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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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  • Need Ontology of software engineering terms

    - by v.rekha
    I'm trying to implement a Question Answering System based on software engineering ontology. This is a [class/university?] project; it will use the java language. Can you please help me locate software engineering ontologies i.e. ontologies [or even taxonomies / folksonomies ?] that include words and concepts found in the domain of Software Engineering. Editor's note: I took the liberty of rewriting this question. It was initially poorly worded and being misunderstood was closed. Maybe it can be reopen, for it is programming related and of interest to some SO contributors. According to this dupe: the poster is looking for an OWL resource: quote, "I need software engineering ontology.owl file."

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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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  • Regular Expression question

    - by Mohammad Kotb
    Hi, In my academic assignment, I want make a regular expression to match a word with the following specifications: word length greater than or equal 1 and less than or equal 8 contains letters, digits, and underscore first digit is a letter only word is not A,X,S,T or PC,SW I tried for this regex but can't continue (My big problem is to make the word not equal to PC and SW) ([a-zA-Z&&[^AXST]])|([a-zA-Z][\w]{0,7}) But in the previous regex I didn't handle the that it is not PC and SW Thanks,

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  • How to draw the "trail" in a maze solving application

    - by snow-spur
    Hello i have designed a maze and i want to draw a path between the cells as the 'person' moves from one cell to the next. So each time i move the cell a line is drawn Also i am using the graphics module The graphics module is an object oriented library Im importing from graphics import* from maze import* my circle which is my cell center = Point(15, 15) c = Circle(center, 12) c.setFill('blue') c.setOutline('yellow') c.draw(win) p1 = Point(c.getCenter().getX(), c.getCenter().getY()) this is my loop if mazez.blockedCount(cloc)> 2: mazez.addDecoration(cloc, "grey") mazez[cloc].deadend = True c.move(-25, 0) p2 = Point(getX(), getY()) line = graphics.Line(p1, p2) cloc.col = cloc.col - 1 Now it says getX not defined every time i press a key is this because of p2???

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  • Asymptotic runtime of list-to-tree function

    - by Deestan
    I have a merge function which takes time O(log n) to combine two trees into one, and a listToTree function which converts an initial list of elements to singleton trees and repeatedly calls merge on each successive pair of trees until only one tree remains. Function signatures and relevant implementations are as follows: merge :: Tree a -> Tree a -> Tree a --// O(log n) where n is size of input trees singleton :: a -> Tree a --// O(1) empty :: Tree a --// O(1) listToTree :: [a] -> Tree a --// Supposedly O(n) listToTree = listToTreeR . (map singleton) listToTreeR :: [Tree a] -> Tree a listToTreeR [] = empty listToTreeR (x:[]) = x listToTreeR xs = listToTreeR (mergePairs xs) mergePairs :: [Tree a] -> [Tree a] mergePairs [] = [] mergePairs (x:[]) = [x] mergePairs (x:y:xs) = merge x y : mergePairs xs This is a slightly simplified version of exercise 3.3 in Purely Functional Data Structures by Chris Okasaki. According to the exercise, I shall now show that listToTree takes O(n) time. Which I can't. :-( There are trivially ceil(log n) recursive calls to listToTreeR, meaning ceil(log n) calls to mergePairs. The running time of mergePairs is dependent on the length of the list, and the sizes of the trees. The length of the list is 2^h-1, and the sizes of the trees are log(n/(2^h)), where h=log n is the first recursive step, and h=1 is the last recursive step. Each call to mergePairs thus takes time (2^h-1) * log(n/(2^h)) I'm having trouble taking this analysis any further. Can anyone give me a hint in the right direction?

