Question

I am having the following problem that I cannot solve.
Let’s say I have an array of 100 bools, all false.
I have a number N < 100 and I want to toggle N of these booleans to true, but so that they appear as regularly spaced as possible.
This is obviously simple when N divides 100, but how to do it for instance if N = 53?
I naively tried :

for(int i =0; i<53; ++i)
    std::cout << i*100/53 <<'\n';

to get 53 equaly distributed integers, and in this case I actualy have distincts integers, but I’m not sure this always works. Any pointers on this?

Answer 1

For the method you suggest, you will always get distinct integers for i * 100/X and (i + 1) * 100/X as long as X < 100 (as the difference is 100/X, which is > 1). So it’s a simple proof by induction that all the integers you obtain this way are unique.

So this is a fine approach. The real question, which only you can answer, is how do you define equally distributed integers?

Perhaps for X > 2, you should always have i[0] and i[99] set to true and divide the space in between equally. For the special case of X = 1, set the middle element (50th or 51st) to true.