319. Bulb Switcher

There are n bulbs that are initially off. You first turn on all the bulbs. Then, you turn off every second bulb. On the third round, you toggle every third bulb (turning on if it's off or turning off if it's on). For the i-th round, you toggle every i bulb. For the n-th round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Input: 3
Output: 1 
Explanation: 
At first, the three bulbs are [off, off, off].
After first round, the three bulbs are [on, on, on].
After second round, the three bulbs are [on, off, on].
After third round, the three bulbs are [on, off, off]. 

So you should return 1, because there is only one bulb is on.

class Solution {
    public int bulbSwitch(int n) {
        return (int)Math.sqrt(n);
    }
}

总结：

prime numbers have two switches, at 1 and at itself, so it's off

non-prime numbers are switch at its factors, but when it can be represented as a * a, then it is at on status because no other factors can turn it off.

like 9 = 3 * 3, and 4 = 2 * 2, so at 4 and 9 the bulb is on.

36 -> 25 -> 16 -> 9 -> 4 -> 1, 36 has 6 on bulbs.