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  • 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 ?

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  • 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

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  • 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?

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  • 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.

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  • 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); }

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  • 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; }

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  • 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.

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  • 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.

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  • 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!!

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  • CMYK + CMYK = ? CMYK / 2 = ?

    - by Pete
    Suppose there are two colors defined in CMYK: color1 = 30, 40, 50, 60 color2 = 50, 60, 70, 80 If they were to be printed what values would the resulting color have? color_new = min(cyan1 + cyan2, 100), min(magenta1 + magenta2, 100), min(yellow1 + yellow2, 100), min(black1 + black2, 100)? Suppose there is a color defined in CMYK: color = 40, 30, 30, 100 It is possible to print a color at partial intensity, i.e. as a tint. What values would have a 50% tint of that color? color_new = cyan / 2, magenta / 2, yellow / 2, black / 2? I'm asking this to better understand the "tintTransform" function in PDF Reference 1.7, 4.5.5 Special Color Spaces, DeviceN Color Spaces Update: To better clarify: I'm not entirely concerned with human perception or how the CMYK dyies react to the paper. If someone specifies 90% tint which, when printed, looks like full intensity colorant, that's ok. In other words, if I asking how to compute 50% of cmyk(40, 30, 30, 100) I'm asking how to compute the new values, regardless of whether the result looks half-dark or not. Update 2: I'm confused now. I checked this in InDesign and Acrobat. For example Pantone 3005 has CMYK 100, 34, 0, 2, and its 25% tint has CMYK 25, 8.5, 0, 0.5. Does it mean I can "monkey around in a linear way"?

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  • Project euler problem 45

    - by Peter
    Hi, I'm not yet a skilled programmer but I thought this was an interesting problem and I thought I'd give it a go. Triangle, pentagonal, and hexagonal numbers are generated by the following formulae: Triangle T_(n)=n(n+1)/2 1, 3, 6, 10, 15, ... Pentagonal P_(n)=n(3n-1)/2 1, 5, 12, 22, 35, ... Hexagonal H_(n)=n(2n-1) 1, 6, 15, 28, 45, ... It can be verified that T_(285) = P_(165) = H_(143) = 40755. Find the next triangle number that is also pentagonal and hexagonal. Is the task description. I know that Hexagonal numbers are a subset of triangle numbers which means that you only have to find a number where Hn=Pn. But I can't seem to get my code to work. I only know java language which is why I'm having trouble finding a solution on the net womewhere. Anyway hope someone can help. Here's my code public class NextNumber { public NextNumber() { next(); } public void next() { int n = 144; int i = 165; int p = i * (3 * i - 1) / 2; int h = n * (2 * n - 1); while(p!=h) { n++; h = n * (2 * n - 1); if (h == p) { System.out.println("the next triangular number is" + h); } else { while (h > p) { i++; p = i * (3 * i - 1) / 2; } if (h == p) { System.out.println("the next triangular number is" + h); break; } else if (p > h) { System.out.println("bummer"); } } } } } I realize it's probably a very slow and ineffecient code but that doesn't concern me much at this point I only care about finding the next number even if it would take my computer years :) . Peter

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  • Choice of programming language for learning data structures and algorithms

    - by bguiz
    Which programming language would you recommend to learn about data structures and algorithms in? Considering the follwing: Personal experience Language features (pointers, OO, etc) Suitability for learning DS & A concepts I ask because there are some books out there that are programming language-agnostic (written from a Mathematical perspective, and use pseudocode). If I learn from one of these I would like to work out the algorithms in a chosen language. Then, there are other books which introduce DS & A concepts with examples in a particular programming laguage - and I would follow these examples as well. Either way, I have to choose a language, and I would like to stick to one throughout. Which one best fits the bill.