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  • C++ Operator overloading - 'recreating the Vector'

    - by Wallter
    I am currently in a collage second level programing course... We are working on operator overloading... to do this we are to rebuild the vector class... I was building the class and found that most of it is based on the [] operator. When I was trying to implement the + operator I run into a weird error that my professor has not seen before (apparently since the class switched IDE's from MinGW to VS express...) (I am using Visual Studio Express 2008 C++ edition...) Vector.h #include <string> #include <iostream> using namespace std; #ifndef _VECTOR_H #define _VECTOR_H const int DEFAULT_VECTOR_SIZE = 5; class Vector { private: int * data; int size; int comp; public: inline Vector (int Comp = 5,int Size = 0) : comp(Comp), size(Size) { if (comp > 0) { data = new int [comp]; } else { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; } } int size_ () const { return size; } int comp_ () const { return comp; } bool push_back (int); bool push_front (int); void expand (); void expand (int); void clear (); const string at (int); int operator[ ](int); Vector& operator+ (Vector&); Vector& operator- (const Vector&); bool operator== (const Vector&); bool operator!= (const Vector&); ~Vector() { delete [] data; } }; ostream& operator<< (ostream&, const Vector&); #endif Vector.cpp #include <iostream> #include <string> #include "Vector.h" using namespace std; const string Vector::at(int i) { this[i]; } void Vector::expand() { expand(size); } void Vector::expand(int n ) { int * newdata = new int [comp * 2]; if (*data != NULL) { for (int i = 0; i <= (comp); i++) { newdata[i] = data[i]; } newdata -= comp; comp += n; delete [] data; *data = *newdata; } else if ( *data == NULL || comp == 0) { data = new int [DEFAULT_VECTOR_SIZE]; comp = DEFAULT_VECTOR_SIZE; size = 0; } } bool Vector::push_back(int n) { if (comp = 0) { expand(); } for (int k = 0; k != 2; k++) { if ( size != comp ){ data[size] = n; size++; return true; } else { expand(); } } return false; } void Vector::clear() { delete [] data; comp = 0; size = 0; } int Vector::operator[] (int place) { return (data[place]); } Vector& Vector::operator+ (Vector& n) { int temp_int = 0; if (size > n.size_() || size == n.size_()) { temp_int = size; } else if (size < n.size_()) { temp_int = n.size_(); } Vector newone(temp_int); int temp_2_int = 0; for ( int j = 0; j <= temp_int && j <= n.size_() && j <= size; j++) { temp_2_int = n[j] + data[j]; newone[j] = temp_2_int; } //////////////////////////////////////////////////////////// return newone; //////////////////////////////////////////////////////////// } ostream& operator<< (ostream& out, const Vector& n) { for (int i = 0; i <= n.size_(); i++) { //////////////////////////////////////////////////////////// out << n[i] << " "; //////////////////////////////////////////////////////////// } return out; } Errors: out << n[i] << " "; error C2678: binary '[' : no operator found which takes a left-hand operand of type 'const Vector' (or there is no acceptable conversion) return newone; error C2106: '=' : left operand must be l-value As stated above, I am a student going into Computer Science as my selected major I would appreciate tips, pointers, and better ways to do stuff :D

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  • Why is my file being cleared if I don't save it?

    - by Kat
    My program is suppose to maintain a collection of Photos in a PhotoAlbum. It begins by reading a folder of photos and adds them to my PhotoAlbum. It then prints a menu that allows the user to list all the photos, add a photo, find a photo, save, and quit the program. Right now if I run my program it will add the 100 photos to the PhotoAlbum, but if I quit the program without saving, it clears the file I am reading from even if I haven't added a photo or done anything to the PhotoAlbum and I'm not sure why. Here is my method for printing to a file: private static void saveFile(PrintWriter writer) { String result; ArrayList<Photo> temp = album.getPhotoAlbum(); for (int i = 0; i < temp.size(); i++){ result = temp.get(i).toString() + "\n"; writer.println(result); } writer.close(); } And where the PrintWriter is instantiated: File file = new File(args[0] + File.separator + "album.dat"); try { PrintWriter fout = new PrintWriter(new FileWriter(file)); fileWriter = fout; } catch (IOException e){ System.out.println("ReadFromFile: Folder " + args[0] + " is not found."); System.exit(0); } And where it is called in my runMenu Method: private static void runMainMenu(Scanner scan) { String input; do { showMainMenu(); input = scan.nextLine().toLowerCase(); switch (input.charAt(0)) { case 'p': System.out.println(album.toString()); break; case 'a': album.addPhoto(readPhoto(scan, t)); break; case 'f': findMenu(scan); break; case 's': saveFile(fileWriter); System.exit(0); break; case 'q': break; default: System.out.println("Invalid entry: " + input.charAt(0)); break; } } while (!input.equalsIgnoreCase("q")); }

