| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | jiangly | 3631 |
| 4 | Kevin114514 | 3574 |
| 5 | maroonrk | 3521 |
| 6 | strapple | 3515 |
| 7 | Radewoosh | 3461 |
| 8 | tourist | 3428 |
| 9 | turmax | 3378 |
| 10 | Um_nik | 3376 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 161 |
| 2 | adamant | 148 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 143 |
| 5 | errorgorn | 140 |
| 6 | cry | 138 |
| 7 | Proof_by_QED | 136 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 10 | soullless | 133 |
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+14
Countries ranked by mean score: Ranking
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+21
Malaysia top 5:
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+8
This can also be extended to the general case to achieve $$$S=14$$$: Solution For nodes of degree $$$2$$$, the sum of labels on the edges is always $$$k$$$, so we need to find a way to differentiate between the two types of nodes (all edges pointing outwards or all edges pointing inwards). Consider adding $$$k+1$$$ to the labels of edges at depth congruent to $$$0$$$ or $$$1 \pmod 4$$$. Notice that a degree $$$2$$$ node has its adjacent edges pointing towards it if and only if it has an even number of adjacent edges that have label $$$\geq k+1$$$. Now that we can figure out the parity of the depth of a degree $$$2$$$ node based on the labels, we can proceed with the same solution as subtasks 3 and 4. |
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