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  • Trying to sentinel loop this program.

    - by roger34
    Okay, I spent all this time making this for class but I have one thing that I can't quite get: I need this to sentinel loop continuously (exiting upon entering x) so that the System.out.println("What type of Employee? Enter 'o' for Office " + "Clerical, 'f' for Factory, or 's' for Saleperson. Enter 'x' to exit." ); line comes back up after they enter the first round of information. Also, I can't leave this up long on the (very) off chance a classmate might see this and steal the code. Full code following: import java.util.Scanner; public class Project1 { public static void main (String args[]){ Scanner inp = new Scanner( System.in ); double totalPay; System.out.println("What type of Employee? Enter 'o' for Office " + "Clerical, 'f' for Factory, or 's' for Saleperson. Enter 'x' to exit." ); String response= inp.nextLine(); while (!response.toLowerCase().equals("o")&&!response.toLowerCase().equals("f") &&!response.toLowerCase().equals("s")&&!response.toLowerCase().equals("x")){ System.out.print("\nInvalid selection,please enter your choice again:\n"); response=inp.nextLine(); } char choice = response.toLowerCase().charAt( 0 ); switch (choice){ case 'o': System.out.println("Enter your hourly rate:"); double officeRate=inp.nextDouble(); System.out.println("Enter the number of hours worked:"); double officeHours=inp.nextDouble(); totalPay = officeCalc(officeRate,officeHours); taxCalc(totalPay); break; case 'f': System.out.println("How many Widgets did you produce during the week?"); double widgets=inp.nextDouble(); totalPay=factoryCalc(widgets); taxCalc(totalPay); break; case 's': System.out.println("What were your total sales for the week?"); double totalSales=inp.nextDouble(); totalPay=salesCalc(totalSales); taxCalc(totalPay); break; } } public static double taxCalc(double totalPay){ double federal=totalPay*.22; double state =totalPay*.055; double netPay = totalPay - federal - state; federal =federal*Math.pow(10,2); federal =Math.round(federal); federal= federal/Math.pow(10,2); state =state*Math.pow(10,2); state =Math.round(state); state= state/Math.pow(10,2); totalPay =totalPay*Math.pow(10,2); totalPay =Math.round(totalPay); totalPay= totalPay/Math.pow(10,2); netPay =netPay*Math.pow(10,2); netPay =Math.round(netPay); netPay= netPay/Math.pow(10,2); System.out.printf("\nTotal Pay \t: %1$.2f.\n", totalPay); System.out.printf("State W/H \t: %1$.2f.\n", state); System.out.printf("Federal W/H : %1$.2f.\n", federal); System.out.printf("Net Pay \t: %1$.2f.\n", netPay); return totalPay; } public static double officeCalc(double officeRate,double officeHours){ double overtime=0; if (officeHours>=40) overtime = officeHours-40; else overtime = 0; if (officeHours >= 40) officeHours = 40; double otRate = officeRate * 1.5; double totalPay= (officeRate * officeHours) + (otRate*overtime); return totalPay; } public static double factoryCalc(double widgets){ double totalPay=widgets*.35 +300; return totalPay; } public static double salesCalc(double totalSales){ double totalPay = totalSales * .05 + 500; return totalPay; } }

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  • Tricky Big-O complexity

    - by timeNomad
    public void foo (int n, int m) { int i = m; while (i > 100) i = i/3; for (int k=i ; k>=0; k--) { for (int j=1; j<n; j*=2) System.out.print(k + "\t" + j); System.out.println(); } } I figured the complexity would be O(logn). That is as a product of the inner loop, the outer loop -- will never be executed more than 100 times, so it can be omitted. What I'm not sure about is the while clause, should it be incorporated into the Big-O complexity? For very large i values it could make an impact, or arithmetic operations, doesn't matter on what scale, count as basic operations and can be omitted?

