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  • C# Dev Challenge Part 1 of n &ndash; Beginner Edition

    - by mbcrump
    I developed this challenge to test one’s knowledge of C Sharp. I am planning on creating several challenges with different skill sets, so don’t get mad if this challenge doesn’t well challenge you... I noticed that most people like short quizzes so this one only contains 5 questions. All of the challenges are clear and concise of what I am asking you to do. No smoke and mirrors here, meaning that none of the code has syntax errors. The purpose of this exercise is to test several OOP concepts and see how much of the C# language you really know. Question #1 – Lets start off Easy… Will the following code snippet compile successfully? What does this question test? - Can this compile without a namespace? Do you have to have an entry point of “static void Main()”? class Test { static int Main() { System.Console.WriteLine("Developer Challenge"); return 0; } } Answer (select text in box below): Yes, it will compile successfully. Question #2 – What is the value of the Console.WriteLine statements? What does this question test? – Do I understand reference types/value types? If a variable is declared with the @ symbol and its not a reserved keyword does the application compile successfully? using System; internal struct MyStruct { public int Value; } internal class MyClass { public int Value; } class Test { static void Main() { MyStruct @struct1 = new MyStruct(); MyStruct @struct2 = @struct1; @struct2.Value = 100; MyClass @ref1 = new MyClass(); MyClass @ref2 = @ref1; @ref2.Value = 100; Console.WriteLine("Value Type: {0} {1}", @struct1.Value, @struct2.Value); Console.WriteLine("Reference Type: {0} {1}", @ref1.Value, @ref2.Value); } } Answer (select text in box below): Value Type: 0 100 Reference Type: 100 100 Question #3 – What is the value of the Console.WriteLine statements? What does this question test? – Can 2 objects reference the same point in memory? using System; class Test { static void Main() { string s1 = "Testing2"; string t1 = s1; Console.WriteLine(s1 == t1); Console.WriteLine((object)s1 == (object)t1); } } Answer (select text in box below): True True Question #4 – What is the value of the Console.WriteLine statements? What does this question test? – How does the “Stack” work – LIFO or FIFO?   using System; using System.Collections; class Test { static void Main() { Stack a = new Stack(5); a.Push("1"); a.Push("2"); a.Push("3"); a.Push("4"); a.Push("5"); foreach (var o in a) { Console.WriteLine(o); } } } Answer (select text in box below): 5 4 3 2 1 Question #5 – What is the value of the Console.WriteLine statements? What does this question test? – Array and General Looping Knowledge. using System; namespace ConsoleApplication5 { class Program { static void Main(string[] args) { int[] J_LIST = new int[5] { 1, 2, 3, 4, 5 }; int K = 10; int L = 5; foreach (var J in J_LIST) { K = K - J; L = K + 2 * J; Console.WriteLine("J = {0, 5} K = {1, 5} L = {2, 5}", J, K, L); } Console.ReadLine(); } } } Answer (select text in box below): J = 1 K = 9 L = 11 J = 2 K = 7 L = 11 J = 3 K = 4 L = 10 J = 4 K = 0 L = 8 J = 5 K = -5 L = 5 Stay Tuned for more challenges!

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  • Help me finish this Python self-challenge.

    - by Hamish Grubijan
    This is not a homework. I saw this article praising Linq library and how great it is for doing combinatorics stuff, and I thought to myself: Python can do it in a more readable fashion. After half hour of dabbing with Python I failed. Please finish where I left off. Also, do it in the most Pythonic and efficient way possible please. from itertools import permutations from operator import mul from functools import reduce glob_lst = [] def divisible(n): return (sum(j*10^i for i,j in enumerate(reversed(glob_lst))) % n == 0) oneToNine = list(range(1, 10)) twoToNine = oneToNine[1:] for perm in permutations(oneToNine, 9): for n in twoToNine: glob_lst = perm[1:n] #print(glob_lst) if not divisible(n): continue else: # Is invoked if the loop succeeds # So, we found the number print(perm) Thanks!

