Sometimes some members of community says that, the announcement of div3 round is always identical. vovuh just pointing out this by saying "copy-pasted announcement"

Better try virtual. Community love Codeforces' professionalism.

refresh,because I haven't seen it on my computer

7 problems this time. Hope this information will help you, but I don't know, how it can help.

this is a mind game bro

Yeah was facing the same issue, to deal with it I had clicked on "remember me for a month" during last login, and now its working fine for me .

Maybe something of this form:

5

1 2 5 2 3

F looks like bipartite

F:

For every node, either all edges can be directed towards it or directed away from it. So, set the orientation of edges on any random node say X, which will also set the orientation for all other nodes as the graph is connected. Check if the orientation is consistent or not. If not, change the orientation of X and check again. If not, no orientation of edges is possible.

E: treat s and t as base-26 integer, then ans = s + (t — s) / 2

F: Here is a pretty simple solution. Use dfs to color every node with white or black so that no two adjacent vertexes have the same color. Then iterate each edge to see whether the color of the vertex on that edge is the same. 52118506

Notice that the function basically makes 2 adjacent numbers equal to one of them. Then, find the most frequent element and make all the numbers in the array equal to this number.

Maintain two sorted arrays. For a new element $$$x$$$, if it can be appended to only one of the arrays, do it. If it can be appended to neither, the answer is no. If it can be appended to both, check the next element $$$y$$$, if $$$y>x$$$, then append $$$x$$$ to the increasing one, otherwise append $$$x$$$ to the decreasing one. One can prove this is always optimal.

Another solution using LIS. Find the LIS and the rest of numbers should be Decreasing, if not we can prove that you only need to swap one number from LIS with one of the rest and there are a few valid cases for swap. 52133834

• If you do the following modification the code will work:

    //fflush(stdin);
    getchar();

• It is undefined behavior to use fflush(stdin) -> More Info

Your code is currently a mess, especially the I/O. Just learn how to do it in C++ instead of mixing C I/O with C++ I/O. Here is an example of your code rewritten in C++. Also

Never use gets(). The problem is that gets can write outside the allocated memory so you should never use it. "The function provides no means to prevent buffer overflow of the destination array, given sufficiently long input string. std::gets was deprecated in C++11 and removed from C++14." link

Btw I suspect that the actual reason why your code fails is because fflush(stdin). Depending on the compiler it can be undefined behavior. If you switch fflush(stdin) with cin.ignore() everything does work.

Can u please tell the mistake in logic for my code: https://mirror.codeforces.com/contest/1144/submission/52126510

Just check if the graph can be Bi-colored, if is not possbile then print 'NO' otherwise for each edge assign '1' or '0' depending on how you have colored the nodes 'u' and 'v'.

It will pass the sys test.

Underflow is a wonderful thing.

If it is correct，can someone tell me how to prove it?

In no way is that a problem with Python. What you've written there is essentially a $$$O(n^2)$$$ algorithm. Every time you mult by 26 or divide by 26 it has to do the calculation on the entire integer. As the integer is already $$$O(n)$$$ big every operation takes $$$O(n)$$$.

There are ways to get around this. Python's int(str,base) allows you to read a number in the base 26 in $$$O(n)$$$. But unfortunately there isn't any built in way to print a large integer in base 26 in $$$O(n)$$$ as far as I know.

The solution is correct in the case where the length of $$$LIS$$$ of $$$A$$$ is equal to (length of $$$X + 1$$$). The reason for this is, as you said, after removing one element from $$$Y$$$, it still remains decreasing. But, when $$$LIS$$$ of $$$A$$$ has length equal to length of $$$X$$$, then the remainder sequence $$$R$$$ (after removing $$$LIS$$$) may not be decreasing. Consider the case where $$$Y = [4, 3, 2]$$$, $$$X = [2, 3, 4, 5]$$$, $$$A = [4, 2, 3, 2, 3, 4, 5]$$$. $$$LIS = [2, 3, 4, 5]$$$, $$$R = [4, 2 ,3]$$$