LeetCode *** 319. Bulb Switcher

题目：

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.

分析：

一个灯泡如果被按下偶数次，那么最终是off，被按下奇数次是on。

同时，第ith round，需要按下every i bulb，那意味着需要找到i的所有包括自身因子数。

以上两条结合即可知道，对于每个不能开平方的数而言，因子数始终为偶数，那么就是求解n之前所有能开平方的数。

代码：

class Solution {
public:
int bulbSwitch(int n) {
return sqrt(n);

}
};