题目描述：
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.
解题思路：
这道题目主要是看每个位置是否是被变化了奇数次。比如对于6，可以写成1x6、2x3，那么经过整个n次变化，这个位置经历了4次变化，所以是灭的状态；而9可以写成1x9、3x3，那么其经过3次变化，所以是点亮的状态；顺着这个思路，能够开方为整数的数字都是点亮的状态，所以只需要对整个n开方，然后向下取整即可。
代码如下：
public class Solution {
public int bulbSwitch(int n) {
return (int)Math.sqrt(n);
}
}