{"id":40208,"date":"2020-02-01T06:04:35","date_gmt":"2020-02-01T03:04:35","guid":{"rendered":"https:\/\/prohoster.info\/blog\/blog_prohoster\/kak-nauchit-preodolevat-trudnosti-a-zaodno-i-pisat-czikly"},"modified":"2020-02-01T06:04:35","modified_gmt":"2020-02-01T03:04:35","slug":"kak-nauchit-preodolevat-trudnosti-a-zaodno-i-pisat-czikly","status":"publish","type":"post","link":"https:\/\/prohoster.info\/sq\/blog\/kak-nauchit-preodolevat-trudnosti-a-zaodno-i-pisat-czikly","title":{"rendered":"Si t\u00eb m\u00ebsojm\u00eb t\u00eb kalojm\u00eb v\u00ebshtir\u00ebsit\u00eb dhe nj\u00ebkoh\u00ebsisht t\u00eb shkruajm\u00eb cikle","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p>Megjith\u00ebse do t\u00eb flitet p\u00ebr nj\u00eb nga temat baz\u00eb, ky artikull \u00ebsht\u00eb shkruar p\u00ebr profesionist\u00eb t\u00eb p\u00ebrvojsh\u00ebm. Q\u00ebllimi \u00ebsht\u00eb t\u00eb tregoj\u00eb se cilat keqkuptime ka fillestar\u00ebt n\u00eb programim. P\u00ebr zhvilluesit praktik\u00eb, k\u00ebto probleme tashm\u00eb jan\u00eb zgjidhur, jan\u00eb harruar ose nuk jan\u00eb v\u00ebn\u00eb re fare. Artikulli mund t\u00eb jet\u00eb i dobish\u00ebm, n\u00ebse ndonj\u00ebher\u00eb ju duhet t\u00eb ndihmoni dik\u00eb me k\u00ebt\u00eb tem\u00eb. N\u00eb artikull, b\u00ebhen paralel\u00eb me materialin e librave t\u00eb ndrysh\u00ebm n\u00eb programim, autor\u00ebt e t\u00eb cil\u00ebve jan\u00eb Schildt, Stroustrup, Okulov.<\/p>\n<p>Tema p\u00ebr ciklet \u00ebsht\u00eb zgjedhur p\u00ebr shkak se aty p\u00ebrzgjidhen mjaft njer\u00ebz n\u00eb m\u00ebsimin e programimit.<\/p>\n<p>Kjo metod\u00eb \u00ebsht\u00eb e orientuar p\u00ebr student\u00ebt e dob\u00ebt. Si rregull, ata t\u00eb fort\u00eb nuk mbeten n\u00eb k\u00ebt\u00eb tem\u00eb dhe nuk ka nevoj\u00eb t\u00eb shpikim metoda t\u00eb ve\u00e7anta p\u00ebr ta. Q\u00ebllimi dyt\u00ebsor i artikullit \u00ebsht\u00eb t\u00eb kalojm\u00eb k\u00ebt\u00eb metod\u00eb nga kategoria \"funksionon p\u00ebr t\u00eb gjith\u00eb student\u00ebt, por vet\u00ebm me nj\u00eb m\u00ebsues\" n\u00eb kategorin\u00eb \"funksionon p\u00ebr t\u00eb gjith\u00eb student\u00ebt, t\u00eb gjith\u00eb m\u00ebsuesit\". Nuk pretendoj p\u00ebr origjinalitet absolut. N\u00ebse tashm\u00eb aplikoni nj\u00eb metod\u00eb t\u00eb ngjashme p\u00ebr m\u00ebsimin e k\u00ebsaj teme, ju lutem shkruani se \u00e7far\u00eb e b\u00ebn variantin tuaj t\u00eb ndrysh\u00ebm. N\u00ebse vendosni ta aplikoni, na tregoni si shkoi gjith\u00e7ka. N\u00ebse nj\u00eb metod\u00eb e ngjashme p\u00ebrshkruhet n\u00eb ndonj\u00eb lib\u00ebr, ju lutem shkruani titullin.