343 |
| 8 | strapple | 3324 |
| 9 | heuristica | 3322 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | AmShZ | 143 |
| 4 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | adamant | 137 |
| 8 | BledDest | 135 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
| Name |
|---|



It can be solve in O(n2) time per testcase if you something similar to two pointers instead of binary search. Let me explain:
At each position we try to increase the answer as much as possible (that is, as long as the 1 larger submatrix still has sum ≤ w). Now notice that the answer can only increase n times — so the whole thing is O(n2) (using prefix sums as you did).
Here's my AC code: link
Thanks. I got it.
That is really AC code?