Search Results

Search found 6716 results on 269 pages for 'distributed algorithm'.

Page 85/269 | < Previous Page | 81 82 83 84 85 86 87 88 89 90 91 92  | Next Page >

  • CODE1 at SPOJ - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code on SPOJ, and it's obviously a math problem. The full problem For those who are lazy to go and read, it's like this: a0, a1, a2, ..., an - sequence of N numbers B - a Complex Number (has both real and imaginary components) X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0, a1, ..an. I don't know how or where to start, because not even N is known, just X and B. The problem is not as easy as expressing a number in a base B, because B is a complex number. How can it be solved?

    Read the article

  • compare function for upper_bound / lower_bound

    - by Martin Beckett
    I want to find the first item in a sorted vector that has a field less than some value x. I need to supply a compare function that compares 'x' with the internal value in MyClass but I can't work out the function declaration. Can't I simply overload '<' but how do I do this when the args are '&MyClass' and 'float' ? float x; std::vector< MyClass >::iterator last = std::upper_bound(myClass.begin(),myClass.end(),x);

    Read the article

  • General Address Parser for Freeform Text

    - by Daemonic
    We have a program that displays map data (think Google Maps, but with much more interactivity and custom layers for our clients). We allow navigation via a set of combo boxes that prefill certain fields with a bunch of data (ie: Country: Canada, the Province field is filled in. Select Ontario, and a list of Counties/Regions is filled in. Select a county/region, and a city is filled in, etc...). While this guarantees accurate addresses, it's a pain for the users if they don't know where a street address or a city are located (ie, which county/region is kitchener in?). So we are looking at trying to do an address parser with a freeform text field. The user could enter something like this (similar to Google Maps, Bing Maps, etc...): 22 Main St, Kitchener, On And we could compartmentalize it into sections and do lookups on the data and get to the point they are looking for (or suggest alternatives). The problem with this is that how do we properly compartmentalize information? How do we break up the sections and find possible matches? I'm guessing we wouldn't be guaranteed that the user would enter data in a format we always expected (obviously). A follow up to this would be how to present the data if we don't find an exact match (or find multiple exact matches... two cities with the same street name in different counties, for example). We have a ton of data available in the mapping data (mapinfo tab format mostly). So we can do quick scans of street names, cities, states, etc. But I'm not sure about the best way to go about approaching this problem. Sure, using Google Maps would be nice, bue most of our clients are in closed in networks where outside access is not usually allowed and most aren't willing to rely on google maps (since it doesn't contain as much information as they need, such as custom map layers). They could, obviously, go to google and get the proper location then move to our software, but this would time consuming and speed of the process can be quite important.

    Read the article

  • Diagonal of polygon is inside or outside?

    - by Himadri
    I have three consecutive points of polygon, say p1,p2,p3. Now I wanted to know whether the orthogonal between p1 and p3 is inside the polygon or outside the polygon. I am doing it by taking three vectors v1,v2 and v3. And the point before the point p1 in polygon say p0. v1 = (p0 - p1) v2 = (p2 - p1) v3 = (p3 - p1) With reference to this question, I am using the method shown in the accepted answer of that question. It is only for counterclockwise. What if my points are clockwise. I am also knowing my whole polygon is clockwise or counterclockwise. And accordingly I select the vectors v1 and v2. But still I am getting some problem. I am showing one case where I am getting problem. This polygon is counterclockwise. and It is starting from the origin of v1 and v2.

    Read the article

  • All words in a trie data-structure

    - by John Smith
    I'm trying to put all words in a trie in a string, a word is detonated by the eow field being true for a certain character in the trie data structure, hence a trie can could have letters than lead up to no word, for ex "abc" is in the trie but "c"'s eow field is false so "abc" is not a word Here is my Data structure struct Trie { bool eow; //when a Trie field isWord = true, hence there is a word char letter; Trie *letters[27]; }; and here is my attemped print-all function, basically trying to return all words in one string seperated by spaces for words string printAll( string word, Trie& data) { if (data.eow == 1) return word + " "; for (int i = 0; i < 26; i++) { if (data.letters[i] != NULL) printAll( word + data.letters[i]->letter, *(data.letters[i])); } return ""; } Its not outputting what i want, any suggestions?

