All About Programming: February 2019

Year-End Stock Sale To Harvest Capital Losses: Beware Wash Sales!

7. Form 1099-B, which you receive from your broker by mid-February, reports all your stock sales from the prior year. It shows (in Box 1g) the amount of any nondeductible loss stemming from a wash sale involving covered securities (at least for those in the same account). Note that many brokerage firms reformat Form 1099-B into their own substitute statement that has columns labeled same as the boxes on the IRS form.

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How do I close the key Binding Resolver panel? - support - Atom Discussion

For others who might stumble on this topic, there are two ways to close the Keybinding Resolver panel:

Press Esc
Press Cmd+. on OS X or Ctrl+. on other platforms

Read full article from How do I close the key Binding Resolver panel? - support - Atom Discussion

6 Ex-Googlers Share How They Landed The Job

Look at virtually any "best places to work" list, and you'll likely find Google at or near the top. A 2016 report based on LinkedIn data found Google was the best at attracting and retaining its 60,000 employees, with benefits like free food, massages, and after-death benefits.

Read full article from 6 Ex-Googlers Share How They Landed The Job

Google Chrome is a Greedy RAM Hog. Here's How to Fix It.

Google Chrome may be most people's favorite browser, but it's far from perfect. One of its biggest flaws is how much of your computer's RAM it eats up.

Most laptops don't come with a huge amount of RAM; once the allotted memory is full, the computer can't process any more actions.

Each new tab that is opened in the browser will consume more RAM. In Chrome, 15 tabs can range from 1 GB to 2 GB of memory used, depending on the media content.

Take a look at how much memory your Chrome tabs and extensions by copying chrome://memory-redirect/ in your browser.

After doing a few tests with the same 15 tabs open in Mozilla Firefox, memory usage is 35 to 45% lower. Firefox appears to be a far more lightweight browser, a drastic improvement over older versions.

If you don't want to switch from Chrome, try downloading one of the following Chrome extensions to help reduce memory usage. I would recommend downloading only one, as more extensions will use more RAM.

Read full article from Google Chrome is a Greedy RAM Hog. Here's How to Fix It.

java - org.apache.lucene.index.IndexNotFoundException: no segments* file found in org.apache.lucene.store.RAMDirectory - Stack Overflow

I am new to Java and Lucene. My code gets a line from a file and stores it in Lucene Index. But when I create an IndexReader to search and read from the index it throws an exception.

My java code is below. On creating the IndexReader it throws an IndexNotFoundException

Read full article from java - org.apache.lucene.index.IndexNotFoundException: no segments* file found in org.apache.lucene.store.RAMDirectory - Stack Overflow

solr - How to fix IndexNotFoundException: no segments* file found? - Stack Overflow

I write my own code to index some data files myself. It was all working. Once I modified the schema.xml file, now i get internal server error that i have no idea how to interpret it.

Internal Server Error    request: http://127.0.0.1:8080/solr/update?wt=javabin&version=2      at org.apache.solr.client.solrj.impl.CommonsHttpSolrServer.request(CommonsHttpSolrServer.java:427)      at org.apache.solr.client.solrj.impl.CommonsHttpSolrServer.request(CommonsHttpSolrServer.java:249)      at org.apache.solr.client.solrj.request.AbstractUpdateRequest.process(AbstractUpdateRequest.java:105)      at org.apache.solr.client.solrj.SolrServer.add(SolrServer.java:69)      at org.apache.solr.client.solrj.SolrServer.add(SolrServer.java:54)      at solr.client.SolrClient.index(SolrClient.java:109)

Read full article from solr - How to fix IndexNotFoundException: no segments* file found? - Stack Overflow

SolrCloud Leader Election Failing - Sease

If you're having trouble getting a sane election, you can try deleting the lowest-numbered entries (as well as any lower-numbered duplicates) and try to foce the election again. Possibly followed by restarting the node with that lowest-numbered entry.

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Find maximum in a stack in O(1) time and O(1) extra space - GeeksforGeeks

Given a stack of integers. The task is to design a special stack such that maximum element can be found in O(1) time and O(1) extra space.

Examples:

Given Stack :  2  5  1  64   --> Maximum    So Output must be 64 when getMax() is called.

Below are the different functions designed to push and pop elements from the stack.
Push(x) : Inserts x at the top of stack.

If stack is empty, insert x into the stack and make maxEle equal to x.
If stack is not empty, compare x with maxEle. Two cases arise:
- If x is less than or equal to maxEle, simply insert x.
- If x is greater than maxEle, insert (2*x – maxEle) into the stack and make maxEle equal to x. For example, let previous maxEle was 3. Now we want to insert 4. We update maxEle as 4 and insert 2*4 – 3 = 5 into the stack.

Read full article from Find maximum in a stack in O(1) time and O(1) extra space - GeeksforGeeks

Print reverse of a Linked List without extra space and modifications - GeeksforGeeks

Given a Linked List, display the linked list in reverse without using recursion, stack or modifications to given list.

Examples:

Input : 1->2->3->4->5->NULL  Output :5->4->3->2->1->NULL    Input :10->5->15->20->24->NULL  Output :24->20->15->5->10->NULL

Read full article from Print reverse of a Linked List without extra space and modifications - GeeksforGeeks

This Site Checks to See If Your Next Chrome Extension Is Safe

When it comes to security, Chrome extensions are the wild west. Many of them ask for broad permissions, and if you pick the wrong one, you'll end up sharing a lot more personal data than you probably intended with a mysterious internet stranger.