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  • Linq query challenge

    - by vdh_ant
    My table structure is as follows: Person 1-M PesonAddress Person 1-M PesonPhone Person 1-M PesonEmail Person 1-M Contract Contract M-M Program Contract M-1 Organization At the end of this query I need a populated object graph where each person has their: PesonAddress's PesonPhone's PesonEmail's PesonPhone's Contract's - and this has its respective Program's Now I had the following query and I thought that it was working great, but it has a couple of problems: from people in ctx.People.Include("PersonAddress") .Include("PersonLandline") .Include("PersonMobile") .Include("PersonEmail") .Include("Contract") .Include("Contract.Program") where people.Contract.Any( contract => (param.OrganizationId == contract.OrganizationId) && contract.Program.Any( contractProgram => (param.ProgramId == contractProgram.ProgramId))) select people; The problem is that it filters the person to the criteria but not the Contracts or the Contract's Programs. It brings back all Contracts that each person has not just the ones that have an OrganizationId of x and the same goes for each of those Contract's Programs respectively. What I want is only the people that have at least one contract with an OrgId of x with and where that contract has a Program with the Id of y... and for the object graph that is returned to have only the contracts that match and programs within that contract that match. I kinda understand why its not working, but I don't know how to change it so it is working... This is my attempt thus far: from people in ctx.People.Include("PersonAddress") .Include("PersonLandline") .Include("PersonMobile") .Include("PersonEmail") .Include("Contract") .Include("Contract.Program") let currentContracts = from contract in people.Contract where (param.OrganizationId == contract.OrganizationId) select contract let currentContractPrograms = from contractProgram in currentContracts let temp = from x in contractProgram.Program where (param.ProgramId == contractProgram.ProgramId) select x where temp.Any() select temp where currentContracts.Any() && currentContractPrograms.Any() select new Person { PersonId = people.PersonId, FirstName = people.FirstName, ..., ...., MiddleName = people.MiddleName, Surname = people.Surname, ..., ...., Gender = people.Gender, DateOfBirth = people.DateOfBirth, ..., ...., Contract = currentContracts, ... }; //This doesn't work But this has several problems (where the Person type is an EF object): I am left to do the mapping by myself, which in this case there is quite a lot to map When ever I try to map a list to a property (i.e. Scholarship = currentScholarships) it says I can't because IEnumerable is trying to be cast to EntityCollection Include doesn't work Hence how do I get this to work. Keeping in mind that I am trying to do this as a compiled query so I think that means anonymous types are out.

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  • Where can you find fun/educational programming challenges?

    - by tj9991
    I've searched around for different challenge sites, and most of them seem to be geared towards difficulty in problem solving logically, rather than trying to use your language of choice to do something you haven't used it for. Their center is around mathematics rather than function design. Some kind of point system for correctly solving challenges, or solving them the most efficient/smallest would be neat as well. Listed sites Project Euler TopCoder UVa Online Judge Challenges with Python Google Code Jam Programming Challenges Less Than Dot ACM's Programing Contest archive USACO problems ITA Software's puzzle page Refactor My Code Ruby Quiz

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  • Encoding / Error Correction Challenge

    - by emi1faber
    Is it mathematically feasible to encode and initial 4 byte message into 8 bytes and if one of the 8 bytes is completely dropped and another is wrong to reconstruct the initial 4 byte message? There would be no way to retransmit nor would the location of the dropped byte be known. If one uses Reed Solomon error correction with 4 "parity" bytes tacked on to the end of the 4 "data" bytes, such as DDDDPPPP, and you end up with DDDEPPP (where E is an error) and a parity byte has been dropped, I don't believe there's a way to reconstruct the initial message (although correct me if I am wrong)... What about multiplying (or performing another mathematical operation) the initial 4 byte message by a constant, then utilizing properties of an inverse mathematical operation to determine what byte was dropped. Or, impose some constraints on the structure of the message so every other byte needs to be odd and the others need to be even. Alternatively, instead of bytes, it could also be 4 decimal digits encoded in some fashion into 8 decimal digits where errors could be detected & corrected under the same circumstances mentioned above - no retransmission and the location of the dropped byte is not known. I'm looking for any crazy ideas anyone might have... Any ideas out there?

