Search Results

Search found 8 results on 1 pages for 'abba'.

Page 1/1 | 1 

• What is a good php 5.3.x shared hosting company?

  - by Abba Bryant

  I am looking for the best shared host - features-wise, not price - for hosting CakePHP and Lithium applications. I would like to be able to use MongoDB / MySQL as well as have access to some of the more common PHP extensions like MCrypt, etc. I currently use dreamhost with a custom PHP 5.3.x build on my sandbox domain - Please do not suggest this as a solution. I want to move away from managing my own PHP build if possible. I need ssh access but email support isn't as big of an issue.

  Read the article

• Java - Counting how many characters show up in another string

  - by Vu Châu

  I am comparing two strings, in Java, to see how many characters from the first string show up in the second string. The following is some expectations: matchingChars("AC", "BA") ? 1 matchingChars("ABBA", "B") ? 2 matchingChars("B", "ABBA") ? 1 My approach is as follows: public int matchingChars(String str1, String str2) { int count = 0; for (int a = 0; a < str1.length(); a++) { for (int b = 0; b < str2.length(); b++) { char str1Char = str1.charAt(a); char str2Char = str2.charAt(b); if (str1Char == str2Char) { count++; str1 = str1.replace(str1Char, '0'); } } } return count; } I know my approach is not the best, but I think it should do it. However, for matchingChars("ABBA", "B") ? 2 My code yields "1" instead of "2". Does anyone have any suggestion or advice? Thank you very much.

  Read the article

• What are good hosting companies for PHP 5.3 Mysql / CouchDb / MongoDB Dev ( Lithium / CakePHP Framew

  - by Abba Bryant

  I am looking for a quality reliable host for some lithium development. I don't mind a shared platform as long as I have some ssh access. I require php 5.3.x, Mysql 5.x, and the usual imageMagick etc. Non-relational DB support up front would be nice but if they let me set one up myself I would be okay with doing it. I don't need a lot in the way of control panel tools. Good ones are appreciated but bad ones I would prefer not to even deal with. I don't anticipate needing much in the way of email but mail support would be nice to have. Cost isn't a big issue. I don't want to pay an arm and a leg but don't mind paying for what I need. Good support and decent uptime would be nice but I don't need an SLO or anything.

  Read the article

• How to write a shellscript date check that will not To run Except 3rd week sat, Sunday and 4th week

  - by abba

  E.g. the schellscript will not run on april 18, 19 and 20 and then for may is 16, 17 and 18 and so on Thanks

  Read the article

• How to generate all variations with repetitions of a string?

  - by Svenstaro

  I want to generate all variations with repetitions of a string in C++ and I'd highly prefer a non-recursive algorithm. I've come up with a recursive algorithm in the past but due to the complexity (r^n) I'd like to see an iterative approach. I'm quite surprised that I wasn't able to find a solution to this problem anywhere on the web or on StackOverflow. I've come up with a Python script that does what I want as well: import itertools variations = itertools.product('ab', repeat=4) for variations in variations: variation_string = "" for letter in variations: variation_string += letter print variation_string Output: aaaa aaab aaba aabb abaa abab abba abbb baaa baab baba babb bbaa bbab bbba bbbb Ideally I'd like a C++ program that can produce the exact output, taking the exact same parameters. This is for learning purposes, it isn't homework. I wish my homework was like that.

  Read the article

• Unix Permissions issue with users belonging to the same group accessing a folder

  - by TK Kocheran

  I have a folder I'd really like to allow another user on this machine access to. I'm using mt-daapd to serve music to the network, so I'd like to enable the mt-daapd user to access my Music directory, /home/rfkrocktk/Music. The master user is rfkrocktk obviously. I've tried to set all of my permissions properly on the directory, but the mt-daapd user can't acces the files. I created a group called media-users and added both rfkrocktk and mt-daapd to it in order to give mt-daapd permission to simply read all of the files in that directory and subdirectories. If I run id on each of my users, here's what's displayed: $ id rfkrocktk > uid=1000(rfkrocktk) gid=1000(rfkrocktk) groups=1000(rfkrocktk),4(adm),20(dialout),24(cdrom),29(audio),46(plugdev),104(lpadmin),115(admin),120(sambashare),124(vboxusers),1001(jupiter),2002(media-users) $ id mt-daapd > uid=123(mt-daapd) gid=65534(nogroup) groups=65534(nogroup),2002(media-users) It definitely seems that both users are a part of the media-users group, so what could be going wrong? If I run ls -l on the actual Music directory to see its permissions, here's the output: drwxr-Sr-- 201 rfkrocktk media-users 12288 2011-01-13 12:26 Music If I run ls -l on the Music directory to get its children, here's the output: drwxr-Sr-- 3 rfkrocktk media-users 4096 2010-12-20 15:31 2DBoy drwxr-Sr-- 3 rfkrocktk media-users 4096 2010-05-25 12:50 ABBA drwxr-Sr-- 3 rfkrocktk media-users 4096 2009-12-28 15:19 Access Denied drwxr-Sr-- 10 rfkrocktk media-users 4096 2009-12-28 15:19 AC-DC drwxr-Sr-- 3 rfkrocktk media-users 4096 2009-12-28 15:19 Aerosmith drwxr-Sr-- 3 rfkrocktk media-users 4096 2010-06-04 10:45 A Flock of Seagulls drwxr-Sr-- 4 rfkrocktk media-users 4096 2010-05-28 18:13 Alestorm drwxr-Sr-- 3 rfkrocktk media-users 4096 2010-06-22 23:29 Amon Amarth drwxr-Sr-- 5 rfkrocktk media-users 4096 2009-12-28 15:19 Anberlin ... From this, it would seem that I should be able to access the folders from mt-daapd, but I can't. Running sudo -i -u mt-daapd ls -l /home/rfkrocktk/Music displays nothing, indicating to me that for whatever reason, mt-daapd doesn't have access to read the folder. What am I doing wrong?