Now, there is a new way to check extensions you're considering downloading to give yourself some piece of mind. CRXcavator, a web tool from security firm Duo Labs, analyzes Chrome extensions and gives you a security report based on the findings from an ongoing review of the Chrome Web Store.

Read full article from This Site Checks to See If Your Next Chrome Extension Is Safe

The Smartest Thing To Do With Your 401(k) When You Leave A Job | InvestingAnswers

Your 401(k) is probably one of the best benefits you have with your job.

But you'd never stay with a company just because of the 401(k), right? If you get a better offer somewhere else, you'll probably want to take it.

So when you do finally leave, what happens to your 401(k)?

The money is yours, of course. It's your retirement account. You can decide what to do with your 401(k) when you change jobs, but you need to know what your options are -- and the consequences.

If you make the wrong choice, it could cost you thousands of dollars.

Read full article from The Smartest Thing To Do With Your 401(k) When You Leave A Job | InvestingAnswers

usability - What keyboard input makes sense for tabbing out of a rich text field? - User Experience Stack Exchange

If your application has more than one editable area, or this area is not the main feature of your application (like a Comments-Section editor, etc...), we settled on using Escape to unfocus the textarea. The focus would then move on a wrapper outside this area, and the textarea would become unfocusable until focus has moved out of the wrapper (at which time the wrapper would get unfocusable again). Alternatively, Escape could disable Tab-as-input and display a message saying Tab can now be used to escape the field (until the user types any other character or press Escape again).

Read full article from usability - What keyboard input makes sense for tabbing out of a rich text field? - User Experience Stack Exchange

Mark Cuban explains how to make a 16% guaranteed return - MarketWatch

"The reason for that is whatever interest you have — it might be a student loan with a 7% interest rate — if you pay off that loan, you're making 7%," Cuban said.

Read full article from Mark Cuban explains how to make a 16% guaranteed return - MarketWatch

Mark Cuban explains how to make a 16% guaranteed return - MarketWatch

Yes, you read that right. You could be getting a 16% return on your money right now, and it's not that hard to do.

Read full article from Mark Cuban explains how to make a 16% guaranteed return - MarketWatch

Google is testing a new feature that speeds up the Chrome browser – here’s how to get it now – BGR

Google is testing a new feature that speeds up the Chrome browser – here's how to get it now – BGR

You have access to this nifty new feature if you're running the latest Canary build, but it's not enabled by default. As Ghacks explains, however, enabling it is beyond easy. Simply type chrome://flags/#disable-best-effort-tasks in Canary's address bar and hit enter, and you'll open a settings page with the "Skip best effort tasks" flag right at the top. Click the drop down menu next to the feature and select "Enabled," then restart Chrome Canary and you're good to go.

Read full article from Google is testing a new feature that speeds up the Chrome browser – here's how to get it now – BGR

Count total set bits in all numbers from 1 to n - GeeksforGeeks

Given a positive integer n, count the total number of set bits in binary representation of all numbers from 1 to n.

POJ 动态规划题目列表 - qijinbiao1 - 博客园

1.这份列表当然不是我原创的，从文库里下载了一份，放到这里便于自己浏览和查找题目。

※最近更新：Poj斜率优化题目

1180，2018，3709

列表一：经典题目题号：
容易：
1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 1322, 1414, 1456, 1458,

1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018,

2029, 2039, 2063, 2081, 2082, 2181, 2184, 2192, 2231, 2279, 2329, 2336, 2346, 2353, 2355, 2356, 2385, 2392, 2424,
不易：
1019, 1037, 1080, 1112, 1141, 1170, 1192, 1239, 1655, 1695, 1707, 1733(区间减法加并查集),

1737, 1837, 1850, 1920(加强版汉罗塔), 1934(全部最长公共子序列), 1964(最大矩形面积，O(n*m)算法), 2138, 2151, 2161, 2178,
推荐：
1015, 1635, 1636(挺好的), 1671, 1682, 1692(优化), 1704, 1717, 1722, 1726, 1732, 1770,

1821, 1853, 1949, 2019, 2127, 2176, 2228, 2287, 2342, 2374, 2378, 2384, 2411

列表二：完整DP题目列表

1015 Jury Compromise
1029 False coin
1036 Gangsters
1037 A decorative fence
1038 Bugs Integrated, Inc.
1042 Gone Fishing
1050 To the Max
1062 昂贵的聘礼
1074 Parallel Expectations

Read full article from POJ 动态规划题目列表 - qijinbiao1 - 博客园

Palindrome Partitioning | DP-17 - GeeksforGeeks

Given a string, a partitioning of the string is a palindrome partitioning if every substring of the partition is a palindrome.

Count binary strings with twice zeros in first half - GeeksforGeeks

We are given an integer N. We need to count the total number of such binary strings of length N such that the number of '0's in the first string of length N/2 is double of the number of '0's in the second string of length N/2.
Note: N is always an even positive integer.

Examples:

Input : N = 4
Output : 2
Explanation: 0010 and 0001 are two binary string such that the number of zero in the first half is double the number of zero in second half.

Read full article from Count binary strings with twice zeros in first half - GeeksforGeeks

dp之背包总结篇 - 紫忆 - 博客园

前言：背包问题在dp中可以说是经典，作为一个acmer，到现在才正式学习dp，可以说是比较失败的。我个人比较认同一点，想要做一个比较成功的acmer，dp、搜索、数学必须精练，比较遗憾的是，对我我自身而言，并没有早早的认识到这个问题，不过现在知道了，还有一年，也不算晚。还有，我建议学背包的童鞋，都看背包九讲......