<\/p>\n<p><noindex><a rel=\"nofollow\" name=\"habracut\"><\/a><\/noindex><br \/>\nK\u00ebt\u00eb metod\u00eb e kam zhvilluar p\u00ebr 4 vjet, duke punuar individualisht me student\u00eb t\u00eb niveleve t\u00eb ndryshme. N\u00eb total rreth pes\u00ebdhjet\u00eb student\u00eb dhe dy mij\u00eb or\u00eb m\u00ebsimi. N\u00eb fillim, n\u00eb k\u00ebt\u00eb tem\u00eb student\u00ebt ngecnin dhe largoheshin. Pas \u00e7do studenti, metoda dhe materialet jan\u00eb korrigjuar. Vitin e fundit, student\u00ebt tashm\u00eb nuk ngecin n\u00eb k\u00ebt\u00eb tem\u00eb, k\u00ebshtu q\u00eb vendosa t\u00eb ndaj p\u00ebrvojat.<\/p>\n<h4>Pse ka kaq shum\u00eb shkronja? Ciklet jan\u00eb elementare!<\/h4>\n<p>\nSi\u00e7 e kam p\u00ebrmendur m\u00eb lart, p\u00ebr zhvilluesit praktik\u00eb dhe p\u00ebr student\u00ebt e fort\u00eb, kompleksiteti i konceptit t\u00eb cikleve mund t\u00eb n\u00ebnvler\u00ebsohet. P\u00ebr shembull, mund t\u00eb organizohet nj\u00eb ligj\u00ebrat\u00eb e gjat\u00eb, t\u00eb shihet kokat q\u00eb l\u00ebvizin dhe syt\u00eb q\u00eb duken inteligjent\u00eb. Por kur p\u00ebrpiqen t\u00eb zgjidhin ndonj\u00eb problem, fillon ng\u00ebr\u00e7i dhe probleme t\u00eb paqart\u00eb. Pas ligj\u00ebrat\u00ebs, student\u00ebt me siguri kan\u00eb krijuar vet\u00ebm nj\u00eb p\u00ebrmbledhje t\u00eb pjesshme t\u00eb kuptimit. Situata p\u00ebrkeq\u00ebsohet nga fakti se student\u00ebt nuk mund t\u00eb shprehin vet\u00eb se n\u00eb \u00e7far\u00eb sakt\u00ebsisht \u00ebsht\u00eb p\u00ebrllogaritja e tyre.<br \/>\nNj\u00ebher\u00eb kuptova se student\u00ebt e shihnin shembujt e mi si hieroglife. Pra, si copa t\u00eb pandara teksti, n\u00eb t\u00eb cilat duhej t\u00eb shtohej ndonj\u00eb shkronj\u00eb \"magjike\" dhe do t\u00eb funksiononte.<br \/>\nHer\u00eb pas here kam v\u00ebn\u00eb re se student\u00ebt mendojn\u00eb se p\u00ebr zgjidhjen e nj\u00eb problemi t\u00eb caktuar nevojitet <i>ndonj\u00eb konstrukcion tjet\u00ebr<\/i> q\u00eb un\u00eb thjesht nuk kam p\u00ebrmendur akoma. Megjithat\u00eb, p\u00ebr zgjidhjen e tij mjaftonte t\u00eb modifikohej pak shembulli.<\/p>\n<p>Prandaj arrita n\u00eb iden\u00eb se duhej t\u00eb fokusohem jo n\u00eb sintaks\u00ebn e shprehjeve, por n\u00eb iden\u00eb e refaktorizimit t\u00eb kodit t\u00eb p\u00ebrs\u00ebritur p\u00ebrmes cikleve. Sapo nx\u00ebn\u00ebsit e kuptojn\u00eb k\u00ebt\u00eb ide, \u00e7do sintaks\u00eb m\u00ebsohet p\u00ebrmes disa ushtrimeve t\u00eb vogla.<\/p>\n<h4>Kush dhe pse iu jap m\u00ebsim<\/h4>\n<p>\nDuke qen\u00eb se nuk ka provime hyr\u00ebse, n\u00eb m\u00ebsime mund t\u00eb ket\u00eb student\u00eb t\u00eb fort\u00eb, por edhe shum\u00eb t\u00eb dob\u00ebt. M\u00eb shum\u00eb rreth student\u00ebve t\u00eb mi mund t\u00eb lexoni n\u00eb artikullin <noindex><a rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/post\/350360\/\">Portreti i d\u00ebgjuesve t\u00eb kurseve t\u00eb pasditeve<\/a><\/noindex><br \/>\nUn\u00eb kam d\u00ebshir\u00eb q\u00eb programimi t\u00eb m\u00ebsohet nga t\u00eb gjith\u00eb ata q\u00eb e duan.