Divide N sticks among M boys as homogeneously as possible (ignoring order)

Question

There are $N$ sticks. $N$ is an integer greater than zero. I want to divide it among $M$ boys. $M$ is also a positive integer. Partitioning $N$ among $M$ is easy, but doing it as evenly as possible is difficult for me to think through. Can someone suggest an algorithm? Similar questions exist on this site, but I could not find an answer that solves this problem, though it is possible that I overlooked something.

EDIT:

The sticks are all homogeneous, the same in every respect. What I mean by "as evenly as possible" is that if there are 6 sticks and 3 boys, the division algorithm should output 2-2-2. If there are 5 sticks among 3 boys, it should output 1-2-2. The disparity between the minimum stick holder and maximum stick holder should be minimized. Ordering does not matter (e.g. 1-2-2 is the same as 2-1-2).

Answer 1

Give some of the boys $\lfloor N/M \rfloor$ sticks (i.e., divide and round down), and some of them $\lceil N/M \rceil$ sticks (divide and round up). Once you fix those two numbers, that uniquely determines how many boys get $\lfloor N/M \rfloor$ sticks and how many get $\lceil N/M \rceil$ sticks -- do some simple arithmetic, try a few examples, use the fact that $\lceil N/M \rceil = \lfloor N/M \rfloor + 1$ if $M$ doesn't evenly divide into $N$, and you'll work out the general formula.

Answer 2

# Divide perfectly homogeneously, as large as possible
for i = 0 to M
    array[i] = N / M    # integer division, resulting in floor

# Divide remainder
for i = 0 to N modulus K
    array[i] += 1

Answer 3

Give them $\lfloor\frac{N}{M}\rfloor$ sticks and then distribute the rest of sticks ($N-\lfloor\frac{N}{M}\rfloor$) among them by giving one stick to each boy until there are no more sticks. By solving it this way, the time complexity will be $\max(\lfloor\frac{N}{M}\rfloor, N \bmod M)$ .