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  • counting twice in a query, once using restrictions

    - by Andrew Heath
    Given the following tables: Table1 [class] [child] math boy1 math boy2 math boy3 art boy1 Table2 [child] [glasses] boy1 yes boy2 yes boy3 no If I want to query for number of children per class, I'd do this: SELECT class, COUNT(child) FROM Table1 GROUP BY class and if I wanted to query for number of children per class wearing glasses, I'd do this: SELECT Table1.class, COUNT(table1.child) FROM Table1 LEFT JOIN Table2 ON Table1.child=Table2.child WHERE Table2.glasses='yes' GROUP BY Table1.class but what I really want to do is: SELECT class, COUNT(child), COUNT(child wearing glasses) and frankly I have no idea how to do that in only one query. help?

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  • How to sort a key of a map

    - by Tsuna Sawada
    How to sort (any kind of sorting) a key of a map(treemap or hashmap) i have a problem and it goes like this. i have a map that has a key of 27527-683, 27525-1179, 27525-1571, 27525-1813, 27525-4911, 27526-1303, 27526-3641, 27525-3989, 27525-4083, 27525-4670, 27526-4102, 27526-558, 27527-2411, 27527-4342 this is the list of keys and the value for each of the key is a list. now, how can i sort this key in ascending order by number. ex. if i want to sort : 1,2,11,20,31,3,10 i want to have as output is : 1,2,3,10,11,20,31 but when i use the autosort of treemap the output goes : 1,10,11,2,20,3,31 how can i sort it in ascending order by numeric? please help me. i can't think of anymore ways because this is my first time handling map and list

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  • Sort Grid Columns of mixed type in EXTJS Grid

    - by Amit
    Hello, I want to sort the extjs columns, I have the column type as float and from the server side i am getting values which can contain "-" value , now what happens the grid is displaying me the NaN value instead of - and the sort is not working anymore. My requirement is to create a custom sort which can sort first based on number and then sort based on string. Thanks to suggest as renderer also not works for me. My Json String is: {metaData:{"totalProperty":"total", "root":"records","fields":[{"header":"Part Number##false","name":"XJE010^VT-007!0","type":"string"},{"header":"Marketing Status##false","name":"STP716^VT-007!0","type":"string"},{"header":"Package##false","name":"XJE016^VT-007!0","type":"string"},{"header":"Automotive Grade##false","name":"STP472^VT-007!0","type":"string"},{"header":"VDSS##false","name":"XJG810^VT-007!0","type":"float"},{"header":"Drain Current (Dc)(I_D) % (A)##false","name":"XJG273^VT-006!0","type":"float"},{"header":"RDS(on) (@VGS=10V) % (&#937;)##false","name":"XJG640^VT-006!3","type":"float"},{"header":"Features##false","name":"GNP023^VT-007!0","type":"string"},{"header":"RDS(on) (@4.5 or 5V) % (&#937;)##false","name":"XJG640^VT-006!6","type":"float"},{"header":"RDS(on) (@2.7V) % (&#937;)##false","name":"XJG640^VT-006!7","type":"float"},{"header":"RDS(on) (@1.8V) % (&#937;)##false","name":"XJG640^VT-006!8","type":"float"},{"header":"Free Samples##false","name":"STP0881^VT-007!0","type":"string"},{"header":"Total Gate Charge(Qg) typ ()##true","name":"STP049^VT-002!0","type":"float"},{"header":"Total Power Dissipation(PD) % (W)##true","name":"XJG820^VT-006!0","type":"float"}]},"success":"true", "total":13,"records":[{"XJE010^VT-007!0":"STB80PF55$$/cn/analog/product/67164.jsp","STP716^VT-007!0":"Active","XJE016^VT-007!0":"D2PAK","STP472^VT-007!0":"_","XJG810^VT-007!0":"-55","XJG273^VT-006!0":"80","XJG640^VT-006!3":".018","GNP023^VT-007!0":"-","XJG640^VT-006!6":"-","XJG640^VT-006!7":"-","XJG640^VT-006!8":"-","STP0881^VT-007!0":"No","STP049^VT-002!0":"190","XJG820^VT-006!0":"300"},{"XJE010^VT-007!0":"STD10PF06$$/cn/analog/product/64543.jsp","STP716^VT-007!0":"Active","XJE016^VT-007!0":"IPAK TO-251 TO 252 DPAK","STP472^VT-007!0":"_","XJG810^VT-007!0":"-60","XJG273^VT-006!0":"-10","XJG640^VT-006!3":".2","GNP023^VT-007!0":"-","XJG640^VT-006!6":"-","XJG640^VT-006!7":"-","XJG640^VT-006!8":"-","STP0881^VT-007!0":"No ... Regards, Amit

