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  • Formating phone numbers

    - by Sven
    Our customers often fill out "incorrect" formated phone-numbers. Do anyone know if there is any lib or standard to convert numbers into a more international style? This is a Swedish example but we have customers around the globe and i don't what to manually handle implementations for everyone. input often is like this: 0555 11122 and the wanted result is something like this: +46(0)555-11122 I can do the formating myself but different countries have different variations and systems so a C/Java/C# lib or a standard method to handle this would be great.

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  • C# Extension Method for String Data Type

    - by Jimbo
    My web application deals with strings that need to be converted to numbers alot - users often put commas, currency symbols etc. in these fields so what I want to do is create a string extension method that cleans the field up and converts it to a decimal. For example: decimal myNumber = "$1,250.85".ToDecimal(); Can anyone help with this? Thanks!

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  • Using system time directly to get random numbers

    - by Richard Mar.
    I had to return a random element from an array so I came up with this placeholder: return codes[(int) (System.currentTimeMillis() % codes.length - 1)]; Now than I think of it, I'm tempted to use it in real code. The Random() seeder uses system time as seed in most languages anyway, so why not use that time directly? As a bonus, I'm free from the worry of non-random lower bits of many RNGs. It this hack coming back to bite me? (The language is Java if that's relevant.)

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  • Random numbers from binomial distribution

    - by Sarah
    I need to generate quickly lots of random numbers from binomial distributions for dramatically different trial sizes (most, however, will be small). I was hoping not to have to code an algorithm by hand (see, e.g., this related discussion from November), because I'm a novice programmer and don't like reinventing wheels. It appears Boost does not supply a generator for binomially distributed variates, but TR1 and GSL do. Is there a good reason to choose one over the other, or is it better that I write something customized to my situation? I don't know if this makes sense, but I'll alternate between generating numbers from uniform distributions and binomial distributions throughout the program, and I'd like for them to share the same seed and to minimize overhead. I'd love some advice or examples for what I should be considering.

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  • Enter a TAB after every Xth character of text

    - by T. Schmidt
    Hi, not necessarily a programmer question but I need to enter a TAB after every 84th character of a text. I'm trying to do it in InDesign but I don’t see an option for this. Is there a good application to do this? Preferably Mac OS X compatlible but Windows XP is fine too. Thank you very much!

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  • Help for creating a random String

    - by Max
    I need to create a random string which should be between the length of 6 to 10 but it sometimes generates only about the length of 3 to 5. Here's my code. Can anyone would be able to find out the problem? :( int lengthOfName = (int)(Math.random() * 4) + 6; String name = ""; /* randomly choosing a name*/ for (int j = 0; j <= lengthOfName; j++) { int freq = (int)(Math.random() * 100) + 1; if(freq <= 6){ name += "a"; }if(freq == 7 && freq == 8){ name += "b"; }if(freq >= 9 && freq <= 11){ name += "c"; }if(freq >= 12 && freq <= 15){ name += "d"; }if(freq >= 16 && freq <= 25){ name += "e"; }if(freq == 26 && freq == 27){ name += "f"; }if(freq == 28 && freq == 29){ name += "g"; }if(freq >= 30 && freq <= 33){ name += "h"; }if(freq >= 34 && freq <= 48){ name += "i"; }if(freq == 49 && freq == 50){ name += "j"; }if(freq >= 51 && freq <= 55){ name += "k"; }if(freq >= 56 && freq <= 60){ name += "l"; }if(freq == 61 && freq == 62){ name += "m"; }if(freq >= 63 && freq <= 70){ name += "n"; }if(freq >= 71 && freq <= 75){ name += "o"; }if(freq == 76 && freq == 77){ name += "p"; }if(freq == 78){ name += "q"; }if(freq >= 79 && freq <= 84){ name += "r"; }if(freq == 85 && freq == 86){ name += "s"; }if(freq == 87 && freq == 88){ name += "t"; }if(freq >= 89 && freq <= 93){ name += "u"; }if(freq == 94){ name += "v"; }if(freq == 95 && freq == 96){ name += "w"; }if(freq == 97){ name += "x"; }if(freq == 98 && freq == 99){ name += "y"; }if(freq == 100){ name += "z"; } }

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  • Set seed on Math.random()

    - by Kevin
    Hi - I need to write some junit tests on java code that calls Math.random(). I know that I can set the seed if I was instantiating my own Random object to produce repeatable results. Is there a way to do this also for Math.random() ?

