[CI] 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.

It’s quite a interesting problem and more like a math problem than a coding one. The direct solution is do the swith round by round and we can get the result at the final round. The time complexity of the direct solution is O(n2). The example code is as follows,

Seems like a perfect answer which definitely is not. When I submitted the above code on LeetCode OJ, the platform threw out an error of RUNNING TIME OUT. So it’s time to get the code optimized.

As you can see, each one of the bulb is turned on or off for n times which is the number of divisors. But still this cannot solve our problem for the time complexity is O(n2) too! What about printing the result to see whether there is a pattern? Let do it.

when n=40, the result is 1001000010000001000000001000000000010000 . And when n=60, the result is 100100001000000100000000100000000001000000000000100000000000 .

What do you see, guys? Yes, the mumber of each group which has only one 1 is an Arithmetic sequence. So we can get the final result without any toggling rounds. A half but passed solution is as follows,

That’s it.

The time complexity of the above code is O(n) and the code can pass the OJ’s test. But is this the most optimized one? No, of course. If you know more maths, the above code can be optimized to the time complexity of O(1) which means there is some formula to get the final result. Can you figure it out, buddy?

By the way, the Ultimate answer is  SQRT(n).

source page: https://leetcode.com/problems/bulb-switcher/

健身计划一期(2015.8.28-2015.10.31)

大家相互监督、相互鼓励,把身体练好!

目标

体重不做高要求,减10斤左右即可,关键是脂肪减下去,肌肉长起来

人员

jasper, bill, xinli

一期周期

2015.8.28-2015.10.31

时间

每周保证四天,每天中午至少一小时

流程

11:40出发,健身完毕后洗澡,吃饭

健身流程

  • 跑步30min以上
    • 慢跑,心率140-160 beat/min
  • 柔韧5min
    • 压腿
    • 扭腰
    • 胳膊
  •  力量15min,包括
    • 臂力:选择合适重量,每组15个,至少两组,做到不能做为止
    • 腹肌:仰卧起坐或够腿,每组15个,做到不能做为止
    • 卧推:选择合适重量,每组20个,至少两组,做到不能做为止
  • 放松活动10min

饮食

  • 健身时多饮水
  • 多吃蛋白质,如蛋清、瘦肉、鸡肉、豆制品、脱脂奶等
  • 少吃碳水化合物,如米饭、馒头、面条等主食,糖类等
  • 不吃油炸、肥肉、坚果等高脂肪含量的食物
  • 不喝饮料(0热量的可以)
  • 晚饭少吃
  • 保证睡眠,良好的睡眠是最轻松的减肥方式

面试程序题-二分查找(1)

一个数组,大小先减后增,请找到增减部分的分界点,要求算法时间复杂度O(logN)。

下面给出了一个递归实现的版本。

这个问题还可以让查找其中的某个元素,也是二分,思路一样。

#include <iostream>
#define MAX 13
using namespace std;
void findBreakpoint(int* a, int starti, int endi){
int i = (endi-starti)/2+starti;
if(starti==i){
cout<<i<<endl;
return;
}
if(a[i]<a[i+1]){
findBreakpoint(a, starti, i);
}else{
findBreakpoint(a, i+1, endi);
}
}

int main(){
int a[MAX] = {12, 11, 10, 8, 5, 6, 7, 9, 11, 12, 13, 20, 50};
findBreakpoint(a, 0, MAX-1);

return 0;
}

一面在准备一些基础的算法,一面在准备语言相关,泛技术渣太苦逼了,在准备两种语言PHP和Java,而我最近一直在做是却是C++,是不是相当奇葩。。。

关于Java类的hashCode和equals方法

前面两篇文章中提到了利用哈希表的数据结构,如HashMap,HashSet等,这些数据结构利用对象的hashCode和equals函数来识别对象是否相同,原则如下:

* hashCode不同的对象一定不同

* hashCode相同的对象不一定相同

比方说HashMap,查找元素时,会先以key的hashCode来查找,如果找到对象,然后再以equals方法来判定是否相等,如果相等,才返回相应的value。

自定义数据结构做key时,需要成对的实现两个函数才可以正确的使用利用哈希表实现的数据结构。