Read full article from dp之背包总结篇 - 紫忆 - 博客园

Teris | Algorithm Notes

This problem asks you to simulate the playing process of teris game. This problem needs no advanced algorithms but needs a lot of work and circumspection. To handle large board we may need to use LinkedList to keep the board status to avoid large amount of cost of removing rows.

Read full article from Teris | Algorithm Notes

LeetCode 514. Freedom Trail – Ysing's blog

由于做出来的时候posted在discuss了，所以我直接复制里面的内容

看Tag上dp上打了个冒号，这似乎不完全算是一个dp solution
基本思路：

对于解一个key，我们可以将这个key的长度作为我们接这道题的step,每一步step接一个char,解到最后一个字符最小step就是我们需要求的答案
所以算法开始，我们需要知道，key对应的每一个char在string ring中的哪一个位置，需要对ring for一遍，将对应的char记录在相应的mp中，这样可以节省很多的搜索时间。
开始搜索key的循环，算法第一步开始的index为 ring[0],需要求转动ring，使达到ring[index]==key[0](第一步)的步数最小，因为可能ring中不止有一个key[0]，所以需要对每一个符合的进行选择，（每一个dp[i][j],记录的实际就是在第i步时，通过起点到达j的最短step,主要是考虑rotate left or right）
依次循环key.size()步后结束循环
对最后一个dp数组for出最小的值,return minStep + key.size()即是最小的步数

Simple dp idea

recording each index of characters in ring,beacuse each characters we have search in this time would be starting index in the next search
How could we solve the problem that rotate the ring to the left or right ? My idea is min((tmp[j] + size -it)%size,(it + size - tmp[j])%size)

Suppose you want to rotate the ring to the right and search 'k', and the size is 5. We could calculate it by this + size -k(index)%size

this -  -  -  -  k

If we want to rotate the ring to the left，what should we do? It is the same problem with above problem,move this to it 's right,and reach k

k -  -  -  -   this

So we could calculate it by k(index) + size -this%size
There are many people use abs() instead of %size,I think it's faster than mine :)

Read full article from LeetCode 514. Freedom Trail – Ysing's blog

LeetCode 928 Minimize Malware Spread II | ReeeStart

这题和Minimize Malware Spread I的区别是:

I: 问如果从initial中取掉一个server, 把这个server的病毒给清了, 那要对哪个server杀毒, 能使得最终Internet里被感染的结点数最少?
II: 问如果从Initial中取掉一个server, 把它从整个网络中拿掉, 要拿掉哪个server, 能使最终被感染的结点数最少?

#Solution

#Solution #1 - Union Find - Without Optimization

最简单的就是重头做N遍union-find, 每次取掉initial中的一个server. 然后每次求中毒的数量.

Time: O(N * NlogN)

#Solution #2 - Union Find Optimization

第二种解法是对上一个方法的优化, 通过观察我们可以得到, 因为只能取掉一个server:

如果排除initials再union-find之后的某个component, 不包含任何initials, 则不受影响.
如果排除initials再union-find之后的某个component, 只包含一个中毒server, 那么取掉这个中毒的可以使这个component不受影响.
如果排除initials再union-find之后的某个component，包含多个中毒servers, 那么取掉谁都一样. 该中毒的还是要中毒.

所以我们最终的答案, 肯定是在那些只包含一个中毒server的component里面找.

Read full article from LeetCode 928 Minimize Malware Spread II | ReeeStart

Leetcode Hard DP · Mowayao's Blog

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2. You have the following 3 operations permitted on a word:

Insert a character
Delete a character
Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros" Output: 3 Explanation: horse -> rorse (replace 'h' with 'r') rorse -> rose (remove 'r') rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution" Output: 5 Explanation: intention -> inention (remove 't') inention -> enention (replace 'i' with 'e') enention -> exention (replace 'n' with 'x') exention -> exection (replace 'n' with 'c') exection -> execution (insert 'u')

Read full article from Leetcode Hard DP · Mowayao's Blog

Maximum profit by buying and selling a share at most k times - GeeksforGeeks

In share trading, a buyer buys shares and sells on a future date. Given the stock price of n days, the trader is allowed to make at most k transactions, where a new transaction can only start after the previous transaction is complete, find out the maximum profit that a share trader could have made.

xkcd: How to

For any task you might want to do, there's a right way, a wrong way, and a way so monumentally bad that no one would ever try it. How To: Absurd Scientific Advice for Common Real-World Problems is a guide to the third kind of approach. It's the world's least useful self-help book.

Read full article from xkcd: How to

How To: Absurd Scientific Advice for Common Real-World Problems

Randall Munroe, proprietor of the excellent XKCD and author of What If? and Thing Explainer, is coming out with a new book in a few months called How To: Absurd Scientific Advice for Common Real-World Problems.

Read full article from How To: Absurd Scientific Advice for Common Real-World Problems

Blue Eyes - A Logic Puzzle

A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.

On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.

The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:

"I can see someone who has blue eyes."

Who leaves the island, and on what night?

There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."

And lastly, the answer is not "no one leaves."

I've done my best to make the wording as precise and unambiguious as possible (after working through the explanation with many people), but if you're confused about anything, please let me know. A word of warning: The answer is not simple. This is an exercise in serious logic, not a lateral thinking riddle. There is not a quick-and-easy answer, and really understanding it takes some effort.