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  • Latest stream cipher considered reasonably secure & easy to implement?

    - by hythlodayr
    (A)RC4 used to fit the bill, since it was so simple to write. But it's also less-than-secure these days. I'm wondering if there's a successor that's: Code is small enough to write & debug within an hour or so, using pseudo code as a template. Still considered secure, as of 2010. Optimized for software. Not encumbered by licensing issues. I can't use crypto libraries, otherwise all of this would be moot. Also, I'll consider block algorithms though I think most are pretty hefty. Thanks.

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  • How do i generate random data with RSA?

    - by acidzombie24
    After loading my RSACryptoServiceProvider rsa object i would like to create a key for my AES object. Since i dont need to store the AES key (i only need it to decrypt on my prv side) i figure i dont need to store it and i can generate it with my public key. I thought doing rsa.Encrypt(byte[] with 4 hardcoded bytes); would generate the data i need. It turns out everytime i call this function even with the same data i get different results. So theres no way for me to recreate the AES key if its different everytime. How can i generate data with RSA in a way that i can recreate anytime i need?

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  • ~/.xsession-errors is 2.7gb big (and growing), on fresh install, caused by gnome-settings-daemon errors

    - by Alex Black
    I've just installed Ubuntu 10.10 x64, activated the recommended Nvidia drivers, and I noticed my hard disk space is disappearing, I narrowed the culprit down to this: alex@alex-home:~$ ls -la .x* -rw------- 1 alex alex 4436076400 2010-11-19 22:35 .xsession-errors -rw------- 1 alex alex 10495 2010-11-19 21:46 .xsession-errors.old Any idea what this file is, why its so big, and why its growing? A few seconds later: alex@alex-home:~$ ls -la .x* -rw------- 1 alex alex 5143604317 2010-11-19 22:36 .xsession-errors -rw------- 1 alex alex 10495 2010-11-19 21:46 .xsession-errors.old tailing it: alex@alex-home:~$ tail .xsession-errors (gnome-settings-daemon:1514): GLib-GObject-CRITICAL **: g_object_unref: assertion `G_IS_OBJECT (object)' failed (gnome-settings-daemon:1514): GLib-GObject-CRITICAL **: g_object_unref: assertion `G_IS_OBJECT (object)' failed (gnome-settings-daemon:1514): GLib-GObject-CRITICAL **: g_object_unref: assertion `G_IS_OBJECT (object)' failed (gnome-settings-daemon:1514): GLib-GObject-CRITICAL **: g_object_unref: assertion `G_IS_OBJECT (object)' failed (gnome-settings-daemon:1514): GLib-GObject-CRITICAL **: g_object_unref: assertion `G_IS_OBJECT (object)' failed Also, the process "gnome-settings" seems to be using 100% cpu: PID USER PR NI VIRT RES SHR S %CPU %MEM TIME+ COMMAND 1514 alex 20 0 268m 10m 7044 R 100 0.1 7:06.10 gnome-settings-

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  • How to approach big developer companies if I have a killer game idea? (for mobile devices)

    - by Balázs Dávid
    I have an idea for a game that has a potential, but I'm not a programmer. How to tell this to development companies without being my idea stolen? All I want from the company is first to watch a 3-minute long video presentation about my idea and if they see fantasy in it then we can talk about the details. I have already sent an e-mail to several big companies that have the expertise needed to code the game, they haven't answer me. Actually the idea is nothing fancy, no 3D, but fun and unique.

