| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3611 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | Radewoosh | 3415 |
| 8 | Um_nik | 3376 |
| 9 | maroonrk | 3361 |
| 10 | XVIII | 3345 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 164 |
| 2 | adamant | 150 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 144 |
| 5 | errorgorn | 141 |
| 6 | cry | 139 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 9 | TheScrasse | 134 |
| Name |
|---|



Consider a graph with 2 * n vertexes, vertex number i corresponds to situation when there are i people that have already been to finals. We have an arc (i, j) in this graph when we can send k of i experienced people in team and after it we have j experienced people (i.e. j = n + i - 2 * k, 0<=k <= i). It's obvious that each infinite path in this graph corresponds to some coach's strategy. One can prove, that one of optimal strategies is repetition of simple cycle in such a graph. That means you should find a cycle with lowest average weight in this graph. You can do it using combination of binary search and Bellman-Ford algorithm for finding cycles of negative weight.
By the way contestants found an O(n^2) solution, by unfortunately nobody have explained me why does it work yet. It would be great if somebody tells us it.