Comments
0

Really!
0

Nice Solution!
On ifsmirnov → On Euler tour trees, 5 years ago
+14

Very Informative, thanks!
+7

People are not that caring these days
0

.
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Can we do problem D using a sliding window? I tried using ordered_set but got TLE.
+3

This is Awesome dude!
On parker.sky001 → Is This Normal?, 5 years ago
-8

Its a Bugaboo
+9

Which site's screen?
On Ari → Codeforces Round #715 Editorial, 5 years ago
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I badly wanted to see a greedy solution for problem C. I spent so much time thinking greedily about it ;-;. Amazing problem anyway!
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Please explain your 2 loop solution?
-43

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-56

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+1

Interesting. Thank you very much for this counter example. I appreciate!
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In case where we found an ancestor of the node, why do we minimise the low time of the node with discovery time of ancestor and not low time of ancestor.

I mean why not

 else // if v was already discovered means that we found an ancestor
      low[u] = min(low[u], low[v]);
+62

You are really very competitive. xD
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Finally I can face my developer friends now.

On Kaey → Codeforces Round #701 (Div. 2), 6 years ago
+15

I see what you did there.
+16

He couldn't complete his sentence. Perhaps some big stone fell on him while he was typing. Sad.
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Got it, thanks!!
0

Can C be done by top-down approach?
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Please predict mine also. Thanks in advance!
+27

To get CF magic option 2020
0

just you.
+2

Spread the greenery!
-11

why would it be infinite G.P. on RHS? I mean it could be finite as well right?
On Monogon → Codeforces Global Round 12, 6 years ago
-24

I dont intend to write a whole contest. Just testing for 1 single question would be more than enough for me.

.Its a (pupil)green flag from my side.
On Monogon → Codeforces Global Round 12, 6 years ago
+45

@antontrygubO_o Is there any chance you are looking for new friends?
0

One optimization you can do to make this code run is that for each digit which is occuring >1 times in a row, just consider its leftmost and rightmost occurence for further calculations. This will not make the answer incorrect because we want the maximum area which will depend only on the extreme coordinates. For e.g. in this case below

1
5
00000
11111
00000
00000
00000

Instead of storing coordinates of 1 as (2,1),(2,2),(2,3),(2,4) & (2,5), store & consider 1 to be present only at (2,1) & at (2,5). This way that n^4 factor would be reduced to 2*n factor.

I know its hard to understand as I have not explained it in that detail but try to think, you will get the feel.

You can see my accepted submission (post contest) which is same as n^4 solution. Just this extra optimization.
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Its best to keep code as genralized as possible.
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Same with me. Don't know how to get past it.
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Welcome.
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Honestly I myself couldn't understand the editorial solution. I solved using a different approach about which you can find here
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We can decrement any suffix by 1. We can decrement any prefix by 1.

Now we are asked if its possible to make each element of given array as 0 by performing above 2 operations.
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In my case, I solved this issue when I removed the input of a integer in code which was not there in the problem statement.
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Never knew that we can hack our own answers!
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Please tell what was it?
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Not really.. Considering the fact that n <**=** 10^9;
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Awesome .. I also thought same way but got offtrack in 3rd step..yeah the main step, :-(
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Hey. Can u pls show its working on 1,3,4,4
0

I have a shorter(/est) solution for Problem B

n,k=map(int,input().split())
if(k==1):
    print(n)
else:
    print(2**(len(bin(n))-2)-1)