  Read the article

• Bash and regex problem : check for tokens entered into a Coke vending machine

  - by Michael Mao

  Hi all: Here is a "challenge question" I've got from Linux system programming lecture. Any of the following strings will give you a Coke if you kick: L = { aaaa, aab, aba, baa, bb, aaaa"a", aaaa"b", aab"a", … ab"b"a, ba"b"a, ab"bbbbbb"a, ... } The letters shown in wrapped double quotes indicate coins that would have fallen through (but those strings are still part of the language in this example). Exercise (a bit hard) show this is the language of a regular expression And this is what I've got so far : #!/usr/bin/bash echo "A bottle of Coke costs you 40 cents" echo -e "Please enter tokens (a = 10 cents, b = 20 cents) in a sequence like 'abba' :\c" read tokens #if [ $tokens = aaaa ]||[ $tokens = aab ]||[ $tokens = bb ] #then # echo "Good! now a coke is yours!" #else echo "Thanks for your money, byebye!" if [[ $token =~ 'aaaa|aab|bb' ]] then echo "Good! now a coke is yours!" else echo "Thanks for your money, byebye!" fi Sadly it doesn't work... always outputs "Thanks for your money, byebye!" I believe something is wrong with syntax... We didn't provided with any good reference book and the only instruction from the professor was to consult "anything you find useful online" and "research the problem yourself" :( I know how could I do it in any programming language such as Java, but get it done with bash script + regex seems not "a bit hard" but in fact "too hard" for anyone with little knowledge on something advanced as "lookahead"(is this the terminology ?) I don't know if there is a way to express the following concept in the language of regex: Valid entry would consist of exactly one of the three components : aaaa, aab and bb, regardless of order, followed by an arbitrary sequence of a or b's So this is what is should be like : (a{4}Ua{2}bUb{2})(aUb)* where the content in first braces is order irrelevant. Thanks a lot in advance for any hints and/or tips :)

  Read the article

• How to calculate this string-dissimilarity function efficiently?

  - by ybungalobill

  Hello, I was looking for a string metric that have the property that moving around large blocks in a string won't affect the distance so much. So "helloworld" is close to "worldhello". Obviously Levenshtein distance and Longest common subsequence don't fulfill this requirement. Using Jaccard distance on the set of n-grams gives good results but has other drawbacks (it's a pseudometric and higher n results in higher penalty for changing single character). [original research] As I thought about it, what I'm looking for is a function f(A,B) such that f(A,B)+1 equals the minimum number of blocks that one have to divide A into (A1 ... An), apply a permutation on the blocks and get B: f("hello", "hello") = 0 f("helloworld", "worldhello") = 1 // hello world -> world hello f("abba", "baba") = 2 // ab b a -> b ab a f("computer", "copmuter") = 3 // co m p uter -> co p m uter This can be extended for A and B that aren't necessarily permutations of each other: any additional character that can't be matched is considered as one additional block. f("computer", "combuter") = 3 // com uter -> com uter, unmatched: p and b. Observing that instead of counting blocks we can count the number of pairs of indices that are taken apart by a permutation, we can write f(A,B) formally as: f(A,B) = min { C(P) | P:|A|?|B|, P is bijective, ?i?dom(P) A[P(i)]=B[P(i)] } C(P) = |A| + |B| - |dom(P)| - |{ i | i,i+1?dom(P) and P(i)+1=P(i+1) }| - 1 The problem is... guess what... ... that I'm not able to calculate this in polynomial time. Can someone suggest a way to do this efficiently? Or perhaps point me to already known metric that exhibits similar properties?

  Read the article

1