Read full article from Blue Eyes - A Logic Puzzle

Codeforces 1107E. Vasya and Binary String（DP） | 张慕晖的博客 | LUX ET VERITAS

给定一个长度为n的只包含0和1的字符串，将该串中连续i个相同字符消去将得到a[i]的收益，问消去整个字符串能得到的最大收益是多少？

分析

在不会做的三道题上鸽了好久，最后还是觉得DP最好写……

这个题解[soln1]特别简明扼要，大概抓住了问题的本质，不过我可能还没有完全理解它：

每一个状态都可以用[start, end, prefix]表示，其中start和end是字符串中的起始和结束index（不妨假设两端都是闭区间）（不是也行），prefix是字符串前面附加的和s[start]相等的字符的数量（包括s[start]）（不包括大概也行）
每一个状态都有两种递推方法：
- 第一种是直接消去字符串前面附加的字符：dp[start, end, prefix] = a[prefix] + dp[start + 1, end, 1]
- 第二种是在字符串其他位置找到一个和s[start]相同的字符，然后消去中间的字符，获得一个更大的前缀：dp[start, end, prefix] = max(dp[start+1, i-1, 1] + dp[i, end, prefix+1]) (start < i <= end, s[i] == s[start])

Read full article from Codeforces 1107E. Vasya and Binary String（DP） | 张慕晖的博客 | LUX ET VERITAS

Google Interview Questions Deconstructed: The Knight’s Dialer

Google Interview Questions Deconstructed: The Knight's Dialer

Before I start, a disclaimer: while interviewing candidates is one of my professional responsibilities, this blog represents my personal observations, my personal anecdotes, and my personal opinions. Please don't mistake this for any sort of official statement by or about Google, Alphabet, or any other person or organization.

Read full article from Google Interview Questions Deconstructed: The Knight's Dialer

Add #. option for ordered list markdown - Meta Stack Exchange

You could use "1." for each item of the ordered list.

1. This is an ordered list  1. The list contains three items  1. This is the last one

This is an ordered list
The list contains three items
This is the last one

Markdown renders the ordered list automatically numbering the single items.

Read full article from Add #. option for ordered list markdown - Meta Stack Exchange

Minimum length of jumps to avoid given array of obstacles - GeeksforGeeks

We are given coordinates of obstacles on a straight line. We start jumping from point 0, we need to reach end avoiding all obstacles. Length of every jump has to be same (For example, if we jump from 0 to 4, then we must make next jump from 4 to 8). We need to find the minimum length of jump so that we can reach end and we avoid all obstacles.

Read full article from Minimum length of jumps to avoid given array of obstacles - GeeksforGeeks

Minimum number of operation required to convert number x into y - GeeksforGeeks

Given a initial number x and two operations which are given below:

Multiply number by 2.
Subtract 1 from the number.

The task is to find out minimum number of operation required to convert number x into y using only above two operations. We can apply these operations any number of times.

Read full article from Minimum number of operation required to convert number x into y - GeeksforGeeks

Number of jumps for a thief to cross walls - GeeksforGeeks

A thief trying to escape from a jail. He has to cross N walls each with varying heights (every height is greater than 0). He climbs X feet every time. But, due to the slippery nature of those walls, every time he slips back by Y feet. Now the task is to calculate the total number of jumps required to cross all walls and escape from the jail.

Read full article from Number of jumps for a thief to cross walls - GeeksforGeeks

Find if two people ever meet after same number of jumps - GeeksforGeeks

Two people start races from two different points p1 and p2. They cover s1 ans s2 meters in a jump. Find if they will ever meet at a point after same number of jumps.

Read full article from Find if two people ever meet after same number of jumps - GeeksforGeeks

Why you failed your tech interview #2 - You messed up your details | FizzBuzzed

Details matter. Getting details wrong and being careless during interviews leads to a lot of failed interviews (that probably should have been passes). In the real world, completely incorrect code can be better than code with subtle bugs. If your code is completely broken, you'll notice with test cases and fix it. If you have subtle errors, the code will likely get through code review and lead to a long and difficult debugging session.

We all make mistakes, but why do some programmers seem to make fewer than others? Are there any tricks you're missing that could help you get the details right?

Here's a classic whiteboard interview gone bad:

Interviewer: "Could you please code (some simple algorithm, binary search, bfs, etc)"

Candidate (after writing some code): "Ok, that should do it."

Interviewer: (squinting at the code): "Hmm well what about this test case?"

Candidate: *Traces what the code would do for the test case and realizes a bug.* "Oh… well let me just fix that."

A small piece of code is scribbled out and changed. The test case now passes…

Interviewer (squinting again): "Hmm well what about this other test case?"

Candidate: *Does some more erasing. Adds another scribble* "Here, now that test case will pass too!"

This 'test case and scribble' pattern continues a few more times until the candidate's code looks nothing like originally intended. The interviewer records the code as written and eventually forwards it to a hiring committee. The hiring committee has only this code as evidence to judge the candidate's ability and assumes the candidate codes spaghetti.

The candidate is rejected.

How can you avoid falling into this trap?

