Divide an integer evenly with a maximum

Question

I need an algorithm for the following:

I'm given a specified target sum n, and a specified limit m. These are both positive integers.
I want to find an integer partition of the target sum n that has as few summands as possible.
Each summand must be less than or equal to the limit m.
Within the above constraints, the summands should be as close together as possible; that is, I want n to be partitioned as evenly as possible.

So, for example, if the target sum is n = 80 and each summand must be at most m = 30, then I need at least three summands, and the most even partition is 26 + 27 + 27.

How would I compute that?

Why not 26, 26, 28? Why is 27 an even integer? Why not 40 2's? — Andy Turner, Oct 13 '16 at 20:36
How about 30, 30, 20? Your requirements are incomplete, as you can see. And it's hard not to think this you're asking someone to do your homework for you... — Christopher Schneider, Oct 13 '16 at 20:39
Flagged as unclear what you are asking...because your requirements are either inconsistent, incorrectly recorded, or your example is wrong.. — Susannah Potts, Oct 13 '16 at 20:40

score 4 · Accepted Answer · answered Oct 13 '16 at 20:44

First, you get the size of the array with the following formula using integer division:

size = (variable + maximum - 1) / maximum

Next you fill the array with the following formulas:

extra = variable % size;
value = variable / size;
for each array value, set to value + 1 as long as there's extra; 
    value when the extra goes to zero.

score 0 · Answer 2 · answered Oct 13 '16 at 20:58

Just a QnD Algorithm and Code...untested.

 double n=107;
    double max = 22;
    int d = (int) Math.ceil(n/max);
    int[] result = new int[d];
    int res=0, i=0,iter=0;
    while(res!=n){
        iter= (int) Math.ceil(n/d);
        while(iter+res>n) iter--;
        res+=iter;
        result[i] = iter;
        System.out.println("i: " + i + " iter: " + iter + " sum: " +res);
        i++;
    }

Divide an integer evenly with a maximum

2 Answers2