<br \/>\nM\u00ebsimet e mia zhvillohen individualisht dhe studenti paguan p\u00ebr \u00e7do sesion. Mund t\u00eb duket se student\u00ebt do t\u00eb optimizojn\u00eb shpenzimet dhe do t\u00eb k\u00ebrkojn\u00eb minimumin. Megjithat\u00eb, njer\u00ebzit ndjekin m\u00ebsime n\u00eb prani t\u00eb nj\u00eb m\u00ebsuesi t\u00eb drejtp\u00ebrdrejt\u00eb jo p\u00ebr njohurit\u00eb vet\u00eb, por p\u00ebr sigurin\u00eb se ata e kan\u00eb kuptuar n\u00eb m\u00ebnyr\u00eb t\u00eb sigurt, p\u00ebr ndjenj\u00ebn e p\u00ebrparimit dhe p\u00ebr miratimin nga eksperti (m\u00ebsuesi). N\u00ebse student\u00ebt nuk ndjejn\u00eb p\u00ebrparim n\u00eb m\u00ebsimin e tyre, ata do t\u00eb largohen. N\u00eb p\u00ebrgjith\u00ebsi, mund t\u00eb organizohet m\u00ebsimi n\u00eb nj\u00eb m\u00ebnyr\u00eb q\u00eb student\u00ebt t\u00eb ndiejn\u00eb p\u00ebrparim n\u00eb rritjen e numrit t\u00eb konstrukcioneve t\u00eb njohura. Pra, fillimisht studiojm\u00eb n\u00eb detaje while, pastaj studiojm\u00eb for, pastaj do while dhe ja ku kemi nj\u00eb kurs p\u00ebr nj\u00eb mij\u00eb e nj\u00eb nat\u00eb, ku p\u00ebr dy muaj studiohen vet\u00ebm ciklet, dhe n\u00eb fund kemi nj\u00eb student q\u00eb ka shkruar n\u00eb p\u00ebrcaktim bibliotek\u00ebn standarde. Megjithat\u00eb, p\u00ebr t\u00eb zgjidhur probleme praktike, nevojitet jo vet\u00ebm njohuri t\u00eb materialit, por edhe pavar\u00ebsi n\u00eb aplikimin e tij dhe n\u00eb k\u00ebrkimin e informacionit t\u00eb ri. Prandaj p\u00ebr kurset n\u00eb klas\u00eb mendoj se \u00ebsht\u00eb e drejt\u00eb parimi \u2014 t\u00eb m\u00ebsohet minimumi dhe t\u00eb inkurajohet studimi i pavarur i nuancave dhe temave t\u00eb lidhura. N\u00eb tem\u00ebn p\u00ebr ciklet, minimumi q\u00eb mendoj \u00ebsht\u00eb konstrukcioni while. Me t\u00eb mund t\u00eb kuptohet principi. Duke edituar principin nj\u00eb ndihm\u00ebs \u00ebsht\u00eb e mundur t\u00eb arrihet p\u00ebrfor dhe do-while n\u00eb m\u00ebnyr\u00eb t\u00eb pavarur.<\/p>\n<p>P\u00ebr t\u00eb arritur p\u00ebrvet\u00ebsimin e materialit nga student\u00ebt e dob\u00ebt, p\u00ebrshkrimi i sintaks\u00ebs nuk mjafton. Duhet t\u00eb ofrohen m\u00eb shum\u00eb detyra t\u00eb thjeshta, por t\u00eb ndryshme dhe t\u00eb jepen shembuj m\u00eb t\u00eb detajuar. M\u00eb n\u00eb fund, shpejt\u00ebsia e p\u00ebrvet\u00ebsimit kufizohet nga aft\u00ebsia e studentit p\u00ebr t\u00eb transformuar shprehjet dhe p\u00ebr t\u00eb k\u00ebrkuar rregullat. P\u00ebr student\u00ebt e mpreht\u00eb, shumica e detyrave do t\u00eb jen\u00eb t\u00eb m\u00ebrzitshme. N\u00eb m\u00ebsime me ta, nuk \u00ebsht\u00eb e nevojshme t\u00eb insistosh n\u00eb zgjidhjen e 100% t\u00eb detyrave. Materialin tim mund ta shihni n\u00eb <noindex><a rel=\"nofollow\" href=\"https:\/\/github.com\/AKryukov92\/programming-basics\">github-in