Isn't it 944D — Game with String?

I think the method mentioned by the tutorial, is just some kind of modified convex hull algorithm.

Say you have only 0's, 1's and 2's in array.

To preserve average you can either pick 0 and 2 and change both to 1 or pick two 1's and change one of them to 0 and other one to 2. Obviously, it makes no sense to take 0, 2 change it to two 1's and then change it back to 0 and 2 again, you won't gain anything.

So now you pick one of the transformations 0,2 into 1,1 or 1,1 into 0,2. Pick one that gives you more numbers to change and perform transformations.

Example:

0 0 0 1 1 1 2 2 2 2 2

Here, you can pick two 1's and get the array:

0 0 0 0 1 2 2 2 2 2 2

but that way you changed only 2 numbers. It's better to pick 0's and 2's, change them into 1's so you will get:

1 1 1 1 1 1 1 1 1 2 2

(notice how you have to left 2's at the end since you don't have more 0's to "balance" the change in terms of average value).

For the next paragraph I will use this definition: cnt(x) — number of times x is present in array

How many changes will you get from transforming 0's and 2's into 1's? min(cnt(0), cnt(2)) * 2

How many changes will you get from transforming 1's into 0's and 2's? floor(cnt(1) / 2) * 2 (fancy way of saying cnt(1) if cnt(1) is even and cnt(1) — 1 if it's odd)

Special cases:

1) There are only 0's in array; there are only 0's and 1's in array — in those cases you cannot change anything (remember about the constraint that your minimal/maximal numbers have to be between minimal/maximal from original array, so you don't have any possibility for change without breaking this constraint or changing average). Obviously the same case for only 1's etc.

2) There are 0's and 2's in array — that just reduces to transforming as many of them as you can into 1's.

My solution:

http://mirror.codeforces.com/contest/931/submission/35947027

(it's not crystal clear, because of the part which actually changes the numbers in array, I guess I should have done it in different way)

Case1 : (xMax-xMin)<2

• Because there is no way to change the numbers such that sum remains the same as before. So we can't change any number and ans will be 'n' followed by the initial given numbers in the next line.

Case 2 : (xMax-xMin)==2

• Suppose xMax = a+1, then xMin = a-1 and the other number whose presence is not guranteed will be 'a'. So there are two ways,

  Way 1 — You replace a+1 and a-1 with 2 a, because sum will remain same.

  Way 2 — You replace 2 a with a+1 and a-1, in this case also sum will remain same.

  The most optimal way to get answer will be to follow one way at all steps, because there is no sense of replacing 2 a with a+1 and a-1, and then again replacing a+1 and a-1 with a.

  Now you just have to check which way, either 1 or 2 is more optimal.

  Best of luck