| # | User | Rating |
|---|---|---|
| 1 | Benq | 3676 |
| 2 | Kevin114514 | 3659 |
| 3 | jiangly | 3640 |
| 4 | turmax | 3450 |
| 5 | tourist | 3382 |
| 6 | strapple | 3360 |
| 7 | Um_nik | 3354 |
| 8 | squareOf105 | 3348 |
| 9 | maroonrk | 3312 |
| 10 | idontwannadothis…e | 3191 |
| # | User | Contrib. |
|---|---|---|
| 1 | Qingyu | 155 |
| 2 | nik_exists | 151 |
| 3 | maspy | 149 |
| 4 | AmShZ | 145 |
| 5 | Um_nik | 143 |
| 6 | Errichto | 139 |
| 7 | BledDest | 136 |
| 8 | adamant | 134 |
| 9 | maroonrk | 132 |
| 10 | cry | 130 |
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+25
Light-Light Decomposition? |
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+43
Your friend is dumb |
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+29
Is it still rated? |
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+16
Kind of defeats the whole purpose of "Top-200" Badge. |
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+27
Were the problems also made by Chat GPT? |
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+16
Stop crying it was almost two years ago |
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+13
As someone that uses codeforces entirely zoomed in (because i don't see a pito), I wouldn't be able to see the snowflakes at all, so I don't really like this idea. Also it completely kills the snowflakes for mobile version. |
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+10
1E feels like a Div2C. I wonder why so few people solved it. |
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-8
1E is not hard |
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+10
Maybe Extending Set of Points? I found it very similar to this one. |
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+33
Does this mean we will be able to choose tourist rank for New Year's Magic? |
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+8
It doesn't TLE, the problem is that you are using a map to store the elements of the current range, and as you make $$$O(n\sqrt n)$$$ insertions and deletions on this map (because mo does $$$O(n\sqrt n)$$$ updates), your code ends up being $$$O(n\sqrt n\ log\ n)$$$ which is too much for $$$n\leq 2*10^5$$$. In fact, there exists a solution with mo for this problem, which got AC in 1528ms, here it is. Tell me if you want me to explain the idea behind that solution (the key is to keep updates in $$$O(1)$$$ and you can still do queries in $$$O(\sqrt n)$$$, as there are only $$$n$$$ queries). |
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0
I wouldn't say it's easy, I'm probably biased towards F, but I personally think that D > E > F. |
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0
The first three are actually YES. (And maybe 6th too, depending on your implementation) |
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