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  • Counting HTML images with Python

    - by user2537246
    I need some feedback on how to count HTML images with Python 3.01 after extracting them, maybe my regular expression are used properly. Here is my code: import re, os import urllib.request def get_image(url): url = 'http://www.google.com' total = 0 try: f = urllib.request.urlopen(url) for line in f.readline(): line = re.compile('<img.*?src="(.*?)">') if total > 0: x = line.count(total) total += x print('Images total:', total) except: pass

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  • Sorting IPv4 Addresses

    - by Kumba
    So I've run into a quandary on sorting IPv4 addresses, and didn't know if there was a set rule in some obscure networking document. Do I do a straight sort on the raw address only (such as converting the IP address to a 32bit number and then sorting), do I factor in the CIDR via some mathematical formula, do I sort via the CIDR only (as if I'm comparing the network size and not the addresses directly)? I.e., normal math, we'd do something like -1 < 0 < 1 to denote the order of precedence. Given say, 10.1.0.0/16, 172.16.0.0/12, 192.168.1.0/24, and 192.168.1.42, what would be the order of precedence?

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  • Sorting a list of variable length integers delimited by decimal points...

    - by brewerdc
    Hey guys, I'm in need of some help. I have a list of delimited integer values that I need to sort. An example: Typical (alpha?) sort: 1.1.32.22 11.2.4 2.1.3.4 2.11.23.1.2 2.3.7 3.12.3.5 Correct (numerical) sort: 1.1.32.22 2.1.3.4 2.3.7 2.11.23.1.2 3.12.3.5 11.2.4 I'm having trouble figuring out how to setup the algorithm to do such a sort with n number of decimal delimiters and m number of integer fields. Any ideas? This has to have been done before. Let me know if you need more information. Thanks a bunch! -Daniel

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  • Programming Contest Question: Counting Polyominos

    - by Martijn Courteaux
    Hi, An example question for a programming contest was to write a program that finds out how much polyominos are possible with a given number of stones. So for two stones (n = 2) there is only one polyominos: XX You might think this is a second solution: X X But it isn't. The polyominos are not unique if you can rotate them. So, for 4 stones (n = 4), there are 7 solutions: X X XX X X X X X X XX X XX XX XX X X X XX X X XX The application has to be able to find the solution for 1 <= n <=10 PS: Using the list of polyominos on Wikipedia isn't allowed ;) EDIT: Of course the question is: How to do this in Java, C/C++, C#

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  • problem in counting two fields in one query

    - by Mac Taylor
    hey guys i need to count new private messages and old one from a table so first thing come to mind is using mysql_num_rows and easy thing to do // check new pms $user_id = $userinfo['user_id']; $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='1' OR pm_unread='1')"; $result = $db->sql_query($sql) ; $new_pms = $db->sql_numrows($result); $db->sql_freeresult($result); // check old pms $sql = "SELECT author_id FROM bb3privmsgs_to WHERE user_id='$user_id' AND (pm_new='0' OR pm_unread='0')"; $result = $db->sql_query($sql) ; $old_pms = $db->sql_numrows($result); $db->sql_freeresult($result); but how can i count these two fields just in one statement and shorter lines ?~

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  • bubble sort on array of c structures not sorting properly