    Read the article

  • CODE1 Spoj - cannot solve it

    - by VaioIsBorn
    I am trying to solve the problem Secret Code and it's obviously math problem. The full problem For those who are lazy to go and read, it's like this: a0,a1,a2,...,an - sequence of N numbers B - some number known to us X = a0 + a1*B + a2*(B^2) + a3*(B^3) + ... + an*(B^n) So if you are given B and X, you should find a0,a1,..an. I don't know how or where to start, because not even N is known, just X and B. Can you help me ?

    Read the article

  • how to determine base of a number?

    - by evil.coder
    Given a integer number and its reresentation in some arbitrary number system. The purpose is to find the base of the number system. For example, number is 10 and representation is 000010, then the base should be 10. Another example: number 21 representation is 0010101 then base is 2. One more example is: number is 6 and representation os 10100 then base is sqrt(2). Does anyone have any idea how to solve such problem?

    Read the article

  • How Can I Improve This Algorithm (LCS)

    - by superguay
    (define (lcs lst1 lst2) (define (except-last-pair list) (if (pair? (cdr list)) (cons (car list) (except-last-pair (cdr list))) '())) (define (car-last-pair list) (if (pair? (cdr list)) (car-last-pair (cdr list)) (car list))) (if (or (null? lst1) (null? lst2)) null (if (= (car-last-pair lst1) (car-last-pair lst2)) (append (lcs (except-last-pair lst1) (except-last-pair lst2)) (cons (car-last-pair lst1) '())) **(if (> (length (lcs lst1 (except-last-pair lst2))) (length (lcs lst2 (except-last-pair lst1)))) (lcs lst1 (except-last-pair lst2)) (lcs lst2 (except-last-pair lst1)))))) I dont want it to run over and over.. Regards, Superguay

    Read the article

  • Grouping rectangles (getting the bounding boxes of rects)

    - by hyn
    What is a good, fast way to get the "final" bounding boxes of a set of random (up to about 40, not many) rectangles? By final I mean that all bounding boxes don't intersect with any other. Brute force way: in a double for loop, for each rect, test for intersection against every other rect. The intersecting rects become a new rect (replaced), indicating the bounding box. Start over and repeat until no intersection is detected. Because the rects are random every time, and the rect count is relatively small, collision detection using spatial hashing seems like overkill. Is there a way to do this more effectively?

    Read the article

  • Given 4 objects, how to figure out whether exactly 2 have a certain property

    - by Cocorico
    Hi guys! I have another question on how to make most elegant solution to this problem, since I cannot afford to go to computer school right so my actual "pure programming" CS knowledge is not perfect or great. This is basically an algorhythm problem (someone please correct me if I am using that wrong, since I don't want to keep saying them and embarass myself) I have 4 objects. Each of them has an species property that can either be a dog, cat, pig or monkey. So a sample situation could be: object1.species=pig object2.species=cat object3.species=pig object4.species=dog Now, if I want to figure out if all 4 are the same species, I know I could just say: if ( (object1.species==object2.species) && (object2.species==object3.species) && (object3.species==object4.species) ) { // They are all the same animal (don't care WHICH animal they are) } But that isn't so elegant right? And if I suddenly want to know if EXACTLY 3 or 2 of them are the same species (don't care WHICH species it is though), suddenly I'm in spaghetti code. I am using Objective C although I don't know if that matters really, since the most elegant solution to this is I assume the same in all languages conceptually? Anyone got good idea? Thanks!!

    Read the article

  • Algorithms behind load-balancers?

    - by Vimvq1987
    I need to study about load-balancers, such as Network Load Balancing, Linux Virtual Server, HAProxy,...There're somethings under-the-hood I need to know: What algorithms/technologies are used in these load-balancers? Which is the most popular? most effective? I expect that these algorithms/technologies will not be too complicated. Are there some resources written about them? Thank you very much for your help.