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  • Context Free Language Question (Pumping Lemma)

    - by Maixy
    I know this isn't directly related to programming, but I was wondering if anyone know how to apply the pumping lemma to the following proof: Show that L={(a^n)(b^n)(c^m) : n!=m} is not a context free language I'm pretty confident with applying pumping lemmas, but this one is really irking me. What do you think?

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  • substitute in a nested list (prolog)

    - by linda
    /* substitute(X,Y,Xs,Ys) is true if the list Ys is the result of substituting Y for all occurrences of X in the list Xs. This is what I have so far: subs(_,_,[],[]). subs(X,Y,[X|L1],[Y|L2]):- subs(X,Y,L1,L2). subs(X,Y,[H|L1],[H|L2]):- X\=H, not(H=[_|_]), subs(X,Y,L1,L2). subs(X,Y,[H|_],[L2]):- X\=H, H=[_|_], subs(X,Y,H,L2). My code works except it omits the elements following the nested list. For example: ?- subs(a,b,[a,[a,c],a],Z). Z = [b, [b, c]] . What should I add to this program?

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  • SwingWorker in Java (beginner question)

    - by Malachi
    I am relatively new to multi-threading and want to execute a background task using a Swingworker thread - the method that is called does not actually return anything but I would like to be notified when it has completed. The code I have so far doesn't appear to be working: private void crawl(ActionEvent evt) { try { SwingWorker<Void, Void> crawler = new SwingWorker<Void, Void>() { @Override protected Void doInBackground() throws Exception { Discoverer discover = new Discoverer(); discover.crawl(); return null; } @Override protected void done() { JOptionPane.showMessageDialog(jfThis, "Finished Crawling", "Success", JOptionPane.INFORMATION_MESSAGE); } }; crawler.execute(); } catch (Exception ex) { JOptionPane.showMessageDialog(this, ex.getMessage(), "Exception", JOptionPane.ERROR_MESSAGE); } } Any feedback/advice would be greatly appreciated as multi-threading is a big area of programming that I am weak in.

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  • Linked List Sorting with Strings In C

    - by user308583
    I have a struct, with a Name and a single Node called nextName It's a Singly Linked list, and my task is to create the list, based on alphabetical order of the strings. So iff i enter Joe Zolt and Arthur i should get my list structured as Joe Than Joe Zolt Than Arthur Joe Zolt I'm having trouble implementing the correct Algorithm, which would put the pointers in the right order. This is What I have as of Now. Temp would be the name the user just entered and is trying to put into the list, namebox is just a copy of my root, being the whole list if(temp != NULL) { struct node* namebox = root; while (namebox!=NULL && (strcmp((namebox)->name,temp->name) <= 0)) { namebox = namebox->nextName; printf("here"); } temp->nextName = namebox; namebox = temp; root = namebox; This Works right now, if i enter names like CCC BBB than AAA I Get Back AAA BBB CCC when i print But if i put AAA BBB CCC , When i print i only get CCC, it cuts the previous off.

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  • Convert VB.NET code to C#

    - by Sorush Rabiee
    Hi people, I have three projects written with VB.NET (2005) and have to convert them to C# code. (I know that i don't need to convert codes of .net languages at all). I have no time to rewrite them, need a tool or script to convert. Note: they are console applications.

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  • Finding the heaviest length-constrained path in a weighted Binary Tree