    - by xmpirate
    I have the following program for books record and I want to sort the records on name of book. the code isn't showing any error but it's not sorting all the records. #include "stdio.h" #include "string.h" #define SIZE 5 struct books{ //define struct char name[100],author[100]; int year,copies; }; struct books book1[SIZE],book2[SIZE],*pointer; //define struct vars void sort(struct books *,int); //define sort func main() { int i; char c; for(i=0;i<SIZE;i++) //scanning values { gets(book1[i].name); gets(book1[i].author); scanf("%d%d",&book1[i].year,&book1[i].copies); while((c = getchar()) != '\n' && c != EOF); } pointer=book1; sort(pointer,SIZE); //sort call i=0; //printing values while(i<SIZE) { printf("##########################################################################\n"); printf("Book: %s\nAuthor: %s\nYear of Publication: %d\nNo of Copies: %d\n",book1[i].name,book1[i].author,book1[i].year,book1[i].copies); printf("##########################################################################\n"); i++; } } void sort(struct books *pointer,int n) { int i,j,sorted=0; struct books temp; for(i=0;(i<n-1)&&(sorted==0);i++) //bubble sort on the book name { sorted=1; for(j=0;j<n-i-1;j++) { if(strcmp((*pointer).name,(*(pointer+1)).name)>0) { //copy to temp val strcpy(temp.name,(*pointer).name); strcpy(temp.author,(*pointer).author); temp.year=(*pointer).year; temp.copies=(*pointer).copies; //copy next val strcpy((*pointer).name,(*(pointer+1)).name); strcpy((*pointer).author,(*(pointer+1)).author); (*pointer).year=(*(pointer+1)).year; (*pointer).copies=(*(pointer+1)).copies; //copy back temp val strcpy((*(pointer+1)).name,temp.name); strcpy((*(pointer+1)).author,temp.author); (*(pointer+1)).year=temp.year; (*(pointer+1)).copies=temp.copies; sorted=0; } *pointer++; } } } My Imput The C Programming Language X Y Z 1934 56 Inferno Dan Brown 1993 453 harry Potter and the soccers stone J K Rowling 2012 150 Ruby On Rails jim aurther nil 2004 130 Learn Python Easy Way gmaps4rails 1967 100 And the output ########################################################################## Book: Inferno Author: Dan Brown Year of Publication: 1993 No of Copies: 453 ########################################################################## ########################################################################## Book: The C Programming Language Author: X Y Z Year of Publication: 1934 No of Copies: 56 ########################################################################## ########################################################################## Book: Ruby On Rails Author: jim aurther nil Year of Publication: 2004 No of Copies: 130 ########################################################################## ########################################################################## Book: Learn Python Easy Way Author: gmaps4rails Year of Publication: 1967 No of Copies: 100 ########################################################################## ########################################################################## Book: Author: Year of Publication: 0 No of Copies: 0 ########################################################################## We can see the above sorting is wrong? What I'm I doing wrong?

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  • Counting characters in an Access database column using SQL

    - by jzr
    I have the following table col1 col2 col3 col4 ==== ==== ==== ===== 1233 4566 ABCD CDEF 1233 4566 ACD1 CDEF 1233 4566 D1AF CDEF I need to count the characters in col3, so from the data in the previous table it would be: char count ==== ===== A 3 B 1 C 2 D 3 F 1 1 2 Is this possible to achieve by using SQL only? At the moment I am thinking of passing a parameter in to SQL query and count the characters one by one and then sum, however I did not start the VBA part yet, and frankly wouldn't want to do that. This is my query at the moment: PARAMETERS X Long; SELECT First(Mid(TABLE.col3,X,1)) AS [col3 Field], Count(Mid(TABLE.col3,X,1)) AS Dcount FROM TEST GROUP BY Mid(TABLE.col3,X,1) HAVING (((Count(Mid([TABLE].[col3],[X],1)))>=1)); Ideas and help are much appreciated, as I don't usually work with Access and SQL.

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  • Counting vowels

    - by user74283
    Can anyone please tell me what is wrong with this script. I am a python newb but i cant seem to figure out what might be causing it not to function. def find_vowels(sentence): """ >>> find_vowels(test) e """ count = 0 vowels = "aeiuoAEIOU" for letter in sentence: if letter in vowels: count += 1 print count if __name__ == '__main__': import doctest doctest.testmod()

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  • Counting Sublist Elements in Prolog

    - by idea_
    How can I count nested list elements in prolog? I have the following predicates defined, which will count a nested list as one element: length([ ], 0). length([H|T],N) :- length(T,M), N is M+1. Usage: ?- length([a,b,c],Out). Out = 3 This works, but I would like to count nested elements as well i.e. length([a,b,[c,d,e],f],Output). ?- length([a,b,[c,d,e],f],Output). Output = 6

