Hope to see one problem related to XOR and problems name as Ehab and ... .

1325B - CopyCopyCopyCopyCopy In the second test case of this problem there is a Pi Number;3,14159.what a coincidence in Pi Day :D.

Depends on the timezone you consider ;-;

kostia orz

yeah...looking forward for "Ehab and..." problem :3

rotavirus orz

No, as long as there is no submission from your account

No, but if you do at least one submission during the contest you will be affected.

As mentioned here.

The 1900+ rating means that you're part of the first division

So, 1900 i believe.

You can give virtual ,as it will be unrated for you anyhow.

I'll ask my coordinator.

Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

Hello. Can you please reschedule the round for 12 hours (Moscow time), in connection with the holding of another Olympiad at 15:00 (Moscow time). Thanks in advance

Hello, aihay.

Ah, i see, Mike didn't discuss contest timing with you this time. Forgive him, he's a bit busy preventing corona from spreading through CF, you know.

I'll make sure this will not happen again! My apologies for this awful inconvenience. CF stuff will be punished, I promise!

Hope you like the new timing! If you have any other problems with CF feel free to DM me!

Best regards, kostia244

Yes, it's rated for you.

Oh, you are so smart!

Why you are still newbie, phps?

https://codeforces.me/contests/writer/mohammedehab2002

Literally every contest he has written has a problem about XOR.

ReAd?Y?!

PWPSD

set smallest numbers to edges to leaf, and others randomly)

If it's a chain, then fill all the edges randomly.

Otherwise, choose a node of which degree is 3 or above, fill three of the edges using 0, 1, 2; others random.

It can be proved to be one of the best answers.

Here's what I did : Sort all vertices by their degree in descending order. Now, starting from the vertice with the maximum degree, try and fill all its unfilled edges with numbers in a serial manner( Maintain a count variable. Fill edges with the count variable and increment it everytime you fill an edge )

I tried to construct bit by bit, starting at lowest bit and the xor input. Then add bits to get the sum, by adding the one lower bit twice to two numbers in the array. And so on...

Did not work for some reason.

max length of the answer is 3

The answer is at most 3 or it is impossible.

If $$$u$$$ and $$$v$$$ don't have same parity or $$$u>v$$$, it is impossible.

If $$$u=v$$$ the answer is $$$0$$$ if $$$u=0$$$, $$$1$$$ otherwise.

Otherwise, try to build a pair which xor gives $$$u$$$ and sum gives $$$v$$$. Let $$$t=(v-u)/2$$$. If $$$u$$$ and $$$t$$$ don't have any bits in common, a pair of solution is $$$u+t$$$ and $$$t$$$.

Otherwise, $$$t$$$, $$$t$$$, $$$u$$$ is a solution.

Look at the bitwise representation of $$$u$$$. Every bit that equals 1 means that there is an odd # of elements with that set bit. Every bit that equals 0 means there is an even # of elements with that set bit (because 1^1^... odd number of times = 1, 1^1^... even number of times = 0). So one of the array elements is at least equal to u. Suppose first element of array = $$$u$$$. Then we have to add two elements such that we don't change the parity of set bits; i.e., we must add two of the same element, call it $$$y$$$. This is 3 total. But then we can consolidate one of the $$$y$$$ into $$$u$$$, iff $$$u$$$ AND $$$y$$$ equals 0.

You can make array of one element if the XOR value equals to SUM,

the one's bits in XOR must appear in odd numbers in array.

so we can make array of two elements if the following conditions comes true,

one of the two numbers has the one's bits in XOR

and the remainder rem = (XOR — SUM),

can be distributed among two numbers (must be half in the first and half in the second to get XOR ZERO)

and the XOR of those two number must be the same of XOR in the input

and you can make array of three elements if the following conditions comes true,

one of the three numbers has the one's bits in XOR

and the remainder rem = (XOR — SUM),

distributed among the other two numbers equally to get (XOR zero) when XOR them,

so when we XOR the input with zero we got XOR.

otherwise there is no answer.

