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  • Flag bit computation and detection

    - by Majid
    Hi all, In some code I'm working on I should take care of ten independent parameters which can take one of two values (0 or 1). This creates 2^10 distinct conditions. Some of the conditions never occur and can be left out, but those which do occur are still A LOT and making a switch to handle all cases is insane. I want to use 10 if statements instead of a huge switch. For this I know I should use flag bits, or rather flag bytes as the language is javascript and its easier to work with a 10 byte string with to represent a 10-bit binary. Now, my problem is, I don't know how to implement this. I have seen this used in APIs where multiple-selectable options are exposed with numbers 1, 2, 4, 8, ... , n^(n-1) which are decimal equivalents of 1, 10, 100, 1000, etc. in binary. So if we make call like bar = foo(7), bar will be an object with whatever options the three rightmost flags enable. I can convert the decimal number into binary and in each if statement check to see if the corresponding digit is set or not. But I wonder, is there a way to determine the n-th digit of a decimal number is zero or one in binary form, without actually doing the conversion?

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  • Why do some questions get closed for no reason? [closed]

    - by IVlad
    Recently there was a question asking about generating all subsets of a set using a stack and a queue, which was closed (and now deleted it seems) as not a real question for no good reason, since it didn't fit into any of these conditions: It's difficult to tell what is being asked here. No, it was clear what was being asked. This question is ambiguous, vague, incomplete, or rhetorical and cannot be reasonably answered in its current form. Not ambiguous, not vague, not incomplete, definitely not rhetorical and could easily be answered if one knew the solution. Now, the exact same thing has happened with this question: http://stackoverflow.com/questions/2791982/a-shortest-path-problem-with-superheroes-and-intergalactic-journeys/2793746#2793746 I am interested in hearing a logical argument for why that question is either ambiguous, vague, incomplete, rhetorical or cannot reasonably be answered in its current form. It seems that (the same bunch of) people like to close questions that they think are homework questions, especially when they think people want to be served the solution on a platter, which is also not the case: Any suggestions or ideas of how this problem might be solved would be most welcomed. Most of the time the people asking these questions are very reasonable and appreciate even the most vague idea, yet their question is closed. Let's go further and assume that it IS a homework problem. So what? When I registered here I didn't see any rule that said not to post homework problems, nor do I see such a rule now. What is wrong with posting homework problems that makes people hunt them down with a passion to close them without even reading the entire question body? This site is full of questions asked by people who get paid to know the things they are asking, yet their questions are considered fine. How is solving someone's homework problem worse? In some places (like where I live), computer science is a mandatory high school subject, and not everyone is interested in it. How is helping at least those people worse than doing someone's JOB? Not answering homework questions is fine and it's everyone's choice, but I consider closing them to be an act of power abuse, selfishness, and an insult to the fellow community members who are also interested in a solution or want feedback on their proposed solution. So my questions are: - Why do questions like the above get closed for reasons that do not apply? Why do you close them? Why don't you? - Why doesn't a vote to reopen a question reopen it automatically? Needing 5 votes for a reopen takes too long, and it's not fair because one reopen vote basically cancels out a close vote, making it 4 close votes (or 5 to 1, which is the same as only 4 people wanting to close the question), which isn't enough to close the question. I think a question should only be closed when CloseVotes - ReopenVotes >= 5. I'm hoping this will stay up, but I realize it probably won't. In either case, I think this is worth saying and discussing, since it IS community-related.

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  • Take the intersection of an arbitrary number of lists in python

    - by thepandaatemyface
    Suppose I have a list of lists of elements which are all the same (i'll use ints in this example) [range(100)[::4], range(100)[::3], range(100)[::2], range(100)[::1]] What would be a nice and/or efficient way to take the intersection of these lists (so you would get every element that is in each of the lists)? For the example that would be: [0, 12, 24, 36, 48, 60, 72, 84, 96]

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  • Byte-Pairing for data compression

    - by user1669533
    Question about Byte-Pairing for data compression. If byte pairing converts two byte values to a single byte value, splitting the file in half, then taking a gig file and recusing it 16 times shrinks it to 62,500,000. My question is, is byte-pairing really efficient? Is the creation of a 5,000,000 iteration loop, to be conservative, efficient? I would like some feed back on and some incisive opinions please. Dave, what I read was: "The US patent office no longer grants patents on perpetual motion machines, but has recently granted at least two patents on a mathematically impossible process: compression of truly random data." I was not inferring the Patent Office was actually considering what I am inquiring about. I was merely commenting on the notion of a "mathematically impossible process." If someone has, in some way created a method of having a "single" data byte as a placeholder of 8 individual bytes of data, that would be a consideration for a patent. Now, about the mathematically impossibility of an 8 to 1 compression method, it is not so much a mathematically impossibility, but a series of rules and conditions that can be created. As long as there is the rule of 8 or 16 bit representation of storing data on a medium, there are ways to manipulate data that mirrors current methods, or creation by a new way of thinking.