Here are a few techniques that can help you get your code correct the first time:

Read full article from Why you failed your tech interview #2 - You messed up your details | FizzBuzzed

All about the “paper” - Ariel_一只猫的旅行 - 博客园

All about the "paper" - Ariel_一只猫的旅行 - 博客园

写论文，good idea很重要

2、看论文。

网友经验：看论文学知识的效率远比看书要快的多，很多人在刚开始看论文的时候，觉得有很多地方不懂，总想着是否要去重新去学习某本书？这是非常错误的，因为论文里涉及的只是某一本书里很小的一部分知识，如果你不懂，可以先记下，以后自然会在别的论文里看到，时间一长，你自然会了解其中的意思，如果去重新学习课本，花费的时间太长，效率也很低。

（本人觉得，某种特定情况下，这话没毛病，确实如此^_^，但是不包含想系统学习掌握某知识的情况，这还是得踏踏实实啃些砖头，这样才系统，才扎实，求同存异O(∩_∩)O~）

3、善于总结

看完一篇很好的论文，我们觉得它好，它到底哪里好？我们需要总结。有时候一篇好的论文也就在某一个方案上做出了一个很有想法的创新，我们可以用几句话或几个公式就能将其总结，把它贴在论文上，或者写在本子里，这样我们以后用到或者回忆的时候会很方便，不用再重复的去看论文内容。

Read full article from All about the "paper" - Ariel_一只猫的旅行 - 博客园

Pathfinding for Tower Defense

In a Tower Defense game, there are many enemies that are all headed to the same place. In many Tower Defense games, there is a predetermined path, or a small number of paths. In some, such as the classic Desktop Tower Defense^[1], you can place towers anywhere, and they act as obstacles that affect the paths taken by enemies. Try it out, and click on the map to toggle walls:

Read full article from Pathfinding for Tower Defense

Going to China? A guide to using a credit card in China | finder.com

When presented with the option of getting your card charged in yuan or US dollars, always opt for the former. You don't want to go the dynamic currency conversion way, which comes into effect when you make payments in US dollars outside of the US. If you do, you'll probably end up with an undesirable exchange rate and a currency conversion fee.

Read full article from Going to China? A guide to using a credit card in China | finder.com

POJ 3614 - Sunscreen - Dilthey - 博客园

把牛按minSPF从小到大排序一下，防晒霜按SPF排序一下。

然后遍历防晒霜lontion[1...L]，对于lotion[i]，把所有满足minSPF小于等于lontion[i].SPF的牛都找出来，入队。

我们将这个队列定义为按cow[i].maxSPF的从小到大排序的优先队列，那么，每次取出的队首，都是当前要求最苛刻（SPF要求范围最小的）那头牛，先满足这些牛，如果当前这种防晒霜能满足这头牛，就用掉一瓶，否则就放弃这头牛。

反复如此，直到所有防晒霜用完。

Read full article from POJ 3614 - Sunscreen - Dilthey - 博客园

AC题目简解-数据结构 - Findxiaoxun - 博客园

要两条路有交叉，(x1,y1)(x2,y2)那么需要满足：(x1-x2)*(y1-y2)<0判断出这是求逆序的问题
树状数组求逆序，先通过自定义的比较器实现按照一种元素排序，然后再逐次查询和add来获取结果

B - Ping pong POJ 3928
分析裁判的位置，则可判断求逆序问题。对每个点的裁判，他能举办的比赛场数为左弱*右强+左强*右弱，通过正序和倒序两次插入和查询累加得最终结果

C - Balanced Lineup POJ 3274
用线段树A的==

D - Mobile phones POJ 1195
二维树状数组，理解了一维的，二维的也很好理解。一维是前面线上的和，二维的是前面矩形的和。算目标范围也很明确，就是减掉两个矩形的加上多减的一个正方形，维护跟一维的也基本一样

E - MooFest POJ 1990
首先，最重要的是理解好题意：1.两头牛只要通讯一次就行。每次的耗费是distance*max(v[i],v[j])2.按power还是按position排序。我选的是power，因为position没有重复值。排序这个想法跟Japan那道题是一样的。
说下思路：
按照power的升序排好后，依次加入树状数组,每加一次，就查一次，累计一次ans。这个是标准的数组数组的解法。
但是，树状数组的索引值是牛的坐标的索引值，这就意味着，能力比当前低的牛可能会排到其后面，而且，它们的距离值是需要的。
这就要求，既能记录到前面有多少头牛，又能记录他们的距离和，后面有多少头牛，和他们的距离和。
而想到，加点的时候，有一个i来控制，也就是说，在i前面有 a头牛的话，后面有i-a头，顺着这个想法，那也可以统计坐标和啊。前面加了i头牛，总共的坐标和distance可以累加出来，而在i的坐标前面有（这里的有和加是两回事）的牛的坐标和为sum(data[i].position-1).dist，那么坐标在第i个的后面，而且power比第i个的小的坐标和就是distance-sum(data[i].position-1).dist
因为是按照power排的序，所以在i前面加的power都比第i个小，也就是说，如果之前加的和第i个通讯的话，取的power指肯定是第i个的
公式:
这里的sum(data[i].position).num表示坐标在data[i].position前面的牛的个数
sum(data[i].position-1).dist i的坐标前面有（这里的有和加是两回事）的牛的坐标和

ans+=(sum(data[i]-1).num*data[i].position - sum(data[i].position-1).dist + (distance-sum(data[i].position-1).dist - (i-1-sum(data[i]-1).num)*data[i].position ))*data[i].power
可以化简为
int x1=sum(data[i]-1).num,x2=sum(data[i].position-1).dist;
ans+=(data[i].position*(x1*2-i+1) - x2*2 + distance)*data[i].power;

