The "minimum" part of this problem was particularly challenging for me. I could see that we need to create m teams as "uniformly" as possible, but I was having a lot of trouble coming up with how to compute that. Hence this post explains the intuition behind distributing n participants into m teams as uniformly as possible.

Note: as a matter of notation, the quantity n / m denotes integer truncating division throughout.

In hindsight, the reason this problem gave me so much trouble is because it demands an understanding of n % m that I did not initially possess. My tried and true understanding of n % m has always been

n = (n/m)*m + n%m.

In english, Equation 1 can be read as "starting with n participants, we can form n / m teams, each having m participants, plus some leftover participants." However, 478B - Случайные команды requires a fixed number of teams (i.e. m), so this interpretation will not do. With this in mind and by a twist of perspective, we can rewrite Equation 1 as

n = m*(n/m) + n%m.

It may look like we've accomplished nothing with this rewrite, but we can now read this in a more useful way! Concretely, Equation 2 can be read as "starting with n participants, we can form m teams, each with n / m participants, plus some leftover participants."

As a result, we now have m teams which are uniform as can be (they all have an equal number of participants!). All that remains is to distribute the leftover participants as uniformly as possible into the m teams. It is not hard to see that the way to accomplish this is to add each leftover participant to a different team. As a result of this process, we wind up with

n = (n%m)*(n/m + 1) + (m - n%m)*(n/m).

This can be read as "starting with n participants, we can form m teams. n % m of those teams have n/m + 1 participants in them (because there are n % m leftover participants). m - n%m of them (i.e. the rest) have n / m participants in them."

As an example, consider n = 8 and m = 3. Beginning with Equation 2, applying the procedure which follows, and arriving at Equation 3, we have

8 = (2 + 2 + 2) + 2 
  = (2+1 + 2+1 + 2) + 0
  = 2*(2+1) + 1*2
  = (8%3)*(8/3 + 1) + (3 - 8%3)*(8/3).