You have to divide the participants into equal teams (rounded)
For example, for "15 10" test - (1, 1, 1, 1, 1, 2, 2, 2, 2, 2)
Good luck :)

Well I don't understand this topic CLEARLY but I can trace some steps towards the final solution.

Let's fix a possible team distribution i.e. number of teams
Then, we need to find the best possible distribution to maximize the product
For eg for 4 teams a1,a2,a3,a4 will be the team distribution
a1+a2+a3+a4=n
now we need to maximize the function f(a1,a2,a3,a4)=a1a2+a1a3+a1a4+a2a3+a2a4+a3a4
under the constraint a1+a2+a3+a4=n, which can be done using lagrangian multiplier.
g(a1,a2,a3,a4)=f(a1,a2,a3,a4)+p(n-a1-a2-a3-a4)
lets take derivative of the function g at a1
g'(a1) = a2+a3+a4 -p
g'(a2) = a1+a3+a4 -p
g'(a3) = a1+a2+a4 - p
g'(a4) = a1+a2+a3 - p
Equate all of them to zero
a1+a2+a3=a2+a3+a4=a1+a3+a4=a1+a2+a4=p
from 1st and second we find a1=a4
from 2nd and 3rd a2=a1
similarly we find a1=a2=a3=a4, all a's must be equal
so a1+a2+a3+a4=n
a1=n/4
In general
a1=n/p, where p is the number of teams.
When p is not divisible, we need to make all of them as equal as possible by distributing n % p.

I don't understand lagrangian multiplier by heart(I have not tried to). But the tool works here.

We can try bruteforcing all possible team numbers from 2 to k and find which is best.

Edited by author 07.11.2016 22:51