You need to specify what is the desired distribution of the random numbers.

If there are no further requirements, I would suggest one of the following:

(1)

• pick random number a[1] in interval 0 .. k
• pick random number a[2] in interval 0 .. k-a[1]
• pick random number a[3] in interval 0 .. k-a[1]-a[2]
• ...
• pick random number a[n] in interval 0 .. k-a[1]-a[2]-...-a[n-1]

If you have upper limit m on the range of the random number, use min(k-a[1]-... m) as upper bound of the interval.

Disadvantages: you will get a lot of small numbers and just a few big ones.

(2)

• pick n random numbers a[1], .., a[n] in interval 0 .. m, m being the upper limit
• s = a[1]+a[2]+...+a[n]
• multiply each a[i] by k/s (if integers are required, round down)

Disadvantages: It is unlikely to get large numbers this way. If integers are required, there will likely be a gap between the sum of numbers and k due to rounding error.

I think you get "nicer" numbers with option (2) but as stated above, it depends on the requirements.

Assuming k is less than 255 * n one solution is to assign k / n to every element of the array, then randomly subtract a value to the array elements.

// for (int i = 0; i < n; i++) array[i] = k / n;
// for (int i = 0; i < n; i++) array[i] -= randbetween(0, array[i]);
for (int i = 0; i < n; i++) array[i] = randbetween(0, k / n);

This has an expected sum of k / 2. By tweaking the randbetween() function you can change the probability of the resulting array sum.

It is easy to create one number within range [0, 255].

It is easy to identify if k > 255*n or k < 0 there is no solution.

If 0 <= k <= 255*n, the solution exists. Here we only talk about n > 1 condition.

You have created n-1 random numbers, and sum of the n-1 numbers is s1, suppose the nth number is x. So s1 + x = k, and x should be [0, 255]. If the n-1 numbers are all within range [0, a], then (n-1)*a + 255 >= k, we get a >= (k-255)/(n-1).

If k > 255, just let a = (k-255)/(n-1). It means s1 is [0, k-255]. Then the nth number x can be any random number within [0, 255]. So the solution is arbitrary select n-1 numbers each within [0, (k-255)/(n-1)] (you know (k-255)/(n-1) <= 255, thus it satisfied your condition), and select one random number within [0, 255].

If k <= 255, arbitrary select n numbers each within [0, k/n] (you know k/n is within [0, 255]).