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  • How to keep a big and complex software product maintainable over the years?

    - by chrmue
    I have been working as a software developer for many years now. It has been my experience that projects get more complex and unmaintainable as more developers get involved in the development of the product. It seems that software at a certain stage of development has the tendency to get "hackier" and "hackier" especially when none of the team members that defined the architecture work at the company any more. I find it frustrating that a developer who has to change something has a hard time getting the big picture of the architecture. Therefore, there is a tendency to fix problems or make changes in a way that works against the original architecture. The result is code that gets more and more complex and even harder to understand. Is there any helpful advice on how to keep source code really maintainable over the years?

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  • SAPPHIRE 2012 : l'informatique à un tournant majeur ? SAP voit une révolution de l'IT professionnelle avec le Big Data, HANA et le Cloud

    SAPPHIRE 2012 : l'informatique professionnelle à un tournant de son Histoire ? SAP veut révolutionner les usages de l'IT avec le Big Data, HANA et le Cloud Au détour des allées du SAPPHIRE 2012, la grand messe annuelle de SAP qui se tient actuellement à Madrid, deux démonstrations sortent du lot. La première vient de McLaren. Bien connue des amateurs de Formule 1, l'entreprise est une grande utilisateur des solutions de SAP. Pour son « showcase », les équipes du constructeur automobile ont collecté des données de courses de l'épreuve de Monza 2010 et les ont compilées pour reproduire in vivo la course de leurs deux pilotes. Cette simulation sert ensuite de base pour ...

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  • How do I create a big multiplayer world in UDK?

    - by Dorpe
    I want to create a big multiplayer world in UDK and I'm having a few difficulties. I created the biggest terrain possible but then any terrain related action I do takes forever. However, I've seen videos of people make same size terrain and working without a problem. My pc is strong enough, so maybe someone can tell me what I'm doing wrong. I want to make it even bigger then the biggest terrain size, so I was thinking of doing level streaming but then I read that streaming is working server side which means if I have a player on every terrain all terrains will still be loaded and I want to save as much memory possible so it will work well online. Thanks for any help you can give.

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  • SAPPHIRE 2012 : SAP veut imposer sa nouvelle image et simplifier l'IT avec ses outils pour le Big Data, la mobilité et le Cloud

    SAPPHIRE 2012 : SAP veut imposer sa nouvelle image et simplifier l'IT Avec des outils pour le Big Data, la mobilité et le Cloud Après plusieurs années de changements radicaux qui l'ont fait passer d'éditeur mono-produit à fournisseur multi-services (BI, SGBD, Cloud, mobilité, In-computing), SAP rentre dans une nouvelle phase : celle de la consolidation. Cette évolution dans la continuité se traduit jusque dans l'arrivée de nouveaux slogans (comme « SAP runs like never before ») inspiré du traditionnel « Runs Better with SAP », placardés sur les murs du SAPPHIRE ? la grande messe annuelle de l'éditeur qui a ouvert ses portes aujourd'hui à Madrid. L'évolution se...

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  • How can I compute the Big-O notation for a given piece of code?

    - by TheNew Rob Mullins
    So I just took a data structure midterm today and I was asked to determine the run time, in Big O notation, of the following nested loop: for (int i = 0; i < n-1; i++) { for(int j = 0; j < i; j++2) { //1 Statement } } I'm having trouble understanding the formula behind determining the run time. I thought that since the inner loop has 1 statement, and using the series equation of: (n * (n - 1)) / 2, I figured it to be: 1n * (n-1) / 2. Thus equaling (n^2 - 1) / 2. And so I generalized the runtime to be O(n^2 / 2). I'm not sure this is right though haha, was I supposed to divide my answer again by 2 since j is being upped in intervals of 2? Or is my answer completely off?

