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0
Could anyone help me find my mistake in this code of problem C? It got WA on 2 but I can't see the exact test data. Here's the code : 106592765 Thx a lot. |
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0
Long time no interactive problem! Look forward to it. |
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0
You mean C right? This is just basic dfs on a tree. |
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On
dreamoon_love_AA →
Codeforces Round #631 (Div. 1 and Div. 2) — Thanks, Denis aramis Shitov!, 6 years ago
0
WOW div2D/div1B is quite tricky. You can find the sequence on OEIS this sequence but only those few numbers in front of the sequence is right for this problem. So tricky/QAQ |
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0
Yeah I didn't see that coming. [facepalm] |
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+11
I don't know why but my rating decreases in every Educational Round...Get educated.... |
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+8
D is hard to understand too. |
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0
Yeah thx now I get it and will be careful next time. Hope they could rejudge this submission but it's kind of unrealistic. SO SAD :( |
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+1
They're both submission to problem B. |
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33....38 works too. |
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0
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0
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+8
The editorial is out. It's the fastest editorial I've ever seen. Orz |
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On
vintage_Vlad_Makeev →
Codeforces Round #626 (Div.1, Div.2, based on Moscow Open Olympiad in Informatics, rated), 6 years ago
0
Only in edu round, you are allowed to copy. |
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+1
It is guaranteed that the sum of n over all test cases does not exceed 5000. So the upper bound is t = 1, n = 5000. |
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+1
5000 * 5000 = 2.5 * 10^7 ..... So I guess time is sufficient. Also it is guaranteed that the sum of n over all test cases does not exceed 5000. |
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+9
To me it's ABodeforces... |
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+1
If a and b have the same result after modulo m, |a — b| mod m = 0. Then it's pigeonhole principle. |
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0
I think you misunderstood the meaning of this problem. You should read it again carefully. |
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0
Dijkstra(O(mlogm)) Count the number of minimum paths as well. |
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0
Will the difficulty of this two contests equals to usual Div.1 & Div.2 respectively? |
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0
Wow, that's not cool at all.I didn't see the code thoroughly. |
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0
Why say that? |
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+8
Can anyone just explain D more clearly ? I'm too stupid to understand the given editorial. |