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  • make tree in scheme

    - by ???
    (define (entry tree) (car tree)) (define (left-branch tree) (cadr tree)) (define (right-branch tree) (caddr tree)) (define (make-tree entry left right) (list entry left right)) (define (mktree order items_list) (cond ((= (length items_list) 1) (make-tree (car items_list) '() '())) (else (insert2 order (car items_list) (mktree order (cdr items_list)))))) (define (insert2 order x t) (cond ((null? t) (make-tree x '() '())) ((order x (entry t)) (make-tree (entry t) (insert2 order x (left-branch t)) (right-branch t))) ((order (entry t) x ) (make-tree (entry t) (left-branch t) (insert2 order x (right-branch t)))) (else t))) The result is: (mktree (lambda (x y) (< x y)) (list 7 3 5 1 9 11)) (11 (9 (1 () (5 (3 () ()) (7 () ()))) ()) ()) But I'm trying to get: (7 (3 (1 () ()) (5 () ())) (9 () (11 () ()))) Where is the problem?

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  • Reverse a singly linked list

    - by Madhan
    I would be wondered if there exists some logic to reverse the linked list using only two pointers. The following is used to reverse the single linked list using three pointers namely p, q, r: struct node { int data; struct node *link; }; void reverse() { struct node *p = first, *q = NULL, *r; while (p != NULL) { r = q; q = p; p = p->link; q->link = r; } q = first; } Is there any other alternate to reverse the linked list? what would be the best logic to reverse a singly linked list, in terms of time complexity?

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  • Graph diffing and versioning tool

    - by hashable
    I am working with a team that edits large DAGs represented as single files. Currently we are unable to work with multiple users concurrently modifying the DAG. Is there a tool (somewhat like the Eclipse SVN plugin) that can do do revision control on the file (manage timestamps/revision stamps) to identify incoming/outgoing/conflicting changes (Node/Link insertion/deletion/modification) and merge changes just like programmers do with source code files? The system should be able to do dependency management also. E.g. an incoming Link must not be accepted when one of the two Nodes is absent. That is, it should not "break" the existing DAG by allowing partial updates. If there is a framework to do this using generic "Node" and "Link" interfaces? Note: I am aware of Protege and its plugins. They currently do not satisfy my requirements.

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  • how to find distinct digit set numbers over a range of integers?

    - by evil.coder
    Suppose i have a unsigned integer, call it low and one another call it high such that highlow. The problem is to find distinct digit set numbers over this range. For example, suppose low is 1 and high is 20 then the answer is 20, because all the numbers in this range are of distinct digit sets. If suppose low is 1 and high is 21, then the answer is 20, because 12 and 21 have same digit set i.e.1, 2. I am not looking for a bruteforce algo., if anyone has a better solution then a usual bruteforce approach, please tell..

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  • Better way to summarize data about stop times?

    - by Vimvq1987
    This question is close to this: http://stackoverflow.com/questions/2947963/find-the-period-of-over-speed Here's my table: Longtitude Latitude Velocity Time 102 401 40 2010-06-01 10:22:34.000 103 403 50 2010-06-01 10:40:00.000 104 405 0 2010-06-01 11:00:03.000 104 405 0 2010-06-01 11:10:05.000 105 406 35 2010-06-01 11:15:30.000 106 403 60 2010-06-01 11:20:00.000 108 404 70 2010-06-01 11:30:05.000 109 405 0 2010-06-01 11:35:00.000 109 405 0 2010-06-01 11:40:00.000 105 407 40 2010-06-01 11:50:00.000 104 406 30 2010-06-01 12:00:00.000 101 409 50 2010-06-01 12:05:30.000 104 405 0 2010-06-01 11:05:30.000 I want to summarize times when vehicle had stopped (velocity = 0), include: it had stopped since "when" to "when" in how much minutes, how many times it stopped and how much time it stopped. I wrote this query to do it: select longtitude, latitude, MIN(time), MAX(time), DATEDIFF(minute, MIN(Time), MAX(time)) as Timespan from table_1 where velocity = 0 group by longtitude,latitude select DATEDIFF(minute, MIN(Time), MAX(time)) as minute into #temp3 from table_1 where velocity = 0 group by longtitude,latitude select COUNT(*) as [number]from #temp select SUM(minute) as [totaltime] from #temp3 drop table #temp This query return: longtitude latitude (No column name) (No column name) Timespan 104 405 2010-06-01 11:00:03.000 2010-06-01 11:10:05.000 10 109 405 2010-06-01 11:35:00.000 2010-06-01 11:40:00.000 5 number 2 totaltime 15 You can see, it works fine, but I really don't like the #temp table. Is there anyway to query this without use a temp table? Thank you.

