319. Bulb Switche

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.代码如下:(超时,但运行结果正确)
 1 public class Solution {
 2     public int bulbSwitch(int n) {
 3         if(n==0)
 4         return 0;
 5         if(n==1)
 6         return 1;
 7
 8         int[] a=new int[n];
 9         for(int i=0;i<n;i++)
10         a[i]=1;
11         for(int j=2;j<=n;j++)
12         {
13             int k=1;
14             while(j*k-1<=n-1)
15             {
16                 if(a[j*k-1]==0)
17                 a[j*k-1]=1;
18                 else
19                 a[j*k-1]=0;
20                 k++;
21             }
22         }
23         int count=0;
24         for(int j=0;j<n;j++)
25         {
26             if(a[j]==1)
27             count++;
28         }
29         return count;
30     }
31 }
时间: 2024-11-25 15:41:07

319. Bulb Switche的相关文章

Leetcode 319. Bulb Switcher

这是一个非常巧妙的题目,网站称之为brain teaser,意思就是代码写起来会非常简介(就是逗你玩:)),题目如下: 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 tu

319. Bulb Switcher

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

Leetcode 319 Bulb Switcher 找规律

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); } };

319 Bulb Switcher 灯泡开关

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,

过中等难度题目.0310

  .   8  String to Integer (atoi)    13.9% Medium   . 151 Reverse Words in a String      15.7% Medium     . 288 Unique Word Abbreviation      15.8% Medium     . 29 Divide Two Integers      16.0% Medium     . 166 Fraction to Recurring Decimal      17.

继续过中等难度.0309

  .   8  String to Integer (atoi)    13.9% Medium   . 151 Reverse Words in a String      15.7% Medium     . 288 Unique Word Abbreviation      15.8% Medium     . 29 Divide Two Integers      16.0% Medium     . 166 Fraction to Recurring Decimal      17.

LeetCode Problems List 题目汇总

No. Title Level Rate 1 Two Sum Medium 17.70% 2 Add Two Numbers Medium 21.10% 3 Longest Substring Without Repeating Characters Medium 20.60% 4 Median of Two Sorted Arrays Hard 17.40% 5 Longest Palindromic Substring Medium 20.70% 6 ZigZag Conversion Ea

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

ZOJ 3203: Light Bulb

ZOJ 3203: Light Bulb 1 ///@author Sycamore, ZJNU 2 ///@date 2017-02-09 3 #include<algorithm> 4 #include<iostream> 5 #include<iomanip> 6 #define eps 1e-10 7 using namespace std; 8 double H, h, L, D; 9 double getL(double a) 10 { 11 return