    - by Hristo
    UPDATE I worked out an algorithm that I think runs in O(n*k) running time. Below is the pseudo-code: routine heaviestKPath( T, k ) // create 2D matrix with n rows and k columns with each element = -8 // we make it size k+1 because the 0th column must be all 0s for a later // function to work properly and simplicity in our algorithm matrix = new array[ T.getVertexCount() ][ k + 1 ] (-8); // set all elements in the first column of this matrix = 0 matrix[ n ][ 0 ] = 0; // fill our matrix by traversing the tree traverseToFillMatrix( T.root, k ); // consider a path that would arc over a node globalMaxWeight = -8; findArcs( T.root, k ); return globalMaxWeight end routine // node = the current node; k = the path length; node.lc = node’s left child; // node.rc = node’s right child; node.idx = node’s index (row) in the matrix; // node.lc.wt/node.rc.wt = weight of the edge to left/right child; routine traverseToFillMatrix( node, k ) if (node == null) return; traverseToFillMatrix(node.lc, k ); // recurse left traverseToFillMatrix(node.rc, k ); // recurse right // in the case that a left/right child doesn’t exist, or both, // let’s assume the code is smart enough to handle these cases matrix[ node.idx ][ 1 ] = max( node.lc.wt, node.rc.wt ); for i = 2 to k { // max returns the heavier of the 2 paths matrix[node.idx][i] = max( matrix[node.lc.idx][i-1] + node.lc.wt, matrix[node.rc.idx][i-1] + node.rc.wt); } end routine // node = the current node, k = the path length routine findArcs( node, k ) if (node == null) return; nodeMax = matrix[node.idx][k]; longPath = path[node.idx][k]; i = 1; j = k-1; while ( i+j == k AND i < k ) { left = node.lc.wt + matrix[node.lc.idx][i-1]; right = node.rc.wt + matrix[node.rc.idx][j-1]; if ( left + right > nodeMax ) { nodeMax = left + right; } i++; j--; } // if this node’s max weight is larger than the global max weight, update if ( globalMaxWeight < nodeMax ) { globalMaxWeight = nodeMax; } findArcs( node.lc, k ); // recurse left findArcs( node.rc, k ); // recurse right end routine Let me know what you think. Feedback is welcome. I think have come up with two naive algorithms that find the heaviest length-constrained path in a weighted Binary Tree. Firstly, the description of the algorithm is as follows: given an n-vertex Binary Tree with weighted edges and some value k, find the heaviest path of length k. For both algorithms, I'll need a reference to all vertices so I'll just do a simple traversal of the Tree to have a reference to all vertices, with each vertex having a reference to its left, right, and parent nodes in the tree. Algorithm 1 For this algorithm, I'm basically planning on running DFS from each node in the Tree, with consideration to the fixed path length. In addition, since the path I'm looking for has the potential of going from left subtree to root to right subtree, I will have to consider 3 choices at each node. But this will result in a O(n*3^k) algorithm and I don't like that. Algorithm 2 I'm essentially thinking about using a modified version of Dijkstra's Algorithm in order to consider a fixed path length. Since I'm looking for heaviest and Dijkstra's Algorithm finds the lightest, I'm planning on negating all edge weights before starting the traversal. Actually... this doesn't make sense since I'd have to run Dijkstra's on each node and that doesn't seem very efficient much better than the above algorithm. So I guess my main questions are several. Firstly, do the algorithms I've described above solve the problem at hand? I'm not totally certain the Dijkstra's version will work as Dijkstra's is meant for positive edge values. Now, I am sure there exist more clever/efficient algorithms for this... what is a better algorithm? I've read about "Using spine decompositions to efficiently solve the length-constrained heaviest path problem for trees" but that is really complicated and I don't understand it at all. Are there other algorithms that tackle this problem, maybe not as efficiently as spine decomposition but easier to understand? Thanks.

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  • JOptionPane opening another JFrame

