319. Bulb Switcher My Submissions QuestionEditorial Solution Total Accepted: 15915 Total Submissions: 39596 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 thi…

[LeetCode]319. Bulb Switcher 解题报告(Python) 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/bulb-switcher/description/ 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 t…

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…

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 e…

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…

题目: 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 eve…

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 i-th round, you toggle every …

初始时有 n 个灯泡关闭. 第 1 轮,你打开所有的灯泡. 第 2 轮,每两个灯泡切换一次开关. 第 3 轮,每三个灯泡切换一次开关(如果关闭,则打开,如果打开则关闭).对于第 i 轮,你每 i 个灯泡切换一次开关. 对于第 n 轮,你只切换最后一个灯泡的开关. 找出 n 轮后有多少个亮着的灯泡.示例:给定 n = 3.状态off表示灯泡关闭,on表示开启.初始时, 灯泡状态 [off, off, off].第一轮后, 灯泡状态 [on, on, on].第二轮后, 灯泡状态 [on, off,…

有n盏关着的灯,第k轮把序号为k倍数的关着的灯打开,开着的灯关闭. class Solution { public: int bulbSwitch(int n) { return (int)sqrt(n*1.0); } };…

智商压制的一道题 这个题有个数学定理: 一般数(非完全平方数)的因子有偶数个 完全平凡数的因子有奇数个 开开关的时候,第i个灯每到它的因子一轮的时候就会拨动一下,也就是每个灯拨动的次数是它的因子数 而拨动偶数次是关,拨动奇数次是开 现在就是求哪些数的因子有奇数个,也就是求n以内的完全平凡数 这里又有一个定理: n以内的完全平方数个数是sprt(n) 所以代码很简单 public int bulbSwitch(int n) { return (int)Math.sqrt(n); }…