Don't see why "find max divizor below M and divide" solution doesn't work.

This may work:

Find all primes <= M. Let size of this array be x. then

for(int i=x-1;i>=0;i--)
{
    while(d<=M && N % prime[i] == 0)
    {
        d *= prime[i];
    }
    if(d > M)
    {
        cout<<d/prime[i]<<" ";
    }
}

should produce di in descending order.

If anyone wants to test a solution (or look through solutions if you have coach mode), that's problem H here: http://mirror.codeforces.com/gym/101490

Edit: actually, this problem wants the actual answer, not just the length of the answer so it's a bit different. My mistake.

if you take the logarithm of N and all of its divisors that are at most M, then now you need to find a subset of minimal size that sums up to N (an element can be picked multiple times). This is basically knapsack with real numbers now. Since you want it greedy... Well I guess it could have some properties:

For instance, some numbers from which you pick a subset are multiples of each other which could help the greedy solution.

I don't think this one helps, but the numbers are pretty low... although the amount of numbers can be large.

If I missed anything please say.

Maybe this works, factorize N, for now, the set D is the prime factorisation of N. If any element of D is > M, print impossible.

Otherwise keep removing the smallest element from the set D, let's call it A and find B the largest element in D such that A * B <= M, remove B and insert their multiplication back in the set. Keep doing this until you can no longer find two elements to multiply in the set.

EDIT: you actually should keep picking the 2 largest elements from the set, but this solution is for the problem you described, not sure how it would work for problem H in the gym