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  • UIPageControl design challenge

    - by Sheehan Alam
    My app has a UITabBarController that loads many different UINavigationControllers. I want a UIPageControl to switch between the different UINavigationControllers. Do I place the UIPageControl in my UINavigationController or in my appDelegate? Suggestions and best practices are welcome.

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  • ActiveRecord Logic Challenge - Smart Ways to Use AR Timestamp

    - by keruilin
    My question is somewhat specific to my app's issue, but the answer should be instructive in terms of use cases for association logic and the record timestamp. I have an NBA pick 'em game where I want to award badges for picking x number of games in a row correctly -- 10, 20, 30. Here are the models, attributes, and associations in-play: User id Pick id result # (values can be 'W', 'L', 'T', or nil. nil means hasn't resolved yet.) resolved # (values can be true, false, or nil.) game_time created_at *Note: There are cases where a pick's result field and resolved field will always be nil. Perhaps the game was cancelled. Badge id Award id user_id badge_id created_at User has many awards. User has many picks. Pick belongs to user. Badge has many awards. Award belongs to user. Award belongs to badge. One of the important rules here to capture in the code is that while a user can be awarded multiple streak badges (e.g., a user can win multiple 10-streak badges), the user CAN'T be awarded another badge for consecutive winning picks that were previously granted an award badge. One way to think of this is that all the dates of the winning picks must come after the date that the streak badge was awarded. For example, let's pretend that a user made 13 winning picks from May 5 to May 8, with the 10th winning pick occurring on May 7, and the last 3 on May 8. The user would be awarded a 10-streak badge on May 7. Now if the user makes another winning pick on May 9, the code must recognize that the user only has a streak of 4 winning picks, not 14, because the user already received an award for the first 10. Now let's assume that the user makes 6 more winning picks. In this case, the code must recognize that all winning picks since May 5 are eligible for a 20-streak badge award, and make the award. Another important rule is that when looking at a winning streak, we don't care about the game time, but rather when the pick was made (created_at). For example, let's say that Team A plays Team B on Sat. And Team C plays Team D on Sun. If the user picks Team C to beat Team D on Thurs, and Team A to beat Team C on Fri, and Team A wins on Sat, but Team C loses on Sun, then the user has a losing streak of 1. So when must the streak-check kick-in? As soon as a pick is a win. If it's a loss or tie, no point in checking. One more note: if the pick is not resolved (false) and the result is nil, that means the game was postponed and must be factored out. With all that said, what is the most efficient, effective and lean way to determine whether a user has a 10-, 20- or 30-win streak?

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  • Hibernate collection mapping challenge

    - by Geln Yang
    Hi, There is a table Item like, code,name 01,parent1 02,parent2 0101,child11 0102,child12 0201,child21 0202,child22 Create a java object and hbm xml to map the table.The Item.parent is a Item whose code is equal to the first two character of its code : class Item{ string code; string name; Item parent; List<Item> children; .... setter/getter.... } <hibernate-mapping> <class name="Item" table="Item"> <id name="code" length="4" type="string"> <generator class="assigned" /> </id> <property name="name" column="name" length="50" not-null="true" /> <!--====================================== --> <many-to-one name="parent" class="Item" not-found="ignore"></many-to-one> <bag name="children"></bag> <!--====================================== --> </class> </hibernate-mapping> How to definition the mapping relationship? Thanks!

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  • Linq query challenge - can this be done?

