LeetCode #319 - Bulb Switcher

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 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.

给定n个灯泡，初始情况下是熄灭的，在第i轮，每隔i个灯泡按一下开关（开着就变成灭，灭着就变成开），求n轮之后有多少灯泡亮着。如果直接按照题目的步骤循环n轮，时间复杂度肯定较高，其实我们任选第k个灯泡分析一下就能发现规律，第k个灯泡必须在k的因数轮才能改变状态，而除了完全平方数的因数个数是奇数，其他数的因数个数都是偶数，也就是说这些灯泡经过偶数次改变还是熄灭的，那么这道题就变成了求1~n一共有多少个完全平方数。

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