leetcode-319-bulb switcher

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.

有n个开关，第i次操作你拨动所有i整数倍序号的开关。问最后有几盏灯留下了。

任何一个开关都会被拨动偶数次，除非它是一个完全平方数，本题目即是求不大于n的完全平方数有几个

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

    }
};