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  • Storing a bucket of numbers in an efficient data structure

    - by BlitzKrieg
    I have a buckets of numbers e.g. - 1 to 4, 5 to 15, 16 to 21, 22 to 34,.... I have roughly 600,000 such buckets. The range of numbers that fall in each of the bucket varies. I need to store these buckets in a suitable data structure so that the lookups for a number is as fast as possible. So my question is what is the suitable data structure and a sorting mechanism for this type of problem. Thanks in advance

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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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  • Express any number as the sum of 4 prime numbers [Doubts]

    - by WarDoGG
    I was give a problem to express any number as sum of 4 prime numbers. Conditions: Not allowed to use any kind of database. Maximum execution time : 3 seconds Numbers only till 100,000 If the splitting is NOT possible, then return -1 What i did : using the sieve of eratosthenes, i calculated all prime numbers till the specified number. looked up a concept called goldbach conjecture which expresses an even number as the summation of 2 primes. However, i am stuck beyond that. Can anyone help me on this one as to what approach u might take ? The sieve of eratosthenes is taking 2 seconds to count primes till 100,000 :(((

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  • How to scale rotated objects properly in Actionscript 3?

    - by Tom
    This is unfortunately a quite complex issue to explain, so please don't get discouraged by the wall of text - it's there for a reason. ;) I'm working on a transformation manager for flash, written with Actionscript 3. Users can place objects on the screen, for example a rectangle. This rectangle can then be selected and transformed: move, scale or rotate. Because flash by default rotates around the top left point of the object, and I want it to rotate around the center, I created a wrapper setup for each display object (eg. a rectangle). This is how the wrappers are setup: //the position wrapper makes sure that we do get the top left position when we access x and y var positionWrapper:Sprite = new Sprite(); positionWrapper.x = renderObject.x; positionWrapper.y = renderObject.y; //set the render objects location to center at the rotation wrappers top left renderObject.x = 0 - renderObject.width / 2; renderObject.y = 0 - renderObject.height / 2; //now create a rotation wrapper, at the center of the display object var rotationWrapper:Sprite = new Sprite(); rotationWrapper.x = renderObject.width / 2; rotationWrapper.y = renderObject.height / 2; //put the rotation wrapper inside the position wrapper and the render object inside the rotation wrapper positionWrapper.addChild(rotationWrapper); rotationWrapper.addChild(renderObject); Now, the x and y of the object can be accessed and set directly: mainWrapper.x or mainWrapper.y. The rotation can be set and accessed from the child of this main wrapper: mainWrapper.getChildAt(0).rotation. Finally, the width and height of the display object can be retreived and set by getting the child of the rotation wrapper and accessing the display object directly. An example on how I access them: //get wrappers and render object var positionWrapper:Sprite = currentSelection["render"]; var rotationWrapper:Sprite = positionWrapper.getChildAt(0) as Sprite; var renderObject:DisplayObject = rotationWrapper.getChildAt(0); This works perfectly for all initial transformations: moving, scaling and rotating. However, the problem arises when you first rotate an object (eg. 45 degrees) and then scale it. The scaled object is getting out of shape and doesn't scale as it should. This for example happens when you scale to the left. Scaling left is basically adding n width to the object and then reduce the x coord of the position wrapper by n too: renderObject.width -= diffX; positionWrapper.x += diffX; This works when the object is not rotated. However, when it is, the position wrapper won't be rotated as it is a parent of the rotation wrapper. This will make the position wrapper move left horizontally while the width of the object is increased diagonally. I hope this makes any sense, if not, please tell me and I'll try to elaborate more. Now, to the question: should I use a different kind of setup, system or structure? Should I maybe use matrixes, if so, how would you keep a static width/height after rotation? Or how do I fix my current wrapper system for scaling after rotation? Any help is appreciated.

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  • Could I do this blind relative to absolute path conversion (for perforce depot paths) better?

