Let's take some regular house with a man, which has to go to the toilet every n minutes, requiring the seat to be up, and a woman, which has to do it every m minutes, requiring a seat to be down. Is there a possibility to create a O(1) algorithm which will output the exact number of toilet seat movements for a given period of X minutes? There are two different additional inputs:
1. The man always leaves the seat up after a visit.
2. The man always puts the seat down after a visit.
Conclusion: in the real life (which involves n being much more than m, with X->infinity), it is proven that there is no difference in a number of seat movements.
But if a man does it more often, then a woman, it will prolong the seat life if he will just leave the seat up, but is this case one of them (or both) should probably see a doctor.
Now I know what is the best for the seat itself, but which person makes more movements - is another question (which should not be asked anyways).
For 2, the answer is 2*floor(X/n). The man will always go to the bathroom with the toilet seat down and leave it down. The woman will never put it down, since it's only up when the man goes to the bathroom.
1 is a little more tricky.
EDIT: Duh. For 1, the answer is 2*floor(X/m). The toilet seat only transitions when the woman goes to the bathroom.
EDIT2: Plus or minus the initial state of the toilet.
EDIT3: My answer to 1 is only correct if m>=n. I'll figure out the rest later.
EDIT4: If n>=2m, then it's 2*floor(X/n), since the seat will only transition when the man goes pee. If n>m, I believe the answer is also 2*floor(X/n), but I need to work out the math.
EDIT5: So, for 2m>n>m, the seat transitions when the man goes pee after the woman and vice versa. The sequence of man/woman visits repeats every least_common_multiple(m, n) minutes, so we only need to concern ourselves with what happens in that time period. The only time the seat would not transition when the man uses it would be if he managed to visit it twice in a row. Given that the woman is visiting more often than the man, between every man visit there is at least one woman visit. (Twice at the beginning or end.)
Answer 1 then becomes: (n>m ? 2*floor(X/n) : 2*floor(X/m)) + (remainder(X/n) > remainder(X/m) ? 1 : 0). Or something like that.