Me toooooo!!!

So do I

I got wrong answer in testcase 18 because of not handling the answer for size 3 properly.

Basically, for size 3 you have to print u, (v — u)/ 2, (v — u) / 2

Instead, I printed u, v / 2, v / 2

I think you have missed this point too.

Though, thankfully i spotted my mistake quickly during the contest time. :3

Nope, i guess it is (a^b^c) + 2*(a&b + b&c + c&a — 2*(a&b&c))

Something like that, couldn't solve though

Solution of D.

You can check this.

https://codeforces.me/blog/entry/74683?#comment-589082

Disclaimer: I don't have an accepted version yet.

Label any 3 edges, which connects a leaf vertice, with 0,1,2

You can label other edges with any numbers avaliable.

If the tree is a chain, all of the labeling way are acceptable.

if you find one node like x that has 3 neighbors then use 0,1,2 for labels of edges between x and neighbors and you can chose labels of other edges in any order and you will get MEX(u,v) = 2

if you can't find node like x then chose all labels of edges in any order and you will get MEX(u,v) = n — 1

Obviously, if u > v or u % 2 != v % 2, there isn't a solution.

If u == 0 && v == 0, 0 should be the answer.

When it comes to u == v, u could be a answer.

Otherwise:

if(!(u&(v-u)/2))
    printf("2\n%lld %lld", (v - u) / 2 + u, (v - u) / 2);
else
   printf("3\n%lld %lld %lld", (v - u) / 2, u, (v - u) / 2);

You can check it :P

You can check this.

https://codeforces.me/blog/entry/74683?#comment-589082

For second sample, count=3, you assigned 3 to (6-2-3)=1. Even 0 must be used for those edges.

You can check this.

https://codeforces.me/blog/entry/74683?#comment-589082

Please help me in solving D. Thank You.

Was waiting for this Thanks

There is no hacking time, when it is finished, it is finished.

Usually, it is the Div 3 and the Educational Rounds which have hacking phases after the coding phase of the contest. In Div 2 and Div 1 contests, there is only the coding phase but not hacking phase.

Global Round in 5 days...Yayyaya

No, there another one within 5 days

You mean CoronaForces > Codeforces

I did this too! The editorial solution made me realize that there exist at least 3 leaves iff there exists a vertex of degree 3. It is a special case of this theorem

Hello, I am the person the system claims Witless_Deer copied from. For the reasons he mentioned (especially reason #1), it is ridiculous to claim that he copied me. Look at our submissions for B to see for yourselves.

Consider the paths between all nodes. If the edge with value 0 is not on such a path, the MEX of that path is 0.

So, consider the paths with the 0. The MEX of these paths is 1, except if the edge with value 1 is also in this path. Then it is at least 2.

So, is it possible that there is no path with edges 0 and 1 in the path?

It turns out, no, that is not possible. Since by the simple fact that the path from two nodes of the edges with value 0 and 1 has both edges in it. So it allways exists.

But now we are close to the solution. How can we place the edge with value 2 in a way that it is not on that path, too?

Well, more or less simple, we place that edge not on the path between the edges 0 and 1. Then it is not.

Last question to answer is, how do we implement this, how can we find suitable vertex and edges? The tutorial answers this clearly: We search an vertex with at least 3 edges, and use these 3. Then the edge with value 2 will not be on path between the edges 0 and 1.

your nickname is very suitable for your comment about math)))

I think you can start using other IDEs such as

https://www.jdoodle.com/online-compiler-c++17/

or

https://ide.geeksforgeeks.org/

These are much more safe than using ideone.com .. There used to be one more IDE from hackerearth: https://code.hackerearth.com/, but it's out of service now.. So please try using any other IDE or local compiler.

It happens when your submissions are evaluated and you are found guilty under Plagiarism check. I think you should reach out to Codeforces Authorities regarding the same.

Have a nice day!

Exactly.. One more thing which I liked was that problem statements were to the point rather than long paragraphs, and properly framed.