然后distance累加再把当前点add进树状数组就ok了

Read full article from AC题目简解-数据结构 - Findxiaoxun - 博客园

跳跳爸的Abc

确定有限状态机（deterministic finite automaton/dfa）是一个数学计算模型，组成部分是一个5元组：

有限的状态集Q
有限的输入符号S，又被称作alphabet（跟我们熟知的英文字母表应该不一样，是个引申）
状态变换函数F，F：S 𝗑 Q -> Q
初始状态s0，s0 ∈ Q
接纳状态集Z，Z ⊆ Q

Read full article from 跳跳爸的Abc

腾讯程序员面试，妹子400行代码实现双人五子棋，成功拿到offer！ - 知乎

1.回答面试官问题要有自信，尽量不要紧张，说话语气诚恳，简单明了

2.自我介绍的时候，还是要告诉面试官自己的优点，缺点（注意不能过于吹嘘自己，缺点最好是那些无关紧要的。

3.要与面试官有目光交流，沟通的时候最好不要东张西望的，或者低头着头。

4.当然在这之前你得有良好的编程功底，要向主考官展示自己的个性，不要让他认为你只是一个普通的程序员。尽管应聘单位要招的是程序员，但主考官也会注重你在团队合作、与人交往方面的能力。也就是说，他们需要的是有血有肉的程序员，而不是一个只知道钻研技术的书呆子。所以，在应聘时，最好把自己多面手的功能显露出来。一般情况下，一个阳光、充满正能量的应聘者，更容易得到这份工作。

Read full article from 腾讯程序员面试，妹子400行代码实现双人五子棋，成功拿到offer！ - 知乎

Only leetcode and luck ? - Blind

I have about 10 yoe and doing leetcode during my free time . Does it only need a good leetcoder to get into FANG and some luck ?. I would want to know how different it will be in an interview for someone who has quite some experience?

Read full article from Only leetcode and luck ? - Blind

DEAR PROFESSIONAL COMPUTER TOUCHERS -- FRIDAY RANT THREAD FOR February 08, 2019 : cscareerquestions

AND NOW FOR SOMETHING ENTIRELY DIFFERENT.

THE BUILDS I LOVE, THE SCRIPTS I DROP, TO BE PART OF, THE APP, CAN'T STOP

THIS IS THE RANT THREAD. IT IS FOR RANTS.

CAPS LOCK ON, DOWNVOTES OFF, FEEL FREE TO BREAK RULE 2 IF SOMEONE LIKES SOMETHING THAT YOU DON'T BUT IF YOU POST SOME RACIST/HOMOPHOBIC/SEXIST BULLSHIT IT'LL BE GONE FASTER THAN A NEW MESSAGING APP AT GOOGLE.

Read full article from DEAR PROFESSIONAL COMPUTER TOUCHERS -- FRIDAY RANT THREAD FOR February 08, 2019 : cscareerquestions

How much leetcode is enough? People from Chinese forum are claiming 500+ : cscareerquestions

So I recently came across this Chinese forum(I'm bilingual) similar in style to r/cscareerquestions. And saw many people there advocating on studying leetcode religiously, often encouraging candidates to go through at least 500+ questions. Paraphrasing their statements: word search12， word break12，word ladder12，LIS，sort color，LRU，insert & delete in O1，rob house123，234sum. You should be able to solve these with your eyes closed. Min/max heap，bucket sort，topological sort，binary pre/in/post/level order，combination/permutation should appear in your dreams when you're asleep.

Read full article from How much leetcode is enough? People from Chinese forum are claiming 500+ : cscareerquestions

LeetCode | 754. Reach a Number 数学原理题解析与证明 - u012737193的博客 - CSDN博客

给你一个数，第n次可以向左走，也可以向右走，问你最少多少次可以走到目的地

一开始我考虑的是广度优先遍历解决，但是这样的话由于target太大，内存超了，于是只好考虑其他的方法，思考过后我发现这题用数学的方法解决其实很简单，首先，要到达目的地一定会有向左和向右，我们假设一定有一个最好的方案，a1,a2,a3…an,假如a1到an都是正数，那么它一定是最好的方案，假如存在负数，那么可以看成把a1,a2…an里面的若干个数变成负数，然后到达target，可以证明，当a1+a2..an-target为偶数的时候，只需要把a1,a2,…an里面的一个数变成负数就能到达目的地，这就是最好的方案

当a1+a2..an-target为奇数的时候，有两种情况，当n为偶数的时候，n+1为奇数，可以证明，当把a1,a2..an里面一个数变成负数之后，只要在加一个数就能到达target，也就是a1+a2…an+an+1一定可以到达target,当n为奇数的时候，可以证明当把a1,a2..an里面一个数变成负数之后只要在加两个个数就能到达target，也就是a1+a2…an+an+1+an+2,所以有以下算法

Read full article from LeetCode | 754. Reach a Number 数学原理题解析与证明 - u012737193的博客 - CSDN博客

Credit Cards That Pay 2% Cash Back on All Purchases

Fidelity® Rewards Visa Signature® Card–This cash back card offers card members a flat 2% cash back rate in the form of rewards points. Points never expire and there's never a limit on how many you can earn. The rewards are deposited into your Fidelity account, which is required to get the card. Six different Fidelity accounts are eligible to own the Fidelity® Rewards Visa Signature® Card and they are:

Read full article from Credit Cards That Pay 2% Cash Back on All Purchases

Best Credit Cards for Travel in China

It's not uncommon for businesses in China to add a surcharge to the total cost of your purchase if you use a credit card instead of cash. In some cases, surcharges are a result of how the processing bank handles the transaction. When using some credit cards, the bank that processes the transaction converts the transaction into your home currency at an unfavorable exchange rate, essentially adding 4% to 6% to your bill.