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  • UVA Online Judge 3n+1 : Right answer is Wrong answer

    - by Samuraisoulification
    Ive been toying with this problem for more than a week now, I have optimized it a lot, I seem to be getting the right answer, since it's the same as when I compare it to other's answers that got accepted, but I keep getting wrong answer. Im not sure what's going on! Anyone have any advice? I think it's a problem with the input or the output, cause Im not exactly sure how this judge thing works. So if anyone could pinpoint the problem, and also give me any advice on my code, Id be very appreciative!!! #include <iostream> #include <cstdlib> #include <stdio.h> #include <vector> using namespace std; class Node{ // node for each number that has teh cycles and number private: int number; int cycles; bool cycleset; // so it knows whether to re-set the cycle public: Node(int num){ number = num; cycles = 0; cycleset = false; } int getnumber(){ return number; } int getcycles(){ return cycles; } void setnumber(int num){ number = num; } void setcycles(int num){ cycles = num; cycleset = true; } bool cycled(){ return cycleset; } }; class Cycler{ private: vector<Node> cycleArray; int biggest; int cycleReal(unsigned int number){ // actually cycles through the number int cycles = 1; if (number != 1) { if (number < 1000000) { // makes sure it's in vector bounds if (!cycleArray[number].cycled()) { // sees if it's been cycled if (number % 2 == 0) { cycles += this->cycleReal((number / 2)); } else { cycles += this->cycleReal((3 * number) + 1); } } else { // if cycled get the number of cycles and don't re-calculate, ends recursion cycles = cycleArray[number].getcycles(); } } else { // continues recursing if it's too big for the vector if (number % 2 == 0) { cycles += this->cycleReal((number / 2)); } else { cycles += this->cycleReal((3 * number) + 1); } } } if(number < 1000000){ // sets cycles table for the number in the vector if (!cycleArray[number].cycled()) { cycleArray[number].setcycles(cycles); } } return cycles; } public: Cycler(){ biggest = 0; for(int i = 0; i < 1000000; i++){ // initialize the vector, set the numbers Node temp(i); cycleArray.push_back(temp); } } int cycle(int start, int end){ // cycles thorugh the inputted numbers. int size = 0; for(int i = start; i < end ; i++){ size = this->cycleReal(i); if(size > biggest){ biggest = size; } } int temp = biggest; biggest = 0; return temp; } int getBiggest(){ return biggest; } }; int main() { Cycler testCycler; int i, j; while(cin>>i>>j){ //read in untill \n int biggest = 0; if(i > j){ biggest = testCycler.cycle(j, i); }else{ biggest = testCycler.cycle(i, j); } cout << i << " " << j << " " << biggest << endl; } return 0; }

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  • Developing an Implementation Plan with Iterations by Russ Pitts