    - by vdh_ant
    My table structure is as follows: Person 1-M PesonAddress Person 1-M PesonPhone Person 1-M PesonEmail Person 1-M Contract Contract M-M Program Contract M-1 Organization At the end of this query I need a populated object graph where each person has their: PesonAddress's PesonPhone's PesonEmail's PesonPhone's Contract's - and this has its respective Program's Now I had the following query and I thought that it was working great, but it has a couple of problems: from people in ctx.People.Include("PersonAddress") .Include("PersonLandline") .Include("PersonMobile") .Include("PersonEmail") .Include("Contract") .Include("Contract.Program") where people.Contract.Any( contract => (param.OrganizationId == contract.OrganizationId) && contract.Program.Any( contractProgram => (param.ProgramId == contractProgram.ProgramId))) select people; The problem is that it filters the person to the criteria but not the Contracts or the Contract's Programs. It brings back all Contracts that each person has not just the ones that have an OrganizationId of x and the same goes for each of those Contract's Programs respectively. What I want is only the people that have at least one contract with an OrgId of x with and where that contract has a Program with the Id of y... and for the object graph that is returned to have only the contracts that match and programs within that contract that match. I kinda understand why its not working, but I don't know how to change it so it is working... This is my attempt thus far: from people in ctx.People.Include("PersonAddress") .Include("PersonLandline") .Include("PersonMobile") .Include("PersonEmail") .Include("Contract") .Include("Contract.Program") let currentContracts = from contract in people.Contract where (param.OrganizationId == contract.OrganizationId) select contract let currentContractPrograms = from contractProgram in currentContracts let temp = from x in contractProgram.Program where (param.ProgramId == contractProgram.ProgramId) select x where temp.Any() select temp where currentContracts.Any() && currentContractPrograms.Any() select new Person { PersonId = people.PersonId, FirstName = people.FirstName, ..., ...., MiddleName = people.MiddleName, Surname = people.Surname, ..., ...., Gender = people.Gender, DateOfBirth = people.DateOfBirth, ..., ...., Contract = currentContracts, ... }; //This doesn't work But this has several problems (where the Person type is an EF object): I am left to do the mapping by myself, which in this case there is quite a lot to map When ever I try to map a list to a property (i.e. Scholarship = currentScholarships) it says I can't because IEnumerable is trying to be cast to EntityCollection Include doesn't work Hence how do I get this to work. Keeping in mind that I am trying to do this as a compiled query so I think that means anonymous types are out.

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  • Application Design in Interface Builder Challenge

    - by Sheehan Alam
    I want to design an app that launches other sub-apps. Main View will contain 4 buttons. Clicking on each button respectively will launch the other sub-apps. Each sub-app will have a UITabBarController which has its own different views. At any point I want the user to be able to go back to the Main View from any of the sub-apps. I am not sure how to design this in IB.

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  • Nested <a> and <span> challenge

    - by PaddyO
    Hi all, Trying in vain to get a nested link working within a nested span. This is a working test page for the code below to explain what I'm trying to do. Any ideas on how to get this working in valid html? I guess it's either a nesting order or style syntax thing but I am at a loss. Any thoughts much appreciated. <div id="greyback"> <ul id="scrollbox"> <li class="listcat">List header</li> <li><a class="menu" href="#freeze">List item 1<span><b>This text has popped up because you have clicked the list item, which has an "a" tag and now has :focus. That "a" tag is the first of two.</b><br><br>What I am trying to do is to set the second "a" tag as a DIFFERENT "embedded" link in this box<span style="color: blue; background-color: yellow;">eg, here<a href="http://www.conservationenterprises.com" target="blank">This is the second (nested) "a" tag in this html nest. It is a link to an external site. Instead of this being an always-visible link, I want it to sit within the yellow box in the first span (click the List item 1 to display).</a></span> </a></span> </li> </a></span> </li> </ul> </div> and the CSS: #scrollbox {margin: 0 auto; margin-top: 1em; margin-bottom: 1em; width:19em; height:auto; max-height: 21em; overflow:auto; border-bottom: 0.1em solid #FFA500; border-top: 0.1em solid #FFA500;} #scrollbox a {float: left; color:#000000; text-decoration:none; width:18em; height: auto; margin-bottom: 0.5em; font-family: Arial, sans-serif; font-size: 0.9em; text-align:left;} #scrollbox a.menu {} #scrollbox a span {display:none; position:absolute; left:0em; top:0;} #scrollbox a span img {float: right; border:0; max-width:7.5em;} #scrollbox a:hover {border: 0; color: #7ebb11; font-size:0.9em;} #scrollbox a:hover span {border: 0; color: #535353;} #scrollbox a span:focus {color: blue;} #scrollbox a:active {border:none; color: #535353; text-decoration: none;} #scrollbox a:focus {border:0em solid #000; outline:0;} #scrollbox a:active span, #scrollbox li a:active span, #scrollbox a:focus span, #scrollbox li a:focus span {display: block; width: 52.5em; min-height: 20em; height: auto; left: 1.5em; top:18em; z-index:10; font-size:0.9em; text-align: left; padding: 1em; padding-bottom: 0em; background-color: #c3FFe3; color: #535353; border: solid #FFA500 0.25em;} #scrollbox li a:active span span, #scrollbox li a:focus span span{display: block; width: auto; height: auto; min-height: 2em; left: 25em; top:10em; z-index:10; font-size:0.9em; text-align: left; padding: 1em; padding-bottom: 0em; background-color: transparent; color: #535353; border: dashed red 1px;} .ul#scrollbox {padding-left: 0.1em;} #scrollbox li {float:left; list-style: none; background: url(blank.png) no-repeat left center; margin-left: 0em; font-family:Arial, sans-serif; font-size: 0.9em;} #scrollbox li.listcat {float: left; text-align:left; width: 18em; margin-left: 0em; margin-top: 0.1em; margin-bottom: 0.3em; padding-top:0.5em; color: green; font-size: 0.9em; font-weight:bold;} Cheers Patrick.

