| # | 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 | 134 |
| 9 | maroonrk | 133 |
| 10 | qwexd | 130 |
|
0
For div2 C, I just used priority_que and set to simulate cuttings, and got AC.10325360 But I made a data: 200000 200000 199999 V 1 V 2 ... V 199999 I run it locally, write the output to a file and got nearly 4 seconds..(turn off the output, and got 1.9 seconds) 199999 pops and 399998 pushs in total.. So that was really lucky.. |
|
0
When i, j point to the 1st and 2nd hold respectively, we know that replacing j with a hold on its left wont make things better. For if you remove j after that, it is harder to remove next holder because the distance gets larger.And if you dont remove j, the situation becomes the same before replacing. Thank u, ffao, my mind is clear with this greed approach now :) |
|
0
Um...To check if d can be achieved, can we do this? let i point to the 1st element and j to the 2nd, remove a[j] and inc j by 1 until a[j+1]-a[i] > d, then i=j, j=j+1... |
|
0
How to solve the chanllenge of 496A? |
|
0
i got got TLE, for i forgot to break in the second loop.. 8727730 But i think in most cases, 3000^2 won`t cause TLE. |
| Name |
|---|