tim<\/a><\/noindex>. Megjithat\u00eb, repository duket m\u00eb shum\u00eb si nj\u00eb grimuar i magjistarit - askush, p\u00ebrve\u00e7 meje, nuk do ta kuptoj\u00eb se \u00e7far\u00eb ndodhet ku, dhe n\u00ebse d\u00ebshton kontrollin, mund t\u00eb \u00e7mendesh<\/p>\n<h4>Metodika \u00ebsht\u00eb e orientuar drejt praktik\u00ebs<\/h4>\n<p>\nTeoria shpjegohet n\u00ebp\u00ebrmjet zgjidhjes s\u00eb problemeve. N\u00eb m\u00ebsimet e bazave t\u00eb programimit, ku studiohen ndarjet dhe ciklet, thjesht nuk \u00ebsht\u00eb e mundur t\u00eb organizosh nj\u00eb ligj\u00ebrat\u00eb t\u00eb dobishme n\u00eb nj\u00eb tem\u00eb p\u00ebr nj\u00eb or\u00eb t\u00eb t\u00ebr\u00eb. 15-20 minuta jan\u00eb t\u00eb mjaftueshme p\u00ebr t\u00eb p\u00ebrshkruar konceptin. V\u00ebshtir\u00ebsit\u00eb kryesore shfaqen gjat\u00eb zgjidhjes s\u00eb detyrave praktike.<br \/>\nM\u00ebsuesit e fillimit mund t\u00eb paraqesin operator\u00ebt, ndarjet, ciklet dhe array-t gjat\u00eb nj\u00eb ligj\u00ebrate. Por student\u00ebt e tyre do t\u00eb p\u00ebrballen me problemin e p\u00ebrvet\u00ebsimit t\u00eb k\u00ebsaj informacioni.<br \/>\nNuk mjafton t\u00eb tregohet materiali, por gjithashtu duhet t\u00eb sigurohemi q\u00eb d\u00ebgjuesit e kuptojn\u00eb at\u00eb.<\/p>\n<p>Fakti i p\u00ebrvet\u00ebsimit t\u00eb tem\u00ebs p\u00ebrcaktohet nga m\u00ebnyra si student\u00ebt p\u00ebrballen me pun\u00ebt e pavarura. <br \/>\nN\u00ebse studentit i ka arritur t\u00eb zgjidh\u00eb nj\u00eb problem n\u00eb tem\u00eb pa ndihm\u00ebn e m\u00ebsuesit, at\u00ebher\u00eb tema \u00ebsht\u00eb p\u00ebrvet\u00ebsuar. P\u00ebr t\u00eb siguruar kontrollin e pavarur, p\u00ebr \u00e7do detyr\u00eb p\u00ebrshkruhet nj\u00eb tabel\u00eb me skenar\u00ebt e testeve. Detyrat kan\u00eb nj\u00eb rend t\u00eb qart\u00eb. Nuk rekomandohet t\u00eb kalohen detyra. N\u00ebse detyra aktuale \u00ebsht\u00eb shum\u00eb e komplikuar, \u00ebsht\u00eb e kote t\u00eb kalosh te e ardhmja. Ajo \u00ebsht\u00eb edhe m\u00eb e v\u00ebshtir\u00eb. P\u00ebr t\u00eb siguruar q\u00eb studenti t\u00eb p\u00ebrballoj\u00eb detyr\u00ebn e v\u00ebshtir\u00eb aktuale, disa teknika shpjegohen p\u00ebrmes shembullit t\u00eb detyr\u00ebs s\u00eb par\u00eb. N\u00eb thelb, e gjith\u00eb p\u00ebrmbajtja e tem\u00ebs p\u00ebrqendrohet n\u00eb teknikat e kalimit t\u00eb v\u00ebshtir\u00ebsive. Ciklet jan\u00eb, m\u00eb tep\u00ebr, nj\u00eb efekt an\u00ebsor.<\/p>\n<p>Detyra e par\u00eb gjithmon\u00eb \u00ebsht\u00eb nj\u00eb shembull. E dyta ndryshon pak dhe realizohet \"pavar\u00ebsisht\" menj\u00ebher\u00eb pas s\u00eb par\u00ebs n\u00ebn mbik\u00ebqyrjen e m\u00ebsuesit. T\u00eb gjitha detyrat e m\u00ebvonshme jan\u00eb t\u00eb orientuara p\u00ebr t\u00eb t\u00ebrhequr v\u00ebmendjen mbi detaje t\u00eb ndryshme, t\u00eb cilat mund t\u00eb shkaktojn\u00eb keqkuptime.