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

只有被开关奇数次的灯泡最后能够保持开启状态，故对每个灯泡i来说，如果当前趟数j能整除灯泡的编号i，那么i就会被开关一次，由于能整除i的数字都是成对出现，比如对5来说有1和5，对6有1和6,2和3，只有对平方数而言，整除它的数会出现两个数字相同的情况，这样的话就会出现被奇数次开关，比如对4而言，能整除它的有1和4，还有2，所以只需要找到n以内的完全平方数的个数即可，对n开方取整可以做到这一点，参考自Math solution。

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