| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | strapple | 3515 |
| 6 | tourist | 3470 |
| 7 | dXqwq | 3436 |
| 8 | Radewoosh | 3415 |
| 9 | Otomachi_Una | 3413 |
| 10 | Um_nik | 3376 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 158 |
| 2 | adamant | 152 |
| 3 | Um_nik | 146 |
| 4 | Dominater069 | 144 |
| 5 | errorgorn | 141 |
| 6 | cry | 139 |
| 7 | Proof_by_QED | 137 |
| 8 | YuukiS | 135 |
| 9 | chromate00 | 134 |
| 9 | TheScrasse | 134 |
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0
Target — 1400. Current Rating — 1190. Total Problems solved — 1009 |
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+1
Nah bro. Initially I solved lots of easy questions and used to leave contest if I couldn't solve A in 45 minutes. But then I realized my mistake and now I have started working on good problems. The sooner you realize your mistake and starting working on it more better you become. Sorry for my poor English. |
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+14
I took nearest power of 2 for every number using log. Hence it will satisfy b[i] divisible by b[i+1] or b[i+1] divisible by b[i]. |
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