| # | User | Rating |
|---|---|---|
| 1 | jiangly | 3810 |
| 2 | Benq | 3676 |
| 3 | Kevin114514 | 3655 |
| 4 | maroonrk | 3463 |
| 5 | strapple | 3447 |
| 6 | Um_nik | 3387 |
| 7 | heuristica | 3322 |
| 8 | turmax | 3317 |
| 9 | tourist | 3307 |
| 10 | jiangbowen | 3291 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 156 |
| 2 | nik_exists | 150 |
| 2 | maspy | 150 |
| 4 | Um_nik | 143 |
| 5 | Errichto | 139 |
| 6 | adamant | 137 |
| 7 | AmShZ | 135 |
| 8 | maroonrk | 133 |
| 9 | BledDest | 132 |
| 10 | qwexd | 129 |
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Technically the array doesn't need to be sorted, since the condition was for a[i] != a[i-1], not a[i] > a[i-1], meaning that same numbers had to be grouped next to each other. Something like 1 1 1 5 5 3 3 3 2 2 9 9 7 7 7, which you can easily count the number of marked elements for. For arrays in which same numbers aren't grouped together, for example like 1 3 5 2 3 1 4 3 1 5, you can just treat every element as a different group |
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