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  • question about counting sort

    - by davit-datuashvili
    hi i have write following code which prints elements in sorted order only one big problem is that it use two additional array here is my code public class occurance{ public static final int n=5; public static void main(String[]args){ // n is maximum possible value what it should be in array suppose n=5 then array may be int a[]=new int[]{3,4,4,2,1,3,5};// as u see all elements are less or equal to n //create array a.length*n int b[]=new int[a.length*n]; int c[]=new int[b.length]; for (int i=0;i<b.length;i++){ b[i]=0; c[i]=0; } for (int i=0;i<a.length;i++){ if (b[a[i]]==1){ c[a[i]]=1; } else{ b[a[i]]=1; } } for (int i=0;i<b.length;i++){ if (b[i]==1) { System.out.println(i); } if (c[i]==1){ System.out.println(i); } } } } // 1 2 3 3 4 4 5 1.i have two question what is complexity of this algorithm?i mean running time 2. how put this elements into other array with sorted order? thanks

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  • Ruby: counters, counting and incrementing

    - by Shyam
    Hi, If you have seen my previous questions, you'd already know I am a big nuby when it comes to Ruby. So, I discovered this website which is intended for C programming, but I thought whatever one can do in C, must be possible in Ruby (and more readable too). The challenge is to print out a bunch of numbers. I discovered this nifty method .upto() and I used a block (and actually understanding its purpose). However, in IRb, I got some unexpected behavior. class MyCounter def run 1.upto(10) { |x| print x.to_s + " " } end end irb(main):033:0> q = MyCounter.new => #<MyCounter:0x5dca0> irb(main):034:0> q.run 1 2 3 4 5 6 7 8 9 10 => 1 I have no idea where the = 1 comes from :S Should I do this otherwise? I am expecting to have this result: 1 2 3 4 5 6 7 8 9 10 Thank you for your answers, comments and feedback!

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  • Code Golf: Counting XML elements in file on Android

    - by CSharperWithJava
    Take a simple XML file formatted like this: <Lists> <List> <Note/> ... <Note/> </List> <List> <Note/> ... <Note/> </List> </Lists> Each node has some attributes that actually hold the data of the file. I need a very quick way to count the number of each type of element, (List and Note). Lists is simply the root and doesn't matter. I can do this with a simple string search or something similar, but I need to make this as fast as possible. Design Parameters: Must be in java (Android application). Must AVOID allocating memory as much as possible. Must return the total number of Note elements and the number of List elements in the file, regardless of location in file. Number of Lists will typically be small (1-4), and number of notes can potentially be very large (upwards of 1000, typically 100) per file. I look forward to your suggestions.

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  • string comparision and counting the key in target [closed]

    - by mesun
    Suppose we want to count the number of times that a key string appears in a target string. We are going to create two different functions to accomplish this task: one iterative, and one recursive. For both functions, you can rely on Python's find function - you should read up on its specifications to see how to provide optional arguments to start the search for a match at a location other than the beginning of the string. For example, find("atgacatgcacaagtatgcat","atgc") #returns the value 5, while find("atgacatgcacaagtatgcat","atgc",6) #returns the value 15, meaning that by starting the search at index 6, #the next match is found at location 15. For the recursive version, you will want to think about how to use your function on a smaller version of the same problem (e.g., on a smaller target string) and then how to combine the result of that computation to solve the original problem. For example, given you can find the first instance of a key string in a target string, how would you combine that result with invocation of the same function on a smaller target string? You may find the string slicing operation useful in getting substrings of string.

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  • Java Counting # of occurrences of a word in a string

    - by Doug
    I have a large text file I am reading from and I need to find out how many times some words come up. For example, the word "the". I'm doing this line by line each line is a string. I need to make sure that I only count legit "the"'s the the in other would not count. This means I know I need to use regular expressions in some way. What I was trying so far is this: numSpace += line.split("[^a-z]the[^a-z]").length; I realize the regular expression may not be correct at the moment but I tried without that and just tried to find occurrences of the word the and I get wrong numbers to. I was under the impression this would split the string up into an array and how many times that array was split up was how many times the word is in the string. Any ideas I would be grateful.