    - by mike_hornbeck
    So I'm continuing my fight with this : http://stackoverflow.com/questions/2923545/creating-java-dialogs/2926126 task. Now my JOptionPane opens new window with envelope overfiew, but I can't change size of this window. Also I wanted to have sender's data in upper left corner, and receiver's data in bottom right. How can I achieve that ? There is also issue with OptionPane itself. After I click 'OK' it opens small window in the upper left corner of the screen. What is this and why it's appearing ? My code: import java.awt.*; import java.awt.Font; import javax.swing.*; public class Main extends JFrame { private static JTextField nameField = new JTextField(20); private static JTextField surnameField = new JTextField(); private static JTextField addr1Field = new JTextField(); private static JTextField addr2Field = new JTextField(); private static JComboBox sizes = new JComboBox(new String[] { "small", "medium", "large", "extra-large" }); public Main(){ JPanel mainPanel = new JPanel(); mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); getContentPane().add(mainPanel); JPanel addrPanel = new JPanel(new GridLayout(0, 1)); addrPanel.setBorder(BorderFactory.createTitledBorder("Receiver")); addrPanel.add(new JLabel("Name")); addrPanel.add(nameField); addrPanel.add(new JLabel("Surname")); addrPanel.add(surnameField); addrPanel.add(new JLabel("Address 1")); addrPanel.add(addr1Field); addrPanel.add(new JLabel("Address 2")); addrPanel.add(addr2Field); mainPanel.add(addrPanel); mainPanel.add(new JLabel(" ")); mainPanel.add(sizes); String[] buttons = { "OK", "Cancel"}; int c = JOptionPane.showOptionDialog( null, mainPanel, "My Panel", JOptionPane.DEFAULT_OPTION, JOptionPane.PLAIN_MESSAGE, null, buttons, buttons[0] ); if(c ==0){ new Envelope(nameField.getText(), surnameField.getText(), addr1Field.getText() , addr2Field.getText(), sizes.getSelectedIndex()); } setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); pack(); setVisible(true); } public static void main(String[] args) { new Main(); } } class Envelope extends JFrame { private final int SMALL=0; private final int MEDIUM=1; private final int LARGE=2; private final int XLARGE=3; public Envelope(String n, String s, String a1, String a2, int i){ Container content = getContentPane(); JPanel mainPanel = new JPanel(); mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); mainPanel.add(new JLabel("John Doe")); mainPanel.add(new JLabel("FooBar str 14")); mainPanel.add(new JLabel("Newark, 45-99")); JPanel dataPanel = new JPanel(); dataPanel.setFont(new Font("sansserif", Font.PLAIN, 32)); //set size from i mainPanel.setLayout(new BoxLayout(mainPanel, BoxLayout.Y_AXIS)); mainPanel.setBackground(Color.ORANGE); mainPanel.add(new JLabel("Mr "+n+" "+s)); mainPanel.add(new JLabel(a1)); mainPanel.add(new JLabel(a2)); content.setSize(450, 600); content.setBackground(Color.ORANGE); content.add(mainPanel, BorderLayout.NORTH); content.add(dataPanel, BorderLayout.SOUTH); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); pack(); setVisible(true); } }

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  • Intel IA-32 Assembly

    - by Kay
    I'm having a bit of difficulty converting the following java code into Intel IA-32 Assembly: class Person() { char name [8]; int age; void printName() {...} static void printAdults(Person [] list) { for(int k = 0; k < 100; k++){ if (list[k].age >= 18) { list[k].printName(); } } } } My attempt is: Person: push ebp; save callers ebp mov ebp, esp; setup new ebp push esi; esi will hold name push ebx; ebx will hold list push ecx; ecx will hold k init: mov esi, [ebp + 8]; mov ebx, [ebp + 12]; mov ecx, 0; k=0 forloop: cmp ecx, 100; jge end; if k>= 100 then break forloop cmp [ebx + 4 * ecx], 18 ; jl auxloop; if list[k].age < 18 then go to auxloop jmp printName; printName: auxloop: inc ecx; jmp forloop; end: pop ecx; pop ebx; pop esi; pop ebp; Is my code correct? NOTE: I'm not allowed to use global variables.

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  • A[i] * A[j] = k in O(nlog(n))

    - by gleb-pendler
    A is an Array of n positive int numbers k given int Algorithm should find if there is a pair of numbers which product gives the result a. A[i] * A[j] = k b. A[i] = A[j] + k if there is such a couple the algorithm should return thier index. thanks in advance.