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  • How do I check the validity of the Canadian Social Insurance Number in C#?

    - by user518307
    I've been given the assignment to write an algorithm in C# that checks the validity of a Canadian Social Insurance Number (SIN). Here are the steps to validate a SIN. Given an example Number: 123 456 782 Remove the check digit (the last digit): 123456782 Extract the even digits (2,4,6,8th digith): 12345678 Double them: 2 4 6 8 | | | | v v v v 4 8 12 16 Add the digits together: 4+8+1+2+1+6 = 22 Add the Odd placed digits: 1+3+5+7 = 16 Total : 38 Validity Algorithm If the total is a multiple of 10, the check digit should be zero. Otherwise, Subtract the Total from the next highest multiple of 10 (40 in this case) The check digit for this SIN must be equal to the difference of the number and the totals from earlier (in this case, 40-38 = 2; check digit is 2, so the number is valid) I'm lost on how to actually implement this in C#, how do I do this?

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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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  • Theory of computation - Using the pumping lemma for CFLs

    - by Tony
    I'm reviewing my notes for my course on theory of computation and I'm having trouble understanding how to complete a certain proof. Here is the question: A = {0^n 1^m 0^n | n>=1, m>=1} Prove that A is not regular. It's pretty obvious that the pumping lemma has to be used for this. So, we have |vy| = 1 |vxy| <= p (p being the pumping length, = 1) uv^ixy^iz exists in A for all i = 0 Trying to think of the correct string to choose seems a bit iffy for this. I was thinking 0^p 1^q 0^p, but I don't know if I can obscurely make a q, and since there is no bound on u, this could make things unruly.. So, how would one go about this?

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  • Write a program that allows the user to enter a string and then prints the letters of the String sep

    - by WM
    The output is always a String, for example H,E,L,L,O,. How could I limit the commas? I want the commas only between letters, for example H,E,L,L,O. import java.util.Scanner; import java.lang.String; public class forLoop { public static void main(String[] args) { Scanner Scan = new Scanner(System.in); System.out.print("Enter a string: "); String Str1 = Scan.next(); String newString=""; String Str2 =""; for (int i=0; i < Str1.length(); i++) { newString = Str1.charAt(i) + ","; Str2 = Str2 + newString; } System.out.print(Str2); } }