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  • jQuery Grouping Similar Items and Counting When Repeated

    - by NessDan
    So I have this structure setup: <ul> <li>http://www.youtube.com/watch?v=dw1Vh9Yzryo</li> (Vid1) <li>http://www.youtube.com/watch?v=bOF3o8B292U</li> (Vid2) <li>http://www.youtube.com/watch?v=yAY4vNJd7A8</li> (Vid3) <li>http://www.youtube.com/watch?v=yAY4vNJd7A8</li> <li>http://www.youtube.com/watch?v=dw1Vh9Yzryo</li> <li>http://www.youtube.com/watch?v=bOF3o8B292U</li> <li>http://www.youtube.com/watch?v=yAY4vNJd7A8</li> <li>http://www.youtube.com/watch?v=dw1Vh9Yzryo</li> </ul> Vid1 is repeated 3 times, Vid2 is repeated 3 times, and Vid3 is repeated 2 times. I want to put them into a structure where I can reference them like this: Vid1 - 3 (Repeated), http://www.youtube.com/get_video?video_id=dw1Vh9Yzryo&fmt=36 (Download) Vid2 - 3 (Repeated), http://www.youtube.com/get_video?video_id=bOF3o8B292U&fmt=36 (Download) Vid3 - 2 (Repeated), http://www.youtube.com/get_video?video_id=yAY4vNJd7A8&fmt=36 (Download) "This video was repeated " + [Vid1][Repeated] + " times and you can download it here: " + [Vid1][Download]; How can I set this structure up? I think I should be using an array to achieve the above but I'm not sure how I would set it up or how to reference certain things in the array. The other question is how can I get how many times something was repeated? The URL I have no problem with. Can anyone help me out?

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  • Counting amount of items in Pythons 'for'

    - by Markum
    Kind of hard to explain, but when I run something like this: fruits = ['apple', 'orange', 'banana', 'strawberry', 'kiwi'] for fruit in fruits: print fruit.capitalize() It gives me this, as expected: Apple Orange Banana Strawberry Kiwi How would I edit that code so that it would "count" the amount of times it's performing the for, and print this? 1 Apple 2 Orange 3 Banana 4 Strawberry 5 Kiwi

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  • Counting point size based on chart area during zooming/unzoomin

    - by Gacek
    Hi folks. I heave a quite simple task. I know (I suppose) it should be easy, but from the reasons I cannot understand, I try to solve it since 2 days and I don't know where I'm making the mistake. So, the problem is as follows: - we have a chart with some points - The chart starts with some known area and points have known size - we would like to "emulate" the zooming effect. So when we zoom to some part of the chart, the size of points is getting proportionally bigger. In other words, the smaller part of the chart we select, the bigger the point should get. So, we have something like that. We know this two parameters: initialArea; // Initial area - area of the whole chart, counted as width*height initialSize; // initial size of the points Now lets assume we are handling some kind of OnZoom event. We selected some part of the chart and would like to count the current size of the points float CountSizeOnZoom() { float currentArea = CountArea(...); // the area is counted for us. float currentSize = initialSize * initialArea / currentArea; return currentSize; } And it works. But the rate of change is too fast. In other words, the points are getting really big too soon. So I would like the currentSize to be invertly proportional to currentArea, but with some scaling coefficient. So I created the second function: float CountSizeOnZoom() { float currentArea = CountArea(...); % the area is counted for us. // Lets assume we want the size of points to change ten times slower, than area of the chart changed. float currentSize = initialSize + 0.1f* initialSize * ((initialArea / currentArea) -1); return currentSize; } Lets do some calculations in mind. if currentArea is smaller than initialArea, initialArea/currentArea > 1 and then we add "something" small and postive to initialSize. Checked, it works. Lets check what happens if we would un-zoom. currentArea will be equal to initialArea, so we would have 0 at the right side (1-1), so new size should be equal to initialSize. Right? Yeah. So lets check it... and it doesn't work. My question is: where is the mistake? Or maybe you have any ideas how to count this scaled size depending on current area in some other way?

