How do i solve this Dynamic Programming Problem link
→ Pay attention
→ Streams
→ Top rated
| # | User | Rating |
|---|---|---|
| 1 | tourist | 3985 |
| 2 | jiangly | 3814 |
| 3 | jqdai0815 | 3682 |
| 4 | Benq | 3529 |
| 5 | orzdevinwang | 3526 |
| 6 | ksun48 | 3517 |
| 7 | Radewoosh | 3410 |
| 8 | hos.lyric | 3399 |
| 9 | ecnerwala | 3392 |
| 9 | Um_nik | 3392 |
→ Top contributors
| # | User | Contrib. |
|---|---|---|
| 1 | cry | 169 |
| 2 | maomao90 | 162 |
| 2 | Um_nik | 162 |
| 4 | atcoder_official | 161 |
| 5 | djm03178 | 158 |
| 6 | -is-this-fft- | 157 |
| 7 | adamant | 155 |
| 8 | Dominater069 | 154 |
| 8 | awoo | 154 |
| 10 | luogu_official | 150 |
→ Find user
→ Recent actions
Codeforces (c) Copyright 2010-2024 Mike Mirzayanov
The only programming contests Web 2.0 platform
Server time: Dec/23/2024 15:19:31 (i1).
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].