you jinxed it (at least for me) :(

Video editorial and screencast are being uploaded to my youtube channel now. There are lots of nice solutions described in the comments here (and the atcoder editorial is usually quite good), but if you would like to see stuff explained more visually, feel free to look out for that :)

always try inserting new comer into two maximum existing values! Submission

Greedy, keep track of which element contribute to the answer and how many times, like the 1st one contribute only 1 time to the answer, rest of the numbers does it twice. So just take the (sum of largest n / 2 element — 1st largest).

I maintained a priority queue, if anyone is interested I can explain the solution , because I think people have posted better solutions.

Consider all 2^h subsets of rows and 2^h subsets of cols and then try every pair of subsets that is brute force for all 2^h*2^w subsets

C is just BruteForces My submission

since it was only 6. I did the Bruuuuuuuute force. Generated all sized subsets then ran a loop for every row and every column and for each combination checked whether it's valid or not.
My Submission-C

You can consider the nodes and edges independently if you count nodes as +1, and edges as -1

It is similar to this question of codeforces

https://atcoder.jp/contests/abc173/submissions?f.Task=&f.Language=&f.Status=&f.User=Gary

F nice formular, but not understandable for those who did not solve it.

Problem D:

I have implemented it in the exact way that you mentioned, but it just happened to fail at only 2 test cases, can you help me in finding out the bug in my submission ?

can you explain F?

In solution F: My understanding:

1. X= contribution of all the nodes in total summation

2. Y=Contribution of each edge in the total summation

3. ans=X-Y

Kindly explain your recurrence(I understood Y ,kindly explain X)?

Isn't this unfair, giving a question which is already available online, I mean the exact same question, most of the participants who googled something similar to find hints might have ended up copying the exact same code without any efforts of their own.

E: case out n==k, and when everything is negative.

Otherwise, you can repeatedly remove the smallest two nonegatives and replace them with the largest two negatives as long the product of the two negatives is bigger.

F: You can consider nodes and edges independently. Each node contributes +1 to the answer, each edge -1. You just need to count the number of ranges each node and each edge is part of.

For F for every segment think that every element is disconnected, and their contribution to answer is the length of segments. Now for each edge see in how many segments it can occur and subtract that number from answer.

just look up all possible way to select rows and columns either by bitmasking or dynamic programming and check number of blacks for each combination

Brute force

Check for all possible subsets, if we exclude these then if the number of black cells is k or not.

Max number of rows = 6

Max number of subsets of rows = 64

Max number of columns = 6

Max number of subsets of columns = 64

Total number of subsets = subsets of rows * subsets of columns

Max total number of subsets = 64 * 64

Complexity to check each subset = O(nm)

Total complexity = O(2^n * 2^m * n * m)

You can just brute force all the possible choices and for each choice count the number of remaning black squares. Check my submission for details: https://atcoder.jp/contests/abc173/submissions/15012234

I was too getting wrong ans on 1 test case . Then I checked the modular operation I was doing and found the error in it

I think it is less code if we separate the numbers into three groups, counting also the 0.

https://atcoder.jp/contests/abc173/submissions/15010921

I think mine is much cleaner. The idea is that you case out n==k and cases in which all numbers are negative.

Otherwise, you can start with the positive numbers, and replace two positive numbers with two negative number repeatedly.

That code is just:

		int nPos=Math.min(pos.length, k);
		int nNeg=k-nPos;
		if (nNeg%2==1) {
			nNeg++;
			nPos--;
		}
		while (nNeg+2<=nNegs && nPos>=2 && 
				pos[nPos-1]*(long)pos[nPos-2] < negs[nNeg]*(long)negs[nNeg+1]) {
			nPos-=2;
			nNeg+=2;
		}
		print(mul(posCS[nPos], negCS[nNeg]));

https://www.geeksforgeeks.org/maximum-product-subsequence-size-k/

Here is good implementation

E also makes me frustrated. :'(

If you want to see submission of a perticular person then, go to standing hover over the username and click on the searching symbol (magnifying lens). You can also do that by writing the username in all submissions. But I like the first one this way I can see the fastest submissions.

https://atcoder.jp/contests/abc173/submissions/14996063

I can explain you if you need help for C

because your if(k==n) part is wrong

Consider this testcase:

5 4
2 1 0 -1 -2

The product of first k numbers and the product of last k numbers are 0, but the correct answer is 4.

Suppose the array is-
-4 -3 -2 -1 0 1 3 4 and k = 4
Product of first k numbers = 24
Product of last k numbers = 0
Your Answer = 24
Actual Answer = (-4)(-3)(3)(4) = 144