    - by wonderfulthunk
    I need to "blindly" (i.e. without access to the filesystem, in this case the source control server) convert some relative paths to absolute paths. So I'm playing with dotdots and indices. For those that are curious I have a log file produced by someone else's tool that sometimes outputs relative paths, and for performance reasons I don't want to access the source control server where the paths are located to check if they're valid and more easily convert them to their absolute path equivalents. I've gone through a number of (probably foolish) iterations trying to get it to work - mostly a few variations of iterating over the array of folders and trying delete_at(index) and delete_at(index-1) but my index kept incrementing while I was deleting elements of the array out from under myself, which didn't work for cases with multiple dotdots. Any tips on improving it in general or specifically the lack of non-consecutive dotdot support would be welcome. Currently this is working with my limited examples, but I think it could be improved. It can't handle non-consecutive '..' directories, and I am probably doing a lot of wasteful (and error-prone) things that I probably don't need to do because I'm a bit of a hack. I've found a lot of examples of converting other types of relative paths using other languages, but none of them seemed to fit my situation. These are my example paths that I need to convert, from: //depot/foo/../bar/single.c //depot/foo/docs/../../other/double.c //depot/foo/usr/bin/../../../else/more/triple.c to: //depot/bar/single.c //depot/other/double.c //depot/else/more/triple.c And my script: begin paths = File.open(ARGV[0]).readlines puts(paths) new_paths = Array.new paths.each { |path| folders = path.split('/') if ( folders.include?('..') ) num_dotdots = 0 first_dotdot = folders.index('..') last_dotdot = folders.rindex('..') folders.each { |item| if ( item == '..' ) num_dotdots += 1 end } if ( first_dotdot and ( num_dotdots > 0 ) ) # this might be redundant? folders.slice!(first_dotdot - num_dotdots..last_dotdot) # dependent on consecutive dotdots only end end folders.map! { |elem| if ( elem !~ /\n/ ) elem = elem + '/' else elem = elem end } new_paths << folders.to_s } puts(new_paths) end

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  • Echo mysql results in a loop?

    - by Roy D. Porter
    I am using turn.js to make a book. Every div within the 'deathnote' div becomes a new page. <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page </div> //ends book What I am doing is trying to get 3 'content' (content being a name and cause of death) divs onto 1 page, and then generate a new page. So here is what i want: <div id="deathnote"> //starts book <div style="background-image:url(images/coverpage.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"></div> //creates new page <div style="background-image:url(images/paper.jpg);"> //creates new page but leaves it open <div> CONTENT </div> <div> CONTENT </div> <div> CONTENT </div> </div> //ends the page </div> //ends book Seems simple enough, however the content is data from a MySQL DB, so i have to echo it in using PHP. Here is what i have so far <div id="deathnote"> <div style="background-image:url(images/coverpage.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <div style="background-image:url(images/paper.jpg);"></div> <?php $pagecount = 0; $db = new mysqli('localhost', 'username', 'passw', 'DB'); if($db->connect_errno > 0){ die('Unable to connect to database [' . $db->connect_error . ']'); } $sql = <<<SQL SELECT * FROM `TABLE` SQL; if(!$result = $db->query($sql)){ die('There was an error running the query [' . $db->error . ']'); } //IGNORE ALL OF THE GARBAGE ABOVE. IT IS SIMPLE CONNECTING SCRIPT THAT I KNOW WORKS //THE METHOD I AM HAVING TROUBLE WITH IS BELOW $pagecount = 0; while($row = $result->fetch_assoc()){ //GETS THE VALUE (and makes sure it isn't nothing echo '<div style="background-image:url(images/paper.jpg);">'; //THIS OPENS A NEW PAGE while ($pagecount !== 3) { //KEEPS COUNT OF HOW MUCH CONTENT DIVS IS ON THE PAGE while($row = $result->fetch_assoc()){ //START A CONTENT DIV echo '<div class="content"><div class="name">' . $row['victim'] . '</div><div class="cod">' . $row['cod'] . '</div></div>'; //END A CONTENT DIV $pagecount++; //UP THE PAGE COUNT } } $pagecount=0; //PUT IT BACK TO 0 echo '</div>'; //END PAGE } $db->close(); ?> <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE </div> At the moment i seem to be causing and infinite loop so the page won't load. The problem resides within the while loops. Any help is greatly appreciated. Thanks in advance guys. :)

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  • Orbital equations, and power required to run them

    - by Adam Davis
    Due to a discussion on the SO IRC today, I'm curious about orbital mechanics, and The equations needed to solve orbital problems The computing power required to solve complex problems The question in particular is calculating when the Earth will plow into the Sun (or vice versa, depending on the frame of reference). I suspect that all the gravitational pulls within our solar system may need to be calculated, which makes me wonder what type of computer cluster is required, or can this be done on a single box? I don't have the experience to do a back of the napkin test here, but perhaps you do? Also, much thx to Gortok for the original inspiration (see comments).