<\/p>\n<p>Shpjegimi i shembullit p\u00ebrb\u00ebn nj\u00eb dialog, n\u00eb t\u00eb cilin studenti duhet t\u00eb th\u00ebrras\u00eb back propagation dhe cross-validation p\u00ebr t\u00eb siguruar mir\u00ebkuptimin e nj\u00eb s\u00ebr\u00eb materialesh.<\/p>\n<p>Do t\u00eb jem banal dhe do t\u00eb deklaroj se shembulli i par\u00eb mbi tem\u00ebn \u00ebsht\u00eb shum\u00eb i r\u00ebnd\u00ebsish\u00ebm. N\u00ebse ekziston materiali p\u00ebr nj\u00eb pun\u00eb t\u00eb gjer\u00eb vetjake, mungesat e shembullit t\u00eb par\u00eb mund t\u00eb korrigjohen. N\u00ebse nuk ka asgj\u00eb tjet\u00ebr p\u00ebrve\u00e7 shembullit, at\u00ebher\u00eb studenti ndoshta nuk do ta kuptoj\u00eb tem\u00ebn.<\/p>\n<h4>While apo for?<\/h4>\n<p>\nNj\u00eb nga \u00e7\u00ebshtjet e diskutueshme \u00ebsht\u00eb zgjedhja e konstrukcionit p\u00ebr shembull: while apo for. Nj\u00ebher\u00eb nj\u00eb mik i imi, zhvillues praktik pa p\u00ebrvoj\u00eb n\u00eb m\u00ebsimdh\u00ebnie, m\u00eb bindte p\u00ebr nj\u00eb or\u00eb se cikli for \u00ebsht\u00eb m\u00eb i thjesht\u00eb p\u00ebr t'u kuptuar. Argumentet p\u00ebrmblidhnin se \"n\u00eb t\u00eb gjith\u00eb \u00ebsht\u00eb e qart\u00eb dhe e vendosur n\u00eb vende\". Megjithat\u00eb, shkaku kryesor i v\u00ebshtir\u00ebsive p\u00ebr fillestar\u00ebt n\u00eb t\u00eb v\u00ebrtet\u00eb \u00ebsht\u00eb vet\u00eb ideja e ciklit, dhe jo shkrimi i tij. N\u00ebse nj\u00eb person nuk e kupton k\u00ebt\u00eb ide, ai do t\u00eb ket\u00eb v\u00ebshtir\u00ebsi me sintaks\u00ebn. Sapo ideja t\u00eb kuptohet, problemet e formatimit t\u00eb kodit zhduken vet\u00eb.<\/p>\n<p>N\u00eb materialet e mia, tema e cikleve ndjek tem\u00ebn mbi rrethit. Ngjashm\u00ebria e jashtme nd\u00ebrmjet if dhe while lejon t\u00eb b\u00ebhet nj\u00eb analogji e drejtp\u00ebrdrejt\u00eb: \"kur kushte n\u00eb titull \u00ebsht\u00eb e v\u00ebrtet\u00eb, at\u00ebher\u00eb ekzekutohet trupi\". Karakteristika e ciklit \u00ebsht\u00eb se trupi ekzekutohet shum\u00eb her\u00eb.<\/p>\n<p>Argumenti im i dyt\u00eb \u00ebsht\u00eb se while k\u00ebrkon m\u00eb pak formatim se for. M\u00eb pak formatim - m\u00eb pak gabime t\u00eb budallaq\u00ebve me munges\u00eb t\u00eb pik\u00ebv\u00ebnieve dhe kllapave. Fillestar\u00ebt ende nuk e kan\u00eb t\u00eb zhvilluar v\u00ebmendjen dhe k\u00ebrkes\u00ebn p\u00ebr detaje sa p\u00ebr t'u shmangur automatikisht gabimet sintaksore.<br \/>\nArgumenti i tret\u00eb \u00ebsht\u00eb se n\u00eb shum\u00eb libra t\u00eb mir\u00eb while shpjegohet i pari.