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  • Limit Connections with semaphores

    - by Robert
    I'm trying to limit the number of connections my server will accept using semaphores, but when running, my code doesn't seem to make this restriction - am I using the semaphore correctly? eg. I have hardcoded the number of permit as 2, but I can connect an unlimited number of clients... public class EServer implements Runnable { private ServerSocket serverSocket; private int numberofConnections = 0; private Semaphore sem = new Semaphore(2); private volatile boolean keepProcessing = true; public EServer(int port) throws IOException { serverSocket = new ServerSocket(port); } @Override public void run() { while (keepProcessing) { try { sem.acquire(); Socket socket = serverSocket.accept(); process(socket, getNextConnectionNumber()); } catch (Exception e) { } finally { sem.release(); } } closeIgnoringException(serverSocket); } private synchronized int getNextConnectionNumber() { return ++numberofConnections; } // processing related methods }

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  • How to embed html table into the body of email

    - by Michael Mao
    Hi all: I am sending info to target email via PHP native mail() method right now. Everything else works fine but the table part troubles me the most. See sample output : Dear Michael Mao : Thank you for purchasing flight tickets with us, here is your receipt : Your tickets will be delivered by mail to the following address : Street Address 1 : sdfsdafsadf sdf Street Address 2 : N/A City : Sydney State : nsw Postcode : 2 Country : Australia Credit Card Number : *************1234 Your purchase details are recorded as : <table><tr><th class="delete">del?</th><th class="from_city">from</th><th class="to_city">to</th><th class="quantity">qty</th><th class="price">unit price</th><th class="price">total price</th></tr><tr class="evenrow" id="Sydney-Lima"><td><input name="isDeleting" type="checkbox"></td><td>Sydney</td><td>Lima</td><td>1</td><td>1030.00</td><td>1030</td></tr><tr class="oddrow" id="Sydney-Perth"><td><input name="isDeleting" type="checkbox"></td><td>Sydney</td><td>Perth</td><td>3</td><td>340.00</td><td>1020</td></tr><tr class="totalprice"><td colspan="5">Grand Total Price</td><td id="grandtotal">2050</td></tr></table> The source of table is directly taken from a webpage, exactly as the same. However, Gmail, Hotmail and most of other emails will ignore to render this as a table. So I am wondering, without using Outlook or other email sending agent software, how could I craft a embedded table for the PHP mail() method to send? Current code snippet corresponds to table generation : $purchaseinfo = $_POST["purchaseinfo"]; //if html tags are not to be filtered in the body of email $stringBuilder .= "<table>" .stripslashes($purchaseinfo) ."</table>"; //must send json response back to caller ajax request if(mail($email, 'Your purchase information on www.hardlyworldtravel.com', $emailbody, $headers)) echo json_encode(array("feedback"=>"successful")); else echo json_encode(array("feedback"=>"error")); Any hints and suggestions are welcomed, thanks a lot in advance.

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  • Constructing a Binary Tree from its traversals

