Following

Why don't u visit ur recent contests and upsolve the easiest unsolved problems? :-)

yes. I agree brother. This is a very good way to keep busy. Happy Coding. :D

Obviously, what could be a better way to spend time ? ;)

Why ? be greedy ... I want at least 7 contests per weak ... (in the rest of the day Ill upsolve the contest)

In the resulting scoreboard, participants are compared first by the number of solved problems, then by the total penalty time — for every solved problem, it is the time taken to solve it (in minutes) and 10 minutes for every unsuccessful submission on it.

Why don't you ask him? How would others know his configuration?

have a look at the other people's dotfiles (.vimrc to be specific) in github

I found that even orginal 1500 rating is hard to reach.:(

CSP...

we ain't got nothing to do

You could always practice previous contests...

(That's very dangerous! Keep yourself safe and don't get killed during this contest)

https://codeforces.me/blog/entry/65477

Maybe you are expected to be an expert

Improve your spelling and everyone will expect you to be an expert

such a bold statement

Not for the ones who don't get to take it.

Because all are home quarantine and love coding.

YUP

It was dp, and I really liked the problem; something unique. I don't think I can explain more, since my own solution was very messy. Hope the pretests were strong.

I tried to do it using LCA, but my solution gave WA on test 56.

Yes.

The observation is that all the parents of the nodes in a query has to be in the same path.

So find the parent of each node in a query and check if they are in the same path from root using LCA.

I used LCA, and passed it.

I only consider 4 situations:

1. all elements are the same : (1 color)1 1 1...

2. the length of the elements is even : (2 colors)1 2 1 2...

and now we just need to think situations where the length of the elements is odd

if t[1]==t[n] then: 1. (2 colors)1 2 1 2...

else if t[1]!=t[n] and at the same time there exist one i(1<=i<=n-1) where t[i]==t[i+1] then:

1. (2 colors)1 2 1 2...

2. otherwise: (3 colors)1 2 1 2 ... 3

Think greedily. Try to use minimum number of colors to color figure 1 upto n-1(you can color these n-1 figures maintaining all constraints using 2 colors only) and then think is you can color figure n with a color already used. If not then give it a new color!!!

Try using m1.codeforces.com next time, I sent all my submissions from there with no problem (while I couldn't submit sometimes from www.codeforces.com)

If you like graphs you can see it is a coloring problem. Your vertices are the different animals and there is an edge between them if they are of a different type and adjacent. Now you want to color this graph, such that all adjacent vertices have different colors. This gives you three cases: 1. If the graph has no edges (ie only one type) you only need one color

1. The graph is not a cycle, then you need two colors, since you can just color any path as 1212...

2. The graph is a cycle, then it is important whether it has an odd or a even length (if you are interested how this generalises look up bipartite graphs)

If the length of the cycle is even, then you can't color it in a 1212...1 fashion, since the first and the last edge would have the same color. Therefore you need three colors and can color it like 1212...23

If the length of the cycle is odd than you only need two colors and can color it like 1212...2

I hope that is understandable :)

If there is only $$$1$$$ type of animal, then the answer is $$$1$$$ color. We can simply print $$$1$$$ $$$n$$$ times.

If $$$n$$$ is even then the answer is $$$2$$$. We simply print alternating $$$1 2 1 2 . . .$$$

So, we are left with the case when $$$n$$$ is odd. If we can find any $$$i$$$ such that $$$type[i]=type[i-1]$$$(cycle included, i.e. $$$type[0]=type[n-1]$$$) then the answer is $$$2$$$, as the parity can be switched. Otherwise, the answer is $$$3$$$.

To print the colors with answer $$$2$$$, you can alternate $$$1$$$ and $$$2$$$ with the exception of $$$i$$$ where $$$type[i]=type[i-1]$$$. Put that as the same color. This changes the parity.

If all elements are same, answer is 1 If all elements are distinct and number of elements is odd, answer is 3 otherwise answer is 2

if all are distinct, simply print 1 2 1 2 ... like this otherwise take every segment containing distinct elements and fill it by 1 2 1 2... those are left, give any colour to them

By observation it can be seen max distinct color required will be 3. Now if n is even 1212.. pattern will work if there are more than one number. For all same obvious 1 color will be required. Now for odd n, consider if there are at least 2 contiguous same number in cyclic array in that case color required will be 2 otherwise 3

You can solve D without graphs as well. It can be solved greedily.

1. If n is even, just print "1 2" n/2 times

2. If n is odd but a[0] == a[n-1] print "1 2" n/2 times with an extra "1" in the end

3. If n is odd but there is a point where arr[i] == arr[i-1], then from 0 to i-1 print "1 2" and from i to n-1, print "2 1"

4. If none is true, print "1 2" n/2 times with an extra "3" in the end

Corner Case (it was for me): If all are same print "1" n times.

It was very unique; I loved it. I solved it using dp, but it was very messy.

thanks it was helpful

please can you tell me why using library sorting I am getting TLE but with radix sort it passes my_sub constrain is just n= 2*10^5 so O(nlogn) is fine right the why?