<\/p>\n<p>N\u00ebse studenti arrin t\u00eb transformoj\u00eb leht\u00ebsisht shprehjet, mund t\u00eb flasim p\u00ebr for kalimthi. Studentit m\u00eb pas do t'i p\u00eblqej\u00eb t\u00eb zgjedh\u00eb at\u00eb q\u00eb i p\u00eblqen m\u00eb shum\u00eb. N\u00ebse transformimet shkaktojn\u00eb v\u00ebshtir\u00ebsi, \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb mos shp\u00ebrdorojm\u00eb v\u00ebmendjen. Nj\u00ebher\u00eb studentit le t\u00eb zgjidh\u00eb gjith\u00e7ka me ndihm\u00ebn e while. Kur tema e cikleve \u00ebsht\u00eb e kuptuar, mund t\u00eb riprojektohen zgjidhjet p\u00ebr t\u00eb ushtruar transformimin e while n\u00eb for.<br \/>\nCiklet me post-kusht jan\u00eb nj\u00eb kafsho e rrall\u00eb. P\u00ebr k\u00ebt\u00eb nuk e humbas fare koh\u00eb. N\u00ebse studenti ka kuptuar idet\u00eb e identifikimit t\u00eb rregullave dhe transformimit t\u00eb shprehjeve, ai do t\u00eb jet\u00eb n\u00eb gjendje t\u00eb kuptoj\u00eb pa ndihm\u00ebn time.<\/p>\n<p>Kur duke demonstruar shembullin e par\u00eb p\u00ebr student\u00ebt e fuqish\u00ebm, theksoj se \u00ebsht\u00eb e r\u00ebnd\u00ebsishme t\u00eb regjistrohet jo vet\u00ebm zgjidhja, por edhe e gjith\u00eb zinxhiri i veprimeve q\u00eb \u00e7uan n\u00eb rezultat. Student\u00ebt e len\u00eb mund t\u00eb injorojn\u00eb shkrimin dhe t\u00eb ruajn\u00eb vet\u00ebm algoritmin p\u00ebrfundimtar p\u00ebr veten e tyre. Duhet t'i bindim ata se nj\u00eb her\u00eb ata do t\u00eb hasin nj\u00eb detyr\u00eb t\u00eb komplikuar. P\u00ebr ta zgjidhur, do t\u00eb duhet t\u00eb ndjekin hapat si n\u00eb k\u00ebt\u00eb shembull. Prandaj \u00ebsht\u00eb e r\u00ebnd\u00ebsishme t\u00eb regjistrohen t\u00eb gjitha etapet. N\u00eb detyrat e ardhshme, mund t\u00eb l\u00eb vet\u00ebm variantin p\u00ebrfundimtar t\u00eb zgjidhjes.<\/p>\n<p>Ideja kryesore e automatizimit \u00ebsht\u00eb q\u00eb ne i besojm\u00eb kompjuterit t\u00eb kryej\u00eb pun\u00eb rutin\u00eb n\u00eb vend t\u00eb njeriut. Nj\u00eb nga teknikat baz\u00eb \u00ebsht\u00eb shkruajtja e cikleve. Ajo aplikohet kur n\u00eb program shkruhen p\u00ebrs\u00ebrit\u00ebsisht disa veprime t\u00eb nj\u00ebjta.<\/p>\n<h4>E qart\u00eb \u00ebsht\u00eb m\u00eb mir\u00eb se e paqart\u00eb<\/h4>\n<p>\nMund t\u00eb duket si nj\u00eb ide e mir\u00eb n\u00eb detyr\u00ebn e par\u00eb p\u00ebr ciklet t\u00eb printohet nj\u00eb fraz\u00eb e nj\u00ebjt\u00eb disa her\u00eb. P\u00ebr shembull:<\/p>\n<blockquote><p>Ura, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!<br \/>\nUra, punon!\n<\/p><\/blockquote>\n<p>\nKy variant \u00ebsht\u00eb i keq sepse n\u00eb printim nuk \u00ebsht\u00eb e dukshme vlera e num\u00ebruesit. Kjo \u00ebsht\u00eb nj\u00eb problem p\u00ebr fillestar\u00ebt. Nuk duhet ta n\u00ebnvler\u00ebsojm\u00eb. N\u00eb fillim, kjo detyr\u00eb ishte e para, dhe detyra p\u00ebr printimin e nj\u00eb vargu numrash n\u00eb rritje ishte e dyta. Duhej t\u00eb introduktoheshin terma t\u00eb shtuar si \"cikli N her\u00eb\" dhe \"cikli nga A deri n\u00eb B\", t\u00eb cilat n\u00eb thelb jan\u00eb