    - by user991710
    I'm trying to construct a binary tree (unbalanced), given its traversals. I'm currently doing preorder + inorder but when I figure this out postorder will be no issue at all. I realize there are some question on the topic already but none of them seemed to answer my question. I've got a recursive method that takes the Preorder and the Inorder of a binary tree to reconstruct it, but is for some reason failing to link the root node with the subsequent children. Note: I don't want a solution. I've been trying to figure this out for a few hours now and even jotted down the recursion on paper and everything seems fine... so I must be missing something subtle. Here's the code: public static <T> BinaryNode<T> prePlusIn( T[] pre, T[] in) { if(pre.length != in.length) throw new IllegalArgumentException(); BinaryNode<T> base = new BinaryNode(); base.element = pre[0]; // * Get root from the preorder traversal. int indexOfRoot = 0; if(pre.length == 0 && in.length == 0) return null; if(pre.length == 1 && in.length == 1 && pre[0].equals(in[0])) return base; // * If both arrays are of size 1, element is a leaf. for(int i = 0; i < in.length -1; i++){ if(in[i].equals(base.element)){ // * Get the index of the root indexOfRoot = i; // in the inorder traversal. break; } // * If we cannot, the tree cannot be constructed as the traversals differ. else throw new IllegalArgumentException(); } // * Now, we recursively set the left and right subtrees of // the above "base" root node to whatever the new preorder // and inorder traversals end up constructing. T[] preleft = Arrays.copyOfRange(pre, 1, indexOfRoot + 1); T[] preright = Arrays.copyOfRange(pre, indexOfRoot + 1, pre.length); T[] inleft = Arrays.copyOfRange(in, 0, indexOfRoot); T[] inright = Arrays.copyOfRange(in, indexOfRoot + 1, in.length); base.left = prePlusIn( preleft, inleft); // * Construct left subtree. base.right = prePlusIn( preright, inright); // * Construc right subtree. return base; // * Return fully constructed tree } Basically, I construct additional arrays that house the pre- and inorder traversals of the left and right subtree (this seems terribly inefficient but I could not think of a better way with no helpers methods). Any ideas would be quite appreciated. Side note: While debugging it seems that the root note never receives the connections to the additional nodes (they remain null). From what I can see though, that should not happen... EDIT: To clarify, the method is throwing the IllegalArgumentException @ line 21 (else branch of the for loop, which should only be thrown if the traversals contain different elements.

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  • class header+ implementation

    - by igor
    what am I doing wrong here? I keep on getting a compilation error when I try to run this in codelab (turings craft) Instructions: Write the implementation (.cpp file) of the GasTank class of the previous exercise. The full specification of the class is: A data member named amount of type double. A constructor that no parameters. The constructor initializes the data member amount to 0. A function named addGas that accepts a parameter of type double . The value of the amount instance variable is increased by the value of the parameter. A function named useGas that accepts a parameter of type double . The value of the amount data member is decreased by the value of the parameter. A function named getGasLevel that accepts no parameters. getGasLevel returns the value of the amount data member. class GasTank{ double amount; GasTank(); void addGas(double); void useGas(double); double getGasLevel();}; GasTank::GasTank(){ amount=0;} double GasTank::addGas(double a){ amount+=a;} double GasTank::useGas(double a){ amount+=a;} double GasTank::getGasLevel(){ return amount;}

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  • Multiple actions upon a case statement in Haskell

    - by Schroedinger
    One last question for the evening, I'm building the main input function of my Haskell program and I have to check for the args that are brought in so I use args <- getArgs case length args of 0 -> putStrLn "No Arguments, exiting" otherwise -> { other methods here} Is there an intelligent way of setting up other methods, or is it in my best interest to write a function that the other case is thrown to within the main? Or is there an even better solution to the issue of cases. I've just got to take in one name.

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  • Which web Tier Framework for a public commercial website with heavy load ?

    - by Maxime ARNSTAMM
    Hello everyone, As a part of an enterprise architecture exercise, i need to find a java-based framework filling these constraints : heavy (i think) load : 5000 concurrent connections widely known : can't be too exotic, the contractors would be too high priced. relatively easy to use : developpement time must be reasonnable must be as compliant as possible with the css/html layout produced by a designer Must look like "web 2.0" from the marketing point of view. What i learned from my limited experience is : jsf : 1, don't know. 2, 3 ok. 4 not ok (at least not without huge effort) wicket : 1, not really. 2, 3 and 4 ok. gwt : 1, don't know. 2, 3 ok. 4 not ok (but more ok than jsf) others : not really "web 2.0" or not really known I'm really junior, so my ideas about those frameworks are probably wrong, that's why i come to you, stackoverflowees. Thanks for helping :)

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