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  • Inverse relationship of two variables

    - by Jam
    this one is maybe pretty stupid.. Or I am just exhausted or something, but I just cant seem to solve it.. Problem : two variables X and Y, value of Y is dependent on value of X. X can have values ranging from some value to some value (lets say from 0 to 250) and y can have different values (lets say from 0.1 to 1.0 or something..) - but it is inverse relatonship (what I mean is: if value of X is e.g. 250, then value of Y would be 0.1 and when X decreases up to 0, value of Y raises up to 1.0.. So how should I do it? lets say I have function: -- double computeValue (double X) { /computation/ return Y; } Also, is there some easy way to somehow make the scaling of the function not so linear? - For example when X raises, Y decreases slower at first but then more rapidly in the end.. (rly dont know how to say it but I hope you guys got it) Thanks in advance for this stupid question :/

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  • Generate 2D cross-section polygon from 3D mesh

    - by nornagon
    I'm writing a game which uses 3D models to draw a scene (top-down orthographic projection), but a 2D physics engine to calculate response to collisions, etc. I have a few 3D assets for which I'd like to be able to automatically generate a hitbox by 'slicing' the 3D mesh with the X-Y plane and creating a polygon from the resultant edges. Google is failing me on this one (and not much helpful material on SO either). Suggestions?

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  • Mapping Hilbert values to 3D points

    - by Alexander Gladysh
    I have a set of Hilbert values (length from the start of the Hilbert curve to the given point). What is the best way to convert these values to 3D points? Original Hilbert curve was not in 3D, so I guess I have to pick by myself the Hilbert curve rank I need. I do have total curve length though (that is, the maximum value in the set). Perhaps there is an existing implementation? Some library that would allow me to work with Hilbert curve / values? Language does not matter much.

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  • Find the shortest path in a graph which visits certain nodes.

    - by dmd
    I have a undirected graph with about 100 nodes and about 200 edges. One node is labelled 'start', one is 'end', and there's about a dozen labelled 'mustpass'. I need to find the shortest path through this graph that starts at 'start', ends at 'end', and passes through all of the 'mustpass' nodes (in any order). ( http://3e.org/local/maize-graph.png / http://3e.org/local/maize-graph.dot.txt is the graph in question - it represents a corn maze in Lancaster, PA)

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  • flood fill algorithm

    - by user335593
    i want to implement the flood fill algorthm...so that when i get the x and y co-od of a point...it should start flooding from that point and fill till it finds a boundary but it is not filling the entire region...say a pentagon this is the code i am using void setpixel(struct fill fillcolor,int x,int y) { glColor3f(fillcolor.r,fillcolor.g,fillcolor.b); glBegin(GL_POINTS); glVertex2i(x,y); glEnd(); glFlush(); } struct fill getpixcol(int x,int y) { struct fill gotpixel; glReadPixels(x,y,1,1,GL_RGB,GL_UNSIGNED_BYTE,pick_col); gotpixel.r =(float) pick_col[0]/255.0; gotpixel.g =(float) pick_col[1]/255.0; gotpixel.b =(float) pick_col[2]/255.0; return(gotpixel); } void floodFill(int x, int y,struct fill fillcolor,struct fill boundarycolor) { struct fill tmp; // if ((x < 0) || (x >= 500)) return; // if ((y < 0) || (y >= 500)) return; tmp=getpixcol(x,y); while (tmp.r!=boundarycolor.r && tmp.g!=boundarycolor.g && tmp.b!=boundarycolor.b) { setpixel(fillcolor,x,y); setpixel(fillcolor,x+1,y); setpixel(fillcolor,x,y+1); setpixel(fillcolor,x,y-1); setpixel(fillcolor,x-1,y); floodFill(x-1,y+1,fillcolor,boundarycolor); floodFill(x-1,y,fillcolor,boundarycolor); floodFill(x-1,y-1,fillcolor,boundarycolor); floodFill(x,y+1,fillcolor,boundarycolor); floodFill(x,y-1,fillcolor,boundarycolor); floodFill(x+1,y+1,fillcolor,boundarycolor); floodFill(x+1,y,fillcolor,boundarycolor); floodFill(x+1,y-1,fillcolor,boundarycolor); } }

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