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.

Analysis:

这道题等同于找出一个数的因子个数，每个因子翻转一次。当因子个数为奇数时，翻转奇数次（ON）； 当为偶数时，翻转偶数次（off）。

而我们知道只有完全平方数（1，4，9….）的因子个数是奇数，因此使用sqrt(n)求出完全平方数个数即为答案。经测试，C++中int是向下取整。

Source Code（C++）:

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