Bulb Switcher

319. Bulb Switcher

Total Accepted: 976 Total Submissions: 2611 Difficulty: Medium

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 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.

1 = 1*1    1次
2 = 1*2    2次
3 = 1*3    2次
4 = 1*4,2*2  3次
5 = 1*5
...
发现只有1,4,9,16这样的平方数被拉了奇数次
因为，任何一个数都是由因式对构成的，那么其必然是偶数次，特殊的，如果两个因子相等，那么因子数就成了奇数

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