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  • Slalom Consulting San Francisco Custom Dev Challenge is live!

    - by PeterTweed
    The Slalom Consulting San Francisco Custom Dev Challenge is live at www.slalomchallenge.com!!!!! Slalom Consulting employs world-class technical consultants who take on ground breaking projects.  Please take the Slalom Custom Dev Challenge to see how you compare to the level of knowledge we look for in our technical consultants.  The online quiz is focussed on General .NET at this time and will be growing to include other technical topics in the future. This application is written in C#, Silverlight and WCF running deployed in the cloud on Windows Azure and working with SQL Azure and Blob Storage.

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  • WeekEnd BeMyApp : spécial Téléthon, le challenge de développement d'applications mobiles concerne cette fois iOS, Android, Web et Bada

    WeekEnd BeMyApp : spécial Téléthon Le nouveau challenge de développement d'applications mobiles concerne iOS, Android, Web et Bada Une édition spéciale du challenge de développement d'applications mobiles sera organisée pour le Téléthon. Les téléspectateurs et internautes pourront suivre en direct les projets et voter pour leur application préférée. Le principe d'un WeekEnd BeMyApp est de développer des applications mobiles en 2 jours non stop. Si vous avez une bonne idée d'application mobile, mais n'y connaissez rien, vous pouvez participer au BeMyApp en la présentant le vendredi soir. Si votre idée plait, des développeurs, designers et marketeurs la dévelop...

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  • Strlen of MAX 16 chars string using bitwise operators

    - by fabrizioM
    The challenge is to find the fastest way to determine in C/C++ the length of a c-string using bitwise operations in C. char thestring[16]; The c-string has a max size of 16 chars and is inside a buffer If the string is equal to 16 chars doesn't have the null byte at the end. I am sure can be done but didn't got it right yet. I am working on this at the moment, but assuming the string is memcpied on a zero-filled buffer. len = buff[0] != 0x0 + buff[1] != 0x0 + buff[2] != 0x0 + buff[3] != 0x0 + buff[4] != 0x0 + buff[5] != 0x0 + buff[6] != 0x0 + buff[7] != 0x0 + buff[8] != 0x0 + buff[9] != 0x0 + buff[10] != 0x0 + buff[11] != 0x0 + buff[12] != 0x0 + buff[13] != 0x0 + buff[14] != 0x0 + buff[15] != 0x0;

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  • How can one make a web-site accessible only when someone has a dongle?