If possible, make sure the transaction is done in the local currency (RMB) with the Chinese bank charging your bank the RMB amount. Your bank will then convert this to your home currency at a fair exchange rate. When in doubt, ask about a surcharge before you hand over your credit card so you can decide if it makes more sense to pay in cash. (For related insight, explore a few key money-saving travel tips for China.)

Read full article from Best Credit Cards for Travel in China

Is Discover a Visa or Mastercard?

Discover is neither a Visa nor a Mastercard. It's more like an American Express card, but they're not exactly the same, either. Discover is similar to Visa, Mastercard and American Express in the sense that all four are card networks. Each helps process payments for the cards on its network, determines where those cards can be used, and provides extra benefits such as rental car insurance. However, Discover is different from Visa and Mastercard in that most credit cards on the Discover network are issued by Discover. Visa and Mastercard aren't issuers, so cards on their networks come from lots of different banks and credit unions. Amex, like Discover, is both a card issuer and a card network.

Here's how Discover differs from Visa or Mastercard:

Discover is both a credit card issuer and a card network. Visa and Mastercard are networks only.
Visa and Mastercard are accepted in more than 200 countries and territories, compared to 185 for Discover.
Visa and Mastercard are accepted at 10.7 million U.S. merchant locations, versus 10.4 million for Discover.
Discover credit cards don't have foreign transaction fees. Visa and Mastercard credit cards charge 0% to 3% of purchases processed internationally.

If you look at the front of your Discover card, you should see "Discover" with an orange dot for the "o." Each network has a logo, so it's easy to tell them apart. And while Discover cards are generally fine to use anywhere in the U.S., you might have trouble in certain countries abroad. So it's probably best to have a backup Visa or Mastercard.

Read full article from Is Discover a Visa or Mastercard?

Mastercard and Discover in China trip - myFICO® Forums - 5289097

Discover however can be processed as their own Union Pay card, and is automatically charged in local currency (which is what you want to do to minimise conversion fee). Only downside is cashiers may be uninformed since the card doesn't display a UnionPay logo. Discover provides you this small printed instruction card you can provide to the cashiers.

Read full article from Mastercard and Discover in China trip - myFICO® Forums - 5289097

Solving Matrix/Graph Problems on LeetCode using Python

There are n cities connected by m flights. Each fight starts from city u and arrives at v with a price w.

Now given all the cities and fights, together with starting city src and the destination dst, your task is to find the cheapest price from src to dst with up to k stops. If there is no such route, output -1.

Read full article from Solving Matrix/Graph Problems on LeetCode using Python

How to Auto-Hide the Address Bar in Google Chrome and Chromium

One of the features that many people love in Chrome is the minimal UI at the top of the browser. Want to minimize it even more? Then you will definitely want to make use of the new feature added to the Experimental Features page.

00:00

05:00

Read full article from How to Auto-Hide the Address Bar in Google Chrome and Chromium

Using the Eclipse Memory Analyzer (MAT) Remotely on Large Heap Dumps

Sometime your java applilcation will fail and generate an enormous heap dump. One that may be too large to be able to transfer to your local machine and to analyze for lack of RAM, time or both.

One solution is to install the MAT tool on the remote server and generate an HTML output of the analysis to download and view locally. This saves the headache of attempting to get X Windows installed on the remote machine and get all of the ssh tunneling sorted out (which is of course an option as well).

First, download and install the stand-alone Eclipse RCP Application. Then transfer to your server and unpack. Then determine how large the heap dump is and, if necessary, modify the MemoryAnalyzer.ini file to instantiate a JVM with enough RAM for your heap dump.

In this example, I have an 11GB heap dump and have modified the last two lines (adding -Xms)

-startup
plugins/org.eclipse.equinox.launcher_1.3.100.v20150511-1540.jar
--launcher.library
plugins/org.eclipse.equinox.launcher.gtk.linux.x86_64_1.1.300.v20150602-1417
-vmargs
-Xmx16g
-Xms16g

Do an initial run to parse the heap dump. This will generate intermediary data that can be used by subsequent runs to make future analysis faster.

./ParseHeapDump.sh /path/to/heap-dump

Read full article from Using the Eclipse Memory Analyzer (MAT) Remotely on Large Heap Dumps

2018 华为软件精英挑战赛（上合赛区决赛：227/300分） - Demon-初来驾到 - CSDN博客

些许遗憾，没能进入上合赛区前四强，不过从中学习到的一些trick，以及比赛套路值得回味，当然还认识了许多赛区大佬，和他们交流的过程中，总能发掘一些新思路来弥补现有模型的不足。对于我们而言，这次比赛，我们被模型给限制住了思路，没能从全局的角度审视数据集中出现的问题，数据集本身存在的漏洞信息并没有被我们有效地捕捉到，相反进入了构建稳定预测模型的死胡同。多点变通，情况或许就不同了。

赛题回顾（参赛选手可直接跳过本节）

背景:
云平台为了满足不同租户的需求,提供了一种可随时自助获取、可弹性伸缩的云服务器,即弹性云服务器(Elastic Cloud Server,ECS)。为容纳更多的租户请求、并尽可能提高资源利用率、降低成本,自动化、智能化的资源调度管理系统非常关键。