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  • javascript robot

    - by sarah
    hey guys! I need help making this robot game in javascript (notepad++) please HELP! I'm really confused by the functions <html> <head><title>Robot Invasion 2199</title></head> <body style="text-align:center" onload="newGame();"> <h2>Robot Invasion 2199</h2> <div style="text-align:center; background:white; margin-right: auto; margin-left:auto;"> <div style=""> <div style="width: auto; border:solid thin red; text-align:center; margin:10px auto 10px auto; padding:1ex 0ex;font-family: monospace" id="scene"></pre> </div> <div><span id="status"></span></div> <form style="text-align:center"> PUT THE CONTROL PANEL HERE!!! </form> </div> <script type="text/javascript"> // GENERAL SUGGESTIONS ABOUT WRITING THIS PROGRAM: // You should test your program before you've finished writing all of the // functions. The newGame, startLevel, and update functions should be your // first priority since they're all involved in displaying the initial state // of the game board. // // Next, work on putting together the control panel for the game so that you // can begin to interact with it. Your next goal should be to get the move // function working so that everything else can be testable. Note that all nine // of the movement buttons (including the pass button) should call the move // function when they are clicked, just with different parameters. // // All the remaining functions can be completed in pretty much any order, and // you'll see the game gradually improve as you write the functions. // // Just remember to keep your cool when writing this program. There are a // bunch of functions to write, but as long as you stay focused on the function // you're writing, each individual part is not that hard. // These variables specify the number of rows and columns in the game board. // Use these variables instead of hard coding the number of rows and columns // in your loops, etc. // i.e. Write: // for(i = 0; i < NUM_ROWS; i++) ... // not: // for(i = 0; i < 15; i++) ... var NUM_ROWS = 15; var NUM_COLS = 25; // Scene is arguably the most important variable in this whole program. It // should be set up as a two-dimensional array (with NUM_ROWS rows and // NUM_COLS columns). This represents the game board, with the scene[i][j] // representing what's in row i, column j. In particular, the entries should // be: // // "." for empty space // "R" for a robot // "S" for a scrap pile // "H" for the hero var scene; // These variables represent the row and column of the hero's location, // respectively. These are more of a conveniece so you don't have to search // for the "H" in the scene array when you need to know where the hero is. var heroRow; var heroCol; // These variables keep track of various aspects of the gameplay. // score is just the number of robots destroyed. // screwdrivers is the number of sonic screwdriver charges left. // fastTeleports is the number of fast teleports remaining. // level is the current level number. // Be sure to reset all of these when a new game starts, and update them at the // appropriate times. var score; var screwdrivers; var fastTeleports; var level; // This function should use a sonic screwdriver if there are still charges // left. The sonic screwdriver turns any robot that is in one of the eight // squares immediately adjacent to the hero into scrap. If there are no charges // left, then this function should instead pop up a dialog box with the message // "Out of sonic screwdrivers!". As with any function that alters the game's // state, this function should call the update function when it has finished. // // Your "Sonic Screwdriver" button should call this function directly. function screwdriver() { // WRITE THIS FUNCTION } // This function should move the hero to a randomly selected location if there // are still fast teleports left. This function MUST NOT move the hero on to // a square that is already occupied by a robot or a scrap pile, although it // can move the hero next to a robot. The number of fast teleports should also // be decreased by one. If there are no fast teleports left, this function // should just pop up a message box saying so. As with any function that alters // the game's state, this function should call the update function when it has // finished. // // HINT: Have a loop that keeps trying random spots until a valid one is found. // HINT: Use the validPosition function to tell if a spot is valid // // Your "Fast Teleport" button s

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  • Spatial domain to frequency domain

    - by John Elway
    I know about Fourier Transforms, but I don't know how to apply it here, and I think that is over the top. I gave my ideas of the responses, but I really don't know what I'm looking for... Supposed that you form a low-pass spatial filter h(x,y) that averages all the eight immediate neighbors of a pixel (x,y) but excludes itself. a. Find the equivalent frequency domain filter H(u,v): My answer is to (a): 1/8*H(u-1, v-1) + 1/8*H(u-1, v) + 1/8*H(u-1, v+1) + 1/8*H(u, v-1) + 0 + 1/8*H(u, v+1) + 1/8*H(u+1, v-1) + 1/8*H(u+1, v) + 1/8*H(u-1, v-1) is this the frequency domain? b. Show that your result is again a low-pass filter. does this have to do with the coefficients being positive?

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  • javascript too much recursion?

    - by Ken
    Hi, I'm trying to make a script that automatically starts uploading after the data has been enter in the database(I need the autoId that the database makes to upload the file). When I run the javascript the scripts runs the php file but it fails calling the other php to upload the file. too much recursion setTimeout(testIfToegevoegd(),500); the script that gives the error send("/projects/backend/nieuwDeeltaak.php",'deeltaakNaam='+f.deeltaaknaam.value+'&beschrijving='+ f.beschrijving.value+'&startDatum='+f.startDatum.value+'&eindDatum='+f.eindDatum.value +'&deeltaakLeider='+f.leiderID.value+'&projectID='+f.projectID.value,id); function testIfToegevoegd(){ if(document.getElementById('resultaat').innerHTML == "<b>De deeltaak werd toegevoegd</b>"){ //stop met testen + upload file document.getElementById('nieuwDeeltaak').target = 'upload_target'; document.forms["nieuwDeeltaak"].submit() }else{ setTimeout(testIfToegevoegd(),500); } } testIfToegevoegd(); sorry for the dutch names we have to use them it is a school project. when I click the button that calls all this for a second time (after the error) it works fine.

