[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 ith round, you toggle everyi 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】

每个灯如果要在最后是开的必然是因为它被开关了奇数次，而一盏灯在第i轮要被开关只有它的index能被i整除，一个数除非是平方数，不然其因数必然是成对，那么最后开着的灯只可能是index是平方数的灯。

【代码】

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