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  • MySQL help, counting information on last records

    - by ee12csvt
    I need some advice I have two tables, one holds unique serial numbers of items (items) and the other holds status changes and other information for these items (details) The Tables are set up as follows Item itemID itemName itemDate details detID itemID modlvl status detDate All items have at least one record in the details table, but over time the status has changed or the modification level has changed (Both of these are identified by numbers which are held in other appropriate tables) and a new record is created each time the status/modlvl changes I want to display a table on my webpage using php that identifies the different mod levels of the items and shows a count of each of the current status of the items EDIT Hi Ronnis, This is an example of the data in the tables and what I want to achieve The current Mod Levels range from 1 to 3 Status representations are 1 In Use 2 In Store 3 Being repaired 4 In Transit 5 For Disposal 6 Disposed 7 Lost Item itemID OrigMod created 1000 1 2009-10-01 22:12:12 1001 1 2009-10-01 22:12:12 1002 1 2009-10-01 22:12:12 1003 1 2009-10-01 22:12:12 1004 1 2009-10-01 22:12:12 1005 1 2009-10-01 22:12:12 1006 1 2009-10-01 22:12:12 1007 1 2009-10-01 22:12:12 1008 1 2009-10-01 22:12:12 1009 1 2009-10-01 22:12:12 1010 1 2009-10-01 22:12:12 Details detID itemID modlvl detDate status 1 1000 1 2009-10-01 1 2 1001 1 2009-10-01 1 3 1002 1 2009-10-01 1 4 1003 1 2009-10-01 1 5 1004 1 2009-10-01 1 6 1005 1 2009-10-01 1 7 1006 1 2009-10-01 1 8 1007 1 2009-10-01 1 9 1008 1 2009-10-01 1 10 1009 1 2009-10-01 1 11 1010 1 2009-10-01 1 12 1001 1 2010-02-01 2 13 1001 1 2010-02-03 4 14 1001 1 2010-03-01 3 15 1000 1 2010-03-14 2 16 1001 2 2010-04-01 4 17 1006 1 2010-04-01 2 18 1001 2 2010-04-03 2 19 1006 1 2010-04-14 4 20 1006 1 2010-05-01 5 21 1002 1 2010-05-02 2 22 1003 1 2010-05-10 2 23 1010 1 2010-06-01 2 24 1006 1 2010-06-18 6 25 1010 1 2010-07-01 7 26 1007 1 2010-07-02 2 27 1007 1 2010-07-04 4 28 1003 1 2010-07-10 2 29 1007 1 2010-07-11 3 30 1007 2 2010-07-12 4 31 1007 2 2010-07-15 2 32 1001 2 2010-08-31 1 33 1001 2 2010-09-10 2 34 1001 2 2010-10-01 4 35 1008 1 2010-10-01 2 36 1001 2 2010-10-05 3 37 1008 1 2010-10-05 4 38 1008 1 2010-10-10 3 39 1001 3 2010-10-20 4 40 1001 3 2010-10-25 2 Using the tables above I want to get this result MoLvl Use Store Repd Transit Displ Dispd Lost Total 1 3 3 1 0 0 1 1 9 2 0 1 0 0 0 0 0 1 3 0 1 0 0 0 0 0 1 Total 3 5 1 0 0 1 1 11

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  • Sort a list of pointers.

    - by YuppieNetworking
    Hello all, Once again I find myself failing at some really simple task in C++. Sometimes I wish I could de-learn all I know from OO in java, since my problems usually start by thinking like Java. Anyways, I have a std::list<BaseObject*> that I want to sort. Let's say that BaseObject is: class BaseObject { protected: int id; public: BaseObject(int i) : id(i) {}; virtual ~BaseObject() {}; }; I can sort the list of pointer to BaseObject with a comparator struct: struct Comparator { bool operator()(const BaseObject* o1, const BaseObject* o2) const { return o1->id < o2->id; } }; And it would look like this: std::list<BaseObject*> mylist; mylist.push_back(new BaseObject(1)); mylist.push_back(new BaseObject(2)); // ... mylist.sort(Comparator()); // intentionally omitted deletes and exception handling Until here, everything is a-ok. However, I introduced some derived classes: class Child : public BaseObject { protected: int var; public: Child(int id1, int n) : BaseObject(id1), var(n) {}; virtual ~Child() {}; }; class GrandChild : public Child { public: GrandChild(int id1, int n) : Child(id1,n) {}; virtual ~GrandChild() {}; }; So now I would like to sort following the following rules: For any Child object c and BaseObject b, b<c To compare BaseObject objects, use its ids, as before. To compare Child objects, compare its vars. If they are equal, fallback to rule 2. GrandChild objects should fallback to the Child behavior (rule 3). I initially thought that I could probably do some casts in Comparator. However, this casts away constness. Then I thought that probably I could compare typeids, but then everything looked messy and it is not even correct. How could I implement this sort, still using list<BaseObject*>::sort ? Thank you

