How do i solve this Dynamic Programming Problem link
→ Pay attention
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 4009 |
| 2 | jiangly | 3839 |
| 3 | Radewoosh | 3646 |
| 4 | jqdai0815 | 3620 |
| 4 | Benq | 3620 |
| 6 | orzdevinwang | 3612 |
| 7 | Geothermal | 3569 |
| 7 | cnnfls_csy | 3569 |
| 9 | ecnerwala | 3494 |
| 10 | Um_nik | 3396 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | Um_nik | 164 |
| 2 | maomao90 | 160 |
| 3 | -is-this-fft- | 159 |
| 4 | atcoder_official | 158 |
| 4 | cry | 158 |
| 4 | awoo | 158 |
| 7 | adamant | 155 |
| 8 | nor | 154 |
| 9 | TheScrasse | 153 |
| 10 | maroonrk | 152 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Oct/01/2024 19:37:00 (h1).
Desktop version, switch to mobile version.
Supported by
Let F[i] = best sum you can achieve when it's your turn and there are i elements left..
You can make a simple observation that if player A takes F[N] points then player B will have Total Points — F[n].
Based on that let Sum[i] be the sum of the last i elements in the stack then F[i] can be one of these 3 cases:
F[i] = A[i] + sum[i-1] -F[i-1].
F[i]= A[i] + A[i-1] + sum[i-2] -F[i-2].
F[i]= A[i] + A[i-1] + A[i-2] + sum[i-3] -F[i-3].