    Read the article

  • A shortest path problem with superheroes and intergalactic journeys

    - by Dman
    You are a super-hero in the year 2222 and you are faced with this great challenge: starting from your home planet Ilop you must try to reach Acinhet or else your planet will be destroyed by evil green little monsters. To do this you are given a map of the universe: there are N planets and M inter-planetary connections ( bidirectional ) that bind these planets. Each connection requires a certain time and a certain amount of fuel in order for you to cover the connection from one planet to another. The total time spent going from one planet to another is obtained by multiplying the time past to cover each connection between all the planets you go through. There are some "key planets", that allow you to refuel if you arrive on those certain "key planets". A "key planet" is the planet with the property that if it disappears the road between at least two planets would be lost.(In the example posted below with the input/output files such a "key planet" is 2 because without it the road to 7 would be lost) When you start your mission you are given the possibility of choosing between K ships each with its own maximum fuel capacity. The goal is to find the SHORTEST TIME CONSUMING path but also choose the ship with the minimum fuel capacity that can cover that shortest path(this means that if more ships can cover the shortest path you choose the one with the minimum fuel capacity). Because the minimum time can be a rather large number (over long long int) you are asked to provide only the last 6 digits of the number. For a better understanding of the task, here is an example of input/output files: INPUT: mission.in 7 8 6 1 4 6 5 9 8 7 10 1 2 7 8 1 4 14 9 1 5 3 1 2 3 1 2 2 7 7 1 3 4 2 2 4 6 4 1 5 6 3 7 On the first line (in order): N M K On the second line :the number for the starting planet and the finishing planet On the third line :K numbers that represent the capacities of the ships you can choose from Then you have M lines, all of them have the same structure: Xi Yi Ti Fi-which means that there is a connection between Xi and Yi and you can cover the distance from Xi to Yi in Ti time and with a Fi fuel consumption. OUTPUT:mission.out 000014 8 1 2 3 4 On the first line:the minimum time and fuel consumption; On the second line :the path Restrictions: 2 = N = 1 000 1 = M = 30 000 1 = K = 10 000 Any suggestions or ideas of how this problem might be solved would be most welcomed.

    Read the article

  • How to find nth element from the end of a singly linked list?

    - by Codenotguru
    The following function is trying to find the nth to last element of a singly linked list. For example: If the elements are 8->10->5->7->2->1->5->4->10->10 then the result is 7th to last node is 7. Can anybody help me on how this code is working or is there a better and simpler approach? LinkedListNode nthToLast(LinkedListNode head, int n) { if (head == null || n < 1) { return null; } LinkedListNode p1 = head; LinkedListNode p2 = head; for (int j = 0; j < n - 1; ++j) { // skip n-1 steps ahead if (p2 == null) { return null; // not found since list size < n } p2 = p2.next; } while (p2.next != null) { p1 = p1.next; p2 = p2.next; } return p1; }

    Read the article

  • red black tree balancing?

    - by Anirudh Kaki
    i am working to generate tango tree, where i need to check whether every sub tree in tango is balanced or not. if its not balanced i need to make it balance? I trying so hard to make entire RB-tree balance but i not getting any proper logic so can any one help me out?? here i am adding code to check how to find my tree is balanced are not but when its not balanced how can i make it balance. static boolean verifyProperty5(rbnode n) { int left = 0, right = 0; if (n != null) { bh++; left = blackHeight(n.left, 0); right = blackHeight(n.right, 0); } if (left == right) { System.out.println("black height is :: " + bh); return true; } else { System.out.println("in balance"); return false; } } public static int blackHeight(rbnode root, int len) { bh = 0; blackHeight(root, path1, len); return bh; } private static void blackHeight(rbnode root, int path1[], int len) { if (root == null) return; if (root.color == "black"){ root.black_count = root.parent.black_count+1; } else{ root.black_count = root.parent.black_count; } if ((root.left == null) && (root.right == null)) { bh = root.black_count; } blackHeight(root.left, path1, len); blackHeight(root.right, path1, len); }

    Read the article

  • STL vector reserve() and copy()

    - by natersoz
    Greetings, I am trying to perform a copy from one vector (vec1) to another vector (vec2) using the following 2 abbreviated lines of code (full test app follows): vec2.reserve( vec1.size() ); copy(vec1.begin(), vec1.end(), vec2.begin()); While the call to vec2 sets the capacity of vector vec2, the copying of data to vec2 seems to not fill in the values from vec1 to vec2. Replacing the copy() function with calls to push_back() works as expected. What am I missing here? Thanks for your help. vectest.cpp test program followed by resulting output follows. Compiler: gcc 3.4.4 on cygwin. Nat /** * vectest.cpp */ #include <iostream> #include <vector> using namespace std; int main() { vector<int> vec1; vector<int> vec2; vec1.push_back(1); vec1.push_back(2); vec1.push_back(3); vec1.push_back(4); vec1.push_back(5); vec1.push_back(6); vec1.push_back(7); vec2.reserve( vec1.size() ); copy(vec1.begin(), vec1.end(), vec2.begin()); cout << "vec1.size() = " << vec1.size() << endl; cout << "vec1.capacity() = " << vec1.capacity() << endl; cout << "vec1: "; for( vector<int>::const_iterator iter = vec1.begin(); iter < vec1.end(); ++iter ) { cout << *iter << " "; } cout << endl; cout << "vec2.size() = " << vec2.size() << endl; cout << "vec2.capacity() = " << vec2.capacity() << endl; cout << "vec2: "; for( vector<int>::const_iterator iter = vec2.begin(); iter < vec2.end(); ++iter ) { cout << *iter << endl; } cout << endl; } output: vec1.size() = 7 vec1.capacity() = 8 vec1: 1 2 3 4 5 6 7 vec2.size() = 0 vec2.capacity() = 7 vec2:

    Read the article

  • calculating offer period for subscription

    - by TheVillageIdiot
    I'm maintaining a web application which deals with some kind of subscriptions. Users can to renew their subscriptions from 2 months before expiry (not earlier than that). Sometimes user does not renew before expiry and get grace period which is of 3 months. Now he can renew in these 3 months of grace period. Now the problem part. In the previous transactions of renew requests I have to show what was the offer period for that particular request (subscription start and subscription end period if renew was granted). Things are pretty simple if user renews before expiry, but I'm not able to get things straight if there is grace period specially when the subscriptions is expiring in last months of the year. Also there sometimes calculations go haywire when subscription is ending in jan or feb. All this is happening because offer period is not saved with the application anywhere (I don't know why). so if subscription is ending in 20 October 2008 and renew application is submitted in 16 January 2009 (because of grace period) the offer period should be 21 October 2008 to 20 October 2009.

    Read the article

  • determine if intersection of a set with conjunction of two other sets is empty

    - by koen
    For any three given sets A, B and C: is there a way to determine (programmatically) whether there is an element of A that is part of the conjunction of B and C? example: A: all numbers greater than 3 B: all numbers lesser than 7 C: all numbers that equal 5 In this case there is an element in set A, being the number 5, that fits. I'm implementing this as specifications, so this numerical range is just an example. A, B, C could be anything.

    Read the article

  • Chain call in clojure?

    - by Konrad Garus
    I'm trying to implement sieve of Eratosthenes in Clojure. One approach I would like to test is this: Get range (2 3 4 5 6 ... N) For 2 <= i <= N Pass my range through filter that removes multiplies of i For i+1th iteration, use result of the previous filtering I know I could do it with loop/recur, but this is causing stack overflow errors (for some reason tail call optimization is not applied). How can I do it iteratively? I mean invoking N calls to the same routine, passing result of ith iteration to i+1th.

    Read the article

  • Zoom image to pixel level

    - by zaf
    For an art project, one of the things I'll be doing is zooming in on an image to a particular pixel. I've been rubbing my chin and would love some advice on how to proceed. Here are the input parameters: Screen: sw - screen width sh - screen height Image: iw - image width ih - image height Pixel: px - x position of pixel in image py - y position of pixel in image Zoom: zf - zoom factor (0.0 to 1.0) Background colour: bc - background colour to use when screen and image aspect ratios are different Outputs: The zoomed image (no anti-aliasing) The screen position/dimensions of the pixel we are zooming to. When zf is 0 the image must fit the screen with correct aspect ratio. When zf is 1 the selected pixel fits the screen with correct aspect ratio. One idea I had was to use something like povray and move the camera towards a big image texture or some library (e.g. pygame) to do the zooming. Anyone think of something more clever with simple pseudo code? To keep it more simple you can make the image and screen have the same aspect ratio. I can live with that. I'll update with more info as its required.

    Read the article

  • RSA Factorization problem

    - by dada
    At class we found this programming problem, and currently, we have no idea how to solve it. The positive integer n is given. It is known that n = p * q, where p and q are primes, p<=q and |q-k*p|<10^5 for some given positive integer k. You must find p and q. Input: 35 1 121 1 1000730021 9 Output: 5 * 7 11 * 11 10007 * 100003 It's not a homework, we are just trying to solve some interesting problems. If you have some ideas, please post them here so we can try something, thanks.

    Read the article

< Previous Page | 81 82 83 84 85 86 87 88 89 90 91 92  | Next Page >