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  • Counting combinations in c or in python

    - by Dennis
    Hello I looked a bit on this topic here but I found nothing that could help me. I need a program in Python or in C that will give me all possible combinations of a and b that will meet the requirement n=2*a+b, for n from 0 to 10. a, b and n are integers. For example if n=0 both a and b must be 0. For n=1 a must be zero and b must be 1, for n=2 a can be 1 and b=0, or a=0 and b=2, etc. I'm not that good with programming. I made this: #include <stdio.h> int main(void){ int a,b,n; for(n = 0; n <= 10; n++){ for(a = 0; a <= 10; a++){ for(b = 0; b <= 10; b++) if(n == 2*a + b) printf("(%d, %d), ", (a,b)); } printf("\n"); } } But it keeps getting strange results like this: (0, -1079628000), (1, -1079628000), (2, -1079628000), (0, -1079628000), (3, -1079628000), (1, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (7, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (8, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), (9, -1079628000), (7, -1079628000), (5, -1079628000), (3, -1079628000), (1, -1079628000), (10, -1079628000), (8, -1079628000), (6, -1079628000), (4, -1079628000), (2, -1079628000), (0, -1079628000), ideone Any idea what is wrong? Also if I could do this for Python it would be even cooler. :D

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  • jquery parent/children selector for counting <li>

    - by Kreker
    Hi. I have this piece of HTML <div id="fileTreeInviati"> <ul class="php-file-tree"> <li class="pft-directory"> <a href="#" class="" name="101">A006 - SOMETEXT (<span name="contaNew"></span>)</a> <img src="./moduli/home/images/info.png" title="Informazioni Azienda" class="imgInfo"/> <ul style="display: none;"> <li class="pft-file ext-png"> <a href="javascript:getInfoFile('4');" class="" id="4">cut.png</a> </li> <li class="pft-file ext-dll"> <a href="javascript:getInfoFile('27');" class="new" id="27">Safari.dll</a> </li> </ul> </li> <li class="pft-directory"> <a href="#" class="" name="102">A012 - SOMETEXT (<span name="contaNew"></span>)</a> <img src="./moduli/home/images/info.png" title="Informazioni Azienda" class="imgInfo"/> <ul style="display: none;"> <li class="pft-file ext-jpg"> <a href="javascript:getInfoFile('19');" class="new" id="19">04.jpg</a> </li> <li class="pft-file ext-dll"> <a href="javascript:getInfoFile('24');" class="new" id="24">Safari.dll</a> </li> </ul> </li> <li class="pft-directory"> <a href="#" class="" name="103">A014 - SOMETEXT (<span name="contaNew"></span>)</a> <img src="./moduli/home/images/info.png" title="Informazioni Azienda" class="imgInfo"/> <ul style="display: none;"> <li class="pft-file ext-txt"> <a href="javascript:getInfoFile('17');" class="new" id="17">acu.txt</a> </li> <li class="pft-file ext-dll"> <a href="javascript:getInfoFile('22');" class="new" id="22">Safari.dll</a> </li> </ul> </li> </ul> I'm working on a js snippet that cycle through all "a" of the "li" and checks if it has the class "new" if yes increment a counter by one. This counter now has to be printed on the relative "li" "span" 3 level before. So I have the number of the element with the "new" class. The js snippet is this $("#fileTreeInviati .php-file-tree .pft-directory li").each(function(){ $(this).children("a").each(function(i,e){ if ($(e).hasClass("new")){ cont++; console.log($(e).text()); $(this).parent().parent().parent().children("a").children("span").text(cont); } }) cont = 0; }); I think I'm almost there but the counter is always 1. I think there is something mess with .children, maybe it can handle only the first occurrence? Thanks for help

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