    - by user535886
    Developing an Implementation Plan with Iterations by Russ Pitts  Ok, so you have come to grips with understanding that applying the iterative concept, as defined by OUM is simply breaking up the project effort you have estimated for each phase into one or more six week calendar duration blocks of work. Idea being the business user(s) or key recipient(s) of work product(s) being developed never go longer than six weeks without having some sort of review or prototyping of the work results for an iteration…”think-a-little”, “do-a-little”, and “show-a-little” in a six week or less timeframe…ideally the business user(s) or key recipients(s) are involved throughout. You also understand the OUM concept that you only plan for that which you have knowledge of. The concept further defined, a project plan initially is developed at a high-level, and becomes more detailed as project knowledge grows. Agreeing to this concept means you also have to admit to the fallacy that one can plan with precision beyond six weeks into a project…Anything beyond six weeks is a best guess in most cases when dealing with software implementation projects. Project planning, as defined by OUM begins with the Implementation Plan view, which is a very high-level perspective of the effort estimated for each of the five OUM phases, as well as the number of iterations within each phase. You might wonder how can you predict the number of iterations for each phase at this early point in the project. Remember project planning is not an exact science, and initially is high-level and abstract in nature, and then becomes more detailed and precise as the project proceeds. So where do you start in defining iterations for each phase for a project? The following are three easy steps to initially define the number of iterations for each phase: Step 1 => Start with identifying the known factors… …Prior to starting a project you should know: · The agreed upon time-period for an iteration (e.g 6 weeks, or 4 weeks, or…) within a phase (recommend keeping iteration time-period consistent within a phase, if not for the entire project) · The number of resources available for the project · The number of total number of man-day (effort) you have estimated for each of the five OUM phases of the project · The number of work days for a week Step 2 => Calculate the man-days of effort required for an iteration within a phase… Lets assume for the sake of this example there are 10 project resources, and you have estimated 2,536 man-days of work effort which will need to occur for the elaboration phase of the project. Let’s also assume a week for this project is defined as 5 business days, and that each iteration in the elaboration phase will last a calendar duration of 6 weeks. A simple calculation is performed to calculate the daily burn rate for a single iteration, which produces a result of… ((Number of resources * days per week) * duration of iteration) = Number of days required per iteration ((10 resources * 5 days/week) * 6 weeks) = 300 man days of effort required per iteration Step 3 => Calculate the number of iterations that can occur within a phase Next calculate the number of iterations that can occur for the amount of man-days of effort estimated for the phase being considered… (number of man-days of effort estimated / number of man-days required per iteration) = # of iterations for phase (2,536 man-days of estimated effort for phase / 300 man days of effort required per iteration) = 8.45 iterations, which should be rounded to a whole number such as 9 iterations* *Note - It is important to note this is an approximate calculation, not an exact science. This particular example is a simple one, which assumes all resources are utilized throughout the phase, including tech resources, etc. (rounding down or up to a whole number based on project factor considerations). It is also best in many cases to round up to higher number, as this provides some calendar scheduling contingency.

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  • Any reliable polygon normal calculation code?

    - by Jenko
    Do you have any reliable face normal calculation code? I'm using this but it fails when faces are 90 degrees upright or similar. // the normal point var x:Number = 0; var y:Number = 0; var z:Number = 0; // if is a triangle with 3 points if (points.length == 3) { // read vertices of triangle var Ax:Number, Bx:Number, Cx:Number; var Ay:Number, By:Number, Cy:Number; var Az:Number, Bz:Number, Cz:Number; Ax = points[0].x; Bx = points[1].x; Cx = points[2].x; Ay = points[0].y; By = points[1].y; Cy = points[2].y; Az = points[0].z; Bz = points[1].z; Cz = points[2].z; // calculate normal of a triangle x = (By - Ay) * (Cz - Az) - (Bz - Az) * (Cy - Ay); y = (Bz - Az) * (Cx - Ax) - (Bx - Ax) * (Cz - Az); z = (Bx - Ax) * (Cy - Ay) - (By - Ay) * (Cx - Ax); // if is a polygon with 4+ points }else if (points.length > 3){ // calculate normal of a polygon using all points var n:int = points.length; x = 0; y = 0; z = 0 // ensure all points above 0 var minx:Number = 0, miny:Number = 0, minz:Number = 0; for (var p:int = 0, pl:int = points.length; p < pl; p++) { var po:_Point3D = points[p] = points[p].clone(); if (po.x < minx) { minx = po.x; } if (po.y < miny) { miny = po.y; } if (po.z < minz) { minz = po.z; } } if (minx > 0 || miny > 0 || minz > 0){ for (p = 0; p < pl; p++) { po = points[p]; po.x -= minx; po.y -= miny; po.z -= minz; } } var cur:int = 1, prev:int = 0, next:int = 2; for (var i:int = 1; i <= n; i++) { // using Newell method x += points[cur].y * (points[next].z - points[prev].z); y += points[cur].z * (points[next].x - points[prev].x); z += points[cur].x * (points[next].y - points[prev].y); cur = (cur+1) % n; next = (next+1) % n; prev = (prev+1) % n; } } // length of the normal var length:Number = Math.sqrt(x * x + y * y + z * z); // if area is 0 if (length == 0) { return null; }else{ // turn large values into a unit vector x = x / length; y = y / length; z = z / length; }

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