I think if you have only two eggs and m tries, the answer would always be (m/2)*(m+1). The pattern would be 14, (14+13), (14 + 13 + 12)...I don't know how the pattern changes if you have more than 2 eggs; that's very tricky!!

Your analogy is the worst case scenario (eggs always break).
This question is asking for the best case scenario (how high can you go if eggs didn't break (meaning you can reuse the eggs as many times as you want)), while still making sure that you can find the exact critical floor should the eggs break.

That is beauty of algorithms. Everytime you do loop you add n time to your execute time. if you make loop in loop it is n^2 time and so on. You need to find better way to make this program a little bit faster.

not legit my friend :).

codewards made not good enough tests for ya ^^ . you will become a real hacker bro ^^