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  • Brackets matching using BIT

    - by amit.codename13
    edit: I was trying to solve a spoj problem. Here is the link to the problem : http://spoj.pl/problems/BRCKTS I can think of two possible data structures for solving the problem one using segment tree and the other using a BIT. I have already implemented the solution using a segment tree. I have read about BIT but i can't figure out how to do a particular thing with it(which i have mentioned below) I am trying to check if brackets are balanced in a given string containing only ('s or )'s. I am using a BIT(Binary indexed tree) for solving the problem. The procedure i am following is as follows: I am taking an array of size equal to the number of characters in the string. I am assigning -1 for ) and 1 for ( to the corresponding array elements. Brackets are balanced in the string only if the following two conditions are true. The cumulative sum of the whole array is zero. Minimum cumulative sum is non negative. i.e the minimum of cumulative sums of all the prefixes of the array is non-negative. Checking condition 1 using a BIT is trivial. I am facing problem in checking condition 2.

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  • 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.

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  • 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.

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  • 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.

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  • I am trying to build a list of limitations of all graph algorithms

    - by Jack
    Single Source shortest Path Dijkstra's - directed and undirected - works only for positive edge weights - cycles ?? Bellman Ford - directed - no cycles should exist All source shortest path Floyd Warshall - no info Minimum Spanning Tree ( no info about edge weights or nature of graph or cycles) Kruskal's Prim's - undirected Baruvka's

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  • MATLAB: Reading floating point numbers and strings from a file

    - by xsound
    I am using the following functions for writing and reading 4098 floating point numbers in MATLAB: Writing: fid = fopen(completepath, 'w'); fprintf(fid, '%1.30f\r\n', y) Reading: data = textread(completepath, '%f', 4098); where y contains 4098 numbers. I now want to write and read 3 strings at the end of this data. How do I read two different datatypes? Please help me. Thanks in advance.

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  • Grouping php array items based on user and created time

    - by Jim
    This is an array of objects showing a user uploading photos: Array ( [12] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:36:41 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photo] => stdClass Object ( [id] => 4181 ) ) [44] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:37:15 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photo] => stdClass Object ( [id] => 4180 ) ) ) However instead of showing: mr smith uploaded one photo mr smith uploaded one photo I'd like to display: mr smith uploaded two photos by grouping similar items, grouping by user ID and them having added them within, let's say 15 minutes of each other. So I'd like to get the array in this sort of shape: Array ( [12] => stdClass Object ( [type] => photo [created] => 2010-05-14 23:36:41 [user] => stdClass Object ( [id] => 760 [username] => mrsmith ) [photos] => Array ( [0] => stdClass Object ( [id] => 4181 ) [1] => stdClass Object ( [id] => 4180 ) ) ) ) preserving the first item of the group and it's created time, and supplementing it with any other groupable photos and then unsetting any items that were grouped (so the final array doesn't have key 44 anymore as it was grouped in with 12). The array contains other actions than just photos, hence the original keys of 12 and 44. I just can't figure out a way to do this efficiently. I used to use MySQL and PHP to do this but am trying to just use pure PHP for caching reasons. Can anyone shed any insights? I thought about going through each item and seeing if I can group it with the previous one in the array but the previous one might not necessarily be relevant or even a photo. I've got total brain freeze :(

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  • 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.

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  • total number of magic square from 9 numbers

    - by Peeyush
    9 numbers need to be arranged in a magic number square. A magic number square is a square of numbers that is arranged such that every row and column has the same sum.(condition for diagonal has been relaxed) For example: 1 2 3 3 2 1 2 2 2 How do we calculate total number of distinct magic square from 9 numbers. Two magic number squares are distinct if they differ in value at one or more positions. For example, there is only one magic number square that can be made of 9 instances of the same number. e.g. for these 9 numbers { 4, 4, 4, 4, 4, 4, 4, 4, 4 }, answer should be 1. Also the complexity should be optimal. Do we need to iterate through all the permutations , discarding if a[0]+a[1]+a[2] %3!=0 such combinations ? moreover how do we remove duplicate magic square?

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