Bulb Switcher

最新推荐文章于 2021-11-15 16:42:47 发布

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本文详细解析了BulbSwitcher问题的解决思路，从初始状态到最终状态的变化过程，揭示了数学原理背后的奇妙之处。通过一行代码解决问题，展示了简洁高效的算法之美。

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 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.

思路

最开始自己感觉是公约数的问题，但是，怎么求公约数个数，比自己小的数的公约数个数，偶数就是关，奇数就是开

BUT

开了下别人的代码，真的是感叹数学真奇妙啊
一行就解决了 return sqrt(n);
因为你看啊，公约数是成对出现的1和本身，但是除了那些平方和，1,4,9–1,2,3，都是单独的，所以这些情况下他们是奇数，返回有几个这样的数字就行了

代码

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