    - by Brian M. Hunt
    Suppose you want to add an extra layer of credentials on top of a SSL-encrypted login/password, but you don't want to increase complexity to the user. Is there a way to add the requirement of the possession of a dongle to web-server authentication schemes with existing cross-platform browser capabilities? In other words, to get access to the web-site, you would need a username, password, and a USB dongle that has been plugged into the client computer. The dongle would presumably do some sort of challenge/response. It'd be ideal if this dongle solution worked with Firefox automatically or with the simple addition of a plugin. Thoughts and suggestions are appreciated.

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  • Triangle numbers problem....show within 4 seconds

    - by Daredevil
    The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be: 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ... Let us list the factors of the first seven triangle numbers: 1: 1 3: 1,3 6: 1,2,3,6 10: 1,2,5,10 15: 1,3,5,15 21: 1,3,7,21 28: 1,2,4,7,14,28 We can see that 28 is the first triangle number to have over five divisors. Given an integer n, display the first triangle number having at least n divisors. Sample Input: 5 Output 28 Input Constraints: 1<=n<=320 I was obviously able to do this question, but I used a naive algorithm: Get n. Find triangle numbers and check their number of factors using the mod operator. But the challenge was to show the output within 4 seconds of input. On high inputs like 190 and above it took almost 15-16 seconds. Then I tried to put the triangle numbers and their number of factors in a 2d array first and then get the input from the user and search the array. But somehow I couldn't do it: I got a lot of processor faults. Please try doing it with this method and paste the code. Or if there are any better ways, please tell me.

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  • Windows Phone : Nokia lance son premier Challenge français sur DVLUP, sa nouvelle plateforme de défis et de quiz dédiés aux développeurs

    Windows Phone : Nokia lance son premier Challenge français sur DVLUP Sa plateforme de défis et de quiz dédiés aux développeursOn vous en avait touché deux mots en avant-première, lors du lancement du Lumia 1020 début octobre. Le « programme de fidélité » de Nokia pour les développeurs est depuis devenu officiel. Baptisée DVLUP, la plateforme était accessible depuis d'autres pays mais pas encore en France.Le portail propose des Challenges - environ un nouveau par semaine - et des quiz. Ces défis...

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  • Programming language shootout: code most like pseudocode for Dijkstra's Algorithm

    - by Casebash
    Okay, so this question here asked which language is most like executable pseudocode, so why not find out by actually writing some code! Here we have a competition where I will award a 100 point bounty (I know its not much, but I am poor after the recalc) to the code which most resembles this pseudocode. I've read through this a few times so I'm pretty sure that this pseudocode below is correct and about as unambiguous as pseudocode can be. Personally, I'm going to have a go in Python and probably Haskell as well, but I'm just learning the later so my attempt will probably be pretty poor. Note: Obviously to implement anything looking like this you'll have to define quite a few library functions. define DirectedGraph G with: Vertices as V, Edges as E define Vertex A, Z declare each e in E as having properties: Boolean fixed with: initial=false Real minSoFar with: initial=0 for A else infinity define PriorityQueue pq with: objects=V initial=A priority v=v.minSoFar create triggers for v in V: when v.minSoFar event reduced then pq.addOrUpdate v when v.fixed event becomesTrue then pq.remove v Repeat until Z.fixed==True: define Vertex U=pq.pop() U.fixed=True for Edge E adjacentTo U with other Vertex V: V.minSoFar=U.minSoFar+length(E) if reducesValue return Z.name, Z.minSoFar

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  • The Skyline Problem.