由于租户对ECS实例(虚拟机,VM)请求的行为具有一定规律,可以通过对历史ECS实例请求的分析,预测到未来一段时间的ECS实例请求,然后对预测的请求分配资源(如图1所示),这样可以找到一个接近最优的分配策略,实现资源最大化利用,同时也能参考预测的结果制定云数据中心的建设计划。

Read full article from 2018 华为软件精英挑战赛（上合赛区决赛：227/300分） - Demon-初来驾到 - CSDN博客

为什么Squeeze Theorem要译作“夹逼定理”？ - 知乎

为什么Squeeze Theorem要译作"夹逼定理"？ - 知乎

这样比较有气势，在做题的时候可以配合着大喊一声：哪里跑！

Read full article from 为什么Squeeze Theorem要译作"夹逼定理"？ - 知乎

写给媳妇儿的算法（一）——二分查找 - 简书

【引言】早晨一女生背着一堆书出了图书馆，结果警报响了，大妈让女生看看是哪本书把警报弄响了，那女生把书倒出来，准备一本一本的测。大妈见状急了，把书分成两份，第一份过了一下，响了。又把这一份分成两份接着测，三回就找到了，大妈用鄙视的眼神看着女生，仿佛在说O(n)和O(logn)都分不清。

Read full article from 写给媳妇儿的算法（一）——二分查找 - 简书

二分和枚举（主要是二分思想） - 做一个奋斗不止的小公举！ - CSDN博客

通常二分的题目都会使left=0，right=可能的最大值，然后再left 和right之间寻找最大值。
而最重要的就是能够把二分题目分析称这个思想。（目前，我的个人理解）

第一个是求最大的最小值，第二个是最小的最大值，好好理解一下

疯牛这道题就是先排序，然后知道最大的距离，在0 和MAX之间寻找答案
judge的想法是：判断这个当前距离，如果有另一个栅栏与他的距离大于当前距离，则假定可以安排牛，然后从当前牛开始继续寻找距离比当前距离大的。。
疯牛
时间限制：1000 ms | 内存限制：65535 KB
难度：4

描述
农夫 John 建造了一座很长的畜栏，它包括N (2 <= N <= 100,000)个隔间，这些小隔间依次编号为x1,…,xN (0 <= xi <= 1,000,000,000).
但是，John的C (2 <= C <= N)头牛们并不喜欢这种布局，而且几头牛放在一个隔间里，他们就要发生争斗。为了不让牛互相伤害。John决定自己给牛分配隔间，使任意两头牛之间的最小距离尽可能的大，那么，这个最大的最小距离是什么呢？

输入
有多组测试数据，以EOF结束。
第一行：空格分隔的两个整数N和C
第二行——第N+1行：分别指出了xi的位置
输出
每组测试数据输出一个整数，满足题意的最大的最小值，注意换行。
样例输入
5 312849
样例输出
3

题意：简单的说就是给你一段长度，在这一段中给出m个点，然后在这m个点中选出k个点，让这k个点之间相邻两个点的之间距离的最小值最大

思路：通过二分枚举这个最小值，然后通过贪心的思想找出满足要求的最大的这个最小值

解析：——二分枚举 + 贪心
这道题用到了刘汝佳算法入门经典上贪心那一节讲的算法,用二分枚举满足条件的最大距离，
依次做相应判断.本题不需要担心最后求出的距离不能适应题目中的隔间间的距离，
因为二分枚举之后是按照贪心发判断的，如果当前距离满足要求，会继续增大枚举的距离，
一直到无法满足要求为止，即最后结果一定满足是隔间间的距离 .

Read full article from 二分和枚举（主要是二分思想） - 做一个奋斗不止的小公举！ - CSDN博客

二分查找法的实现和应用汇总 - Ider - 博客园

二分查找法的O(log n)让它成为十分高效的算法。不过它的缺陷却也是那么明显的。就在它的限定之上：

必须有序，我们很难保证我们的数组都是有序的。当然可以在构建数组的时候进行排序，可是又落到了第二个瓶颈上：它必须是数组。

数组读取效率是O(1)，可是它的插入和删除某个元素的效率却是O(n)。因而导致构建有序数组变成低效的事情。

Read full article from 二分查找法的实现和应用汇总 - Ider - 博客园

378 Kth Smallest Element in a Sorted Matrix - 简书

378 Kth Smallest Element in a Sorted Matrix

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

Read full article from 378 Kth Smallest Element in a Sorted Matrix - 简书

Solution to Leetcode problem 971 Flip Binary Tree To Match Preorder Traversal

This is the third problem of the Leetcode contest 118. It is a problem on binary trees (My favorite kind of problem!). It is using the following 2 definitions:

"A node in the binary tree can be flipped by swapping the left child and the right child of that node"
Preorder traversal of a node: we report the current node's value, then preorder-traverse the left child, then preorder-traverse the right child

The question is: Given a binary tree T, with nodes with unique values, and a array A, is there a way to flip nodes of T such that a preorder traversal of T matches A's values in order?

Read full article from Solution to Leetcode problem 971 Flip Binary Tree To Match Preorder Traversal

#Solution

#Solution #1 - Union Find - Without Optimization

#Solution #2 - Union Find Optimization

分析

赛题回顾（参赛选手可直接跳过本节）

378 Kth Smallest Element in a Sorted Matrix

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