| # | User | Rating |
|---|---|---|
| 1 | Benq | 3792 |
| 2 | VivaciousAubergine | 3647 |
| 3 | Kevin114514 | 3603 |
| 4 | jiangly | 3583 |
| 5 | turmax | 3559 |
| 6 | tourist | 3541 |
| 7 | strapple | 3515 |
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| 9 | dXqwq | 3436 |
| 10 | Otomachi_Una | 3413 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 157 |
| 2 | adamant | 153 |
| 3 | Um_nik | 147 |
| 4 | Proof_by_QED | 146 |
| 5 | Dominater069 | 145 |
| 6 | errorgorn | 141 |
| 7 | cry | 139 |
| 8 | YuukiS | 135 |
| 9 | TheScrasse | 134 |
| 10 | chromate00 | 133 |
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system testing started ! hold your nerves ! |
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+1
we can do it without convex hull also. to find shortest distance of each line segment from center first find distance of both points from center and then check if the line having slope equal to the perpendicular of this lines egment and passing through center intersects the line segment or not. if yes, calculate intersection point as well as its distance from centre. take minimum of all calculated distances.this is the smallest radius. |
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still not able to see testcases and submissions :( |
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I vote for the FIRST solution. Second one will lead to a lot of confusion and will be complex overall . Moreover, a rating graph with two different sections having different colors will surely NOT look good ! |
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to_string is missing in MinGW due to a bug. So it doesn't work on codeforces. You can still use THIS patch in your code to do the same. |