    - by zeroDivisible
    I just came across this little problem on UVA's Online Judge and thought, that it may be a good candidate for a little code-golf. The problem: You are to design a program to assist an architect in drawing the skyline of a city given the locations of the buildings in the city. To make the problem tractable, all buildings are rectangular in shape and they share a common bottom (the city they are built in is very flat). The city is also viewed as two-dimensional. A building is specified by an ordered triple (Li, Hi, Ri) where Li and Ri are left and right coordinates, respectively, of building i and Hi is the height of the building. In the diagram below buildings are shown on the left with triples (1,11,5), (2,6,7), (3,13,9), (12,7,16), (14,3,25), (19,18,22), (23,13,29), (24,4,28) and the skyline, shown on the right, is represented by the sequence: 1, 11, 3, 13, 9, 0, 12, 7, 16, 3, 19, 18, 22, 3, 23, 13, 29, 0 The output should consist of the vector that describes the skyline as shown in the example above. In the skyline vector (v1, v2, v3, ... vn) , the vi such that i is an even number represent a horizontal line (height). The vi such that i is an odd number represent a vertical line (x-coordinate). The skyline vector should represent the "path" taken, for example, by a bug starting at the minimum x-coordinate and traveling horizontally and vertically over all the lines that define the skyline. Thus the last entry in the skyline vector will be a 0. The coordinates must be separated by a blank space. If I will not count declaration of provided (test) buildings and including all spaces and tab characters, my solution, in Python, is 223 characters long. Here is the condensed version: B=[[1,11,5],[2,6,7],[3,13,9],[12,7,16],[14,3,25],[19,18,22],[23,13,29],[24,4,28]] # Solution. R=range v=[0 for e in R(max([y[2] for y in B])+1)] for b in B: for x in R(b[0], b[2]): if b[1]>v[x]: v[x]=b[1] p=1 k=0 for x in R(len(v)): V=v[x] if p and V==0: continue elif V!=k: p=0 print "%s %s" % (str(x), str(V)), k=V I think that I didn't made any mistake but if so - feel free to criticize me. EDIT I don't have much reputation, so I will pay only 100 for a bounty - I am curious, if anyone could try to solve this in less than .. lets say, 80 characters. Solution posted by cobbal is 101 characters long and currently it is the best one. ANOTHER EDIT I thought, that 80 characters is a sick limit for this kind of problem. cobbal, with his 46 character solution totaly amazed me - though I must admit, that I spent some time reading his explanation before I partially understood what he had written.

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  • Grouping consecutive identical items: IEnumerable<T> to IEnumerable<IEnumerable<T>>

    - by Romain Verdier
    I've got an interresting problem: Given an IEnumerable<string>, is it possible to yield a sequence of IEnumerable<string> that groups identical adjacent strings in one pass? Let me explain. Considering the following IEnumerable<string> (pseudo representation): {"a","b","b","b","c","c","d"} How to get an IEnumerable<IEnumerable<string>> that would yield something of the form: { // IEnumerable<IEnumerable<string>> {"a"}, // IEnumerable<string> {"a","b","b"}, // IEnumerable<string> {"c","c"}, // IEnumerable<string> {"d"} // IEnumerable<string> } The method prototype would be: public IEnumerable<IEnumerable<string>> Group(IEnumerable<string> items) { // todo } Important notes : Only one iteration over the original sequence No intermediary collections allocations (we can assume millions of strings in the original sequence, and millions consecutives identicals strings in each group) Keeping enumerators and defered execution behavior Is it possible, and how would you write it?

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  • Where do you go to tickle your brain (to get programming challenges)?

    - by Prakash
    I am sure we all have some place to go to get our brain teased! Sometimes i visit Project Euler is a series of challenging mathematical/computer programming problems that will require more than just mathematical insights to solve. Although mathematics will help you arrive at elegant and efficient methods, the use of a computer and programming skills will be required to solve most problems Where do you all go?

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  • How to print "Hello, world!" (in every possible way)

    - by Attila Oláh
    Here's what I', trying to do: 1 language: (Python < 3): print "Hello, world!" 2 languages: (Python < 3 & Windows Shell, aka .bat file): rem=""" echo "Hello, world!" exit """ print "Hello, world!" Next step could be something like bash. Since the above one raises an exception, I tried to make it not raise exceptions, like this: rem=""" echo "Hello, world!" exit """ exit="" exit print "Hello, world!" The only issue is, of course, it won't print the hello world. And I really want it to print that hello world for me. Anyone can help with this? Also, any other language would do it, just don't break the previous ones (i.e. the answer still has to be valid Python code and print out the nice hello world greeting when run with Python.) Any ideas are welcome. I'm making this a community wiki so feel free to add ideas to the list.

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