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  • Dynamic programming: Largest diamond(rhombus) block

    - by Darksody
    I have a small program to do in Java. I have a 2D array filled with 0 and 1, and I must find the largest rhombus (as in square rotated by 90 degrees) and their numbers. Example: 0 1 0 0 0 1 1 0 1 1 1 0 1 0 1 1 1 1 0 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 1 1 Result: 1 1 1 1 1 1 1 1 1 1 1 1 1 The problem is similar to this SO question. If you have any idea, post it here.

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  • Validating an integer or String without try-catch

    - by Phil
    Ok, I'm lost. I am required to figure out how to validate an integer and String, but for some stupid reason, I can't use the Try-Catch method. I know this is the easiest way and so all the solutions on the internet are using it. I'm writing in Java. The deal is this, I need someone to put in an numerical ID and String name. If either one of the two inputs are invalid I must tell them they made a mistake. Can someone help me?

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  • Problem in finalizing the link list in C#

    - by Yasin
    My code was almost finished that a maddening bug came up! when i nullify the last node to finalize the link list , it actually dumps all the links and make the first node link Null ! when i trace it , its working totally fine creating the list in the loop but after the loop is done that happens and it destroys the rest of the link by doing so, and i don't understand why something so obvious is becoming problematic! (last line) struct poly { public int coef; public int pow; public poly* link;} ; poly* start ; poly* start2; poly* p; poly* second; poly* result; poly* ptr; poly* start3; poly* q; poly* q2; private void button1_Click(object sender, EventArgs e) { string holder = ""; IntPtr newP = Marshal.AllocHGlobal(sizeof(poly)); q = (poly*)newP.ToPointer(); start = q; int i = 0; while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } q->coef = int.Parse(holder); i++; holder = ""; while (this.textBox1.Text[i] != ';') { holder += this.textBox1.Text[i]; i++; } q->pow = int.Parse(holder); holder = ""; p = start; //creation of the first node finished! i++; for (; i < this.textBox1.Text.Length; i++) { newP = Marshal.AllocHGlobal(sizeof(poly)); poly* test = (poly*)newP.ToPointer(); while (this.textBox1.Text[i] != ',') { holder += this.textBox1.Text[i]; i++; } test->coef = int.Parse(holder); holder = ""; i++; while (this.textBox1.Text[i] != ';' && i < this.textBox1.Text.Length - 1) { holder += this.textBox1.Text[i]; if (i < this.textBox1.Text.Length - 1) i++; } test->pow = int.Parse(holder); holder = ""; p->link = test; //the addresses are correct and the list is complete } p->link = null; //only the first node exists now with a null link! }

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  • Log 2 N generic comparison tree

    - by Morano88
    Hey! I'm working on an algorithm for Redundant Binary Representation (RBR) where every two bits represent a digit. I designed the comparator that takes 4 bits and gives out 2 bits. I want to make the comparison in log 2 n so If I have X and Y .. I compare every 2 bits of X with every 2 bits of Y. This is smooth if the number of bits of X or Y equals n where (n = 2^X) i.e n = 2,4,8,16,32,... etc. Like this : However, If my input let us say is 6 or 10 .. then it becomes not smooth and I have to take into account some odd situations like this : I have a shallow experience in algorithms .. is there a generic way to do it .. so at the end I get only 2 bits no matter what the input is ? I just need hints or pseudo-code. If my question is not appropriate here .. so feel free to flag it or tell me to remove it. I'm using VHDL by the way!

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