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 ith round, you toggle every i bulb. For the nth round, you only toggle the last bulb. Find how many bulbs are on after n rounds.

Example:

Given n = 3.

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.

对于解这道题目如果直接想不到什么方法，那就举一个具体的例子分析，我们假设灯泡个数n等于10，N代表on， F代表off，这个执行的过程如下：
[img]http://dl2.iteye.com/upload/attachment/0115/3694/e931ff24-be94-3481-821b-682f2bc1765f.png[/img]
从中可以看出只有1，4， 9 三个灯泡被toggle了偶数次(把所有灯打开之后的次数), 其余都是被toggle了奇数次，因此我们只需要得到灯泡个数中有几个平方数就可以了。代码很简单：

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