nj\u00ebsoj. P\u00ebr t\u00eb mos krijuar entitete t\u00eb panevojshme, vendosa t\u00eb tregoj vet\u00ebm shembullin me printimin e vargut t\u00eb numrave. Disa arritin ta mbajn\u00eb n\u00eb mendje num\u00ebruesin dhe ta modelojn\u00eb sjelljen e programit n\u00eb mendjen e tyre pa p\u00ebrgatitje. Disa student\u00eb p\u00ebrballen p\u00ebr her\u00eb t\u00eb par\u00eb me modelimin \"n\u00eb mendje\" pik\u00ebrisht n\u00eb tem\u00ebn p\u00ebr ciklet.<br \/>\nPas disa praktikave, detyr\u00ebn p\u00ebr t\u00eb p\u00ebrs\u00ebritur tekstin e nj\u00ebjt\u00eb e jap p\u00ebr zgjidhje t\u00eb pavarur. N\u00ebse e japim fillimisht num\u00ebruesin e duksh\u00ebm, e pastaj t\u00eb paduksh\u00ebm, student\u00ebt hasin m\u00eb pak probleme. Ndonj\u00ebher\u00eb \u00ebsht\u00eb mjaft nj\u00eb nxitje \"mos e shkruaj num\u00ebruesin n\u00eb ekran\".<\/p>\n<h4>Si shpjegohet nga t\u00eb tjer\u00ebt?<\/h4>\n<p>\nN\u00eb shumic\u00ebn e materialeve edukative n\u00eb internet, sintaksa e ciklit jepet si pjes\u00eb e \"lektur\u00ebs\". P\u00ebr shembull, n\u00eb developer.mozilla.org (n\u00eb k\u00ebt\u00eb moment) bashk\u00eb me ciklin while p\u00ebrmenden edhe disa konstruksione t\u00eb tjera. N\u00eb k\u00ebt\u00eb m\u00ebnyr\u00eb, jepen ekskluzivisht vet\u00eb konstrukcionet n\u00eb form\u00eb shabllonesh. Rezultati i ekzekutimit t\u00eb tyre p\u00ebrshkruhet me fjal\u00eb, nd\u00ebrsa ilustirimi mungon. Sipas mendimit tim, nj\u00eb qasje e till\u00eb e tem\u00ebs zvog\u00eblon n\u00eb zero dobishm\u00ebrin\u00eb e materialeve t\u00eb tilla. Nx\u00ebn\u00ebsi mund ta rib\u00ebr\u00eb kodin dhe ta ekzekutoj\u00eb vet\u00eb, por nj\u00eb etalon p\u00ebr krahasim ende nevojitet. Si mund t\u00eb kuptoj\u00eb q\u00eb shembulli \u00ebsht\u00eb riparuar sakt\u00eb, n\u00ebse nuk ka \u00e7far\u00eb t\u00eb krahasoj\u00eb rezultatin?<br \/>\nKur jepet vet\u00ebm shablloni, pa shembuj, studentit i b\u00ebhet edhe m\u00eb e v\u00ebshtir\u00eb. Si mund t\u00eb kuptoj\u00eb q\u00eb fragmentet e kodit jan\u00eb vendosur sakt\u00eb n\u00eb shabllon? Mund t\u00eb provoj\u00eb ta shkruaj\u00eb <i>ndonj\u00ebher\u00eb<\/i>, pastaj ta ekzekutoj\u00eb. Por n\u00ebse nuk ka etalon p\u00ebr krahasimin e rezultatit, ekzekutimi gjithashtu nuk ndihmon.<\/p>\n<p>\u0412 \u043a\u0443\u0440\u0441\u0435 \u043f\u043e C++ \u043d\u0430 \u0438\u043d\u0442\u0443\u0438\u0442\u0435 \u0441\u0438\u043d\u0442\u0430\u043a\u0441\u0438\u0441 \u0446\u0438\u043a\u043b\u0430 \u0437\u0430\u043a\u043e\u043f\u0430\u043d \u0432 \u0442\u0440\u0435\u0442\u044c\u0435\u0439 \u0441\u0442\u0440\u0430\u043d\u0438\u0446\u0435 \u043b\u0435\u043a\u0446\u0438\u0438 4 \u043f\u043e \u0442\u0435\u043c\u0435 \u00ab\u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440\u044b\u00bb. \u041f\u0440\u0438 \u043e\u0431\u044a\u044f\u0441\u043d\u0435\u043d\u0438\u0438 \u0441\u0438\u043d\u0442\u0430\u043a\u0441\u0438\u0441\u0430 \u0446\u0438\u043a\u043b\u043e\u0432 \u0434\u0435\u043b\u0430\u044e\u0442 \u043e\u0441\u043e\u0431\u044b\u0439 \u0443\u043f\u043e\u0440 \u043d\u0430 \u0442\u0435\u0440\u043c\u0438\u043d \u00ab\u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440\u00bb. \u0422\u0435\u0440\u043c\u0438\u043d \u043f\u043e\u0434\u0430\u0435\u0442\u0441\u044f \u0432 \u0432\u0438\u0434\u0435 \u043d\u0430\u0431\u043e\u0440\u0430 \u0444\u0430\u043a\u0442\u043e\u0432 \u0432\u0440\u043e\u0434\u0435 \u00ab\u0441\u0438\u043c\u0432\u043e\u043b; \u044d\u0442\u043e \u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440\u00bb, &quot;{} \u044d\u0442\u043e \u0441\u043e\u0441\u0442\u0430\u0432\u043d\u043e\u0439 \u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440&quot;, \u00ab\u0442\u0435\u043b\u043e \u0446\u0438\u043a\u043b\u0430 \u0434\u043e\u043b\u0436\u043d\u043e \u0431\u044b\u0442\u044c \u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440\u043e\u043c\u00bb. \u041c\u043d\u0435 \u0442\u0430\u043a\u043e\u0439 \u043f\u043e\u0434\u0445\u043e\u0434 \u043d\u0435 \u043d\u0440\u0430\u0432\u0438\u0442\u0441\u044f \u0442\u0435\u043c, \u0447\u0442\u043e \u043e\u043d \u043a\u0430\u043a \u0431\u044b \u043f\u0440\u044f\u0447\u0435\u0442 \u0432\u0430\u0436\u043d\u044b\u0435 \u0432\u0437\u0430\u0438\u043c\u043e\u0441\u0432\u044f\u0437\u0438 \u0437\u0430 \u043e\u0434\u043d\u0438\u043c \u0442\u0435\u0440\u043c\u0438\u043d\u043e\u043c. \u0420\u0430\u0437\u0431\u043e\u0440 \u0438\u0441\u0445\u043e\u0434\u043d\u043e\u0433\u043e \u043a\u043e\u0434\u0430 \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u044b \u043d\u0430 \u0442\u0435\u0440\u043c\u044b \u043d\u0430 \u0442\u0430\u043a\u043e\u043c \u0443\u0440\u043e\u0432\u043d\u0435 \u043d\u0443\u0436\u0435\u043d \u0440\u0430\u0437\u0440\u0430\u0431\u043e\u0442\u0447\u0438\u043a\u0430\u043c \u043a\u043e\u043c\u043f\u0438\u043b\u044f\u0442\u043e\u0440\u043e\u0432 \u0434\u043b\u044f \u0440\u0435\u0430\u043b\u0438\u0437\u0430\u0446\u0438\u0438 \u0441\u043f\u0435\u0446\u0438\u0444\u0438\u043a\u0430\u0446\u0438\u0438 \u044f\u0437\u044b\u043a\u0430, \u043d\u043e \u043d\u0438\u043a\u0430\u043a \u043d\u0435 \u0441\u0442\u0443\u0434\u0435\u043d\u0442\u0430\u043c \u0432 \u043f\u0435\u0440\u0432\u043e\u043c \u043f\u0440\u0438\u0431\u043b\u0438\u0436\u0435\u043d\u0438\u0438. \u041d\u043e\u0432\u0438\u0447\u043a\u0438 \u0432 \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0438\u0440\u043e\u0432\u0430\u043d\u0438\u0438 \u0440\u0435\u0434\u043a\u043e \u043e\u0431\u043b\u0430\u0434\u0430\u044e\u0442 \u0434\u043e\u0441\u0442\u0430\u0442\u043e\u0447\u043d\u043e\u0439 \u0434\u043e\u0442\u043e\u0448\u043d\u043e\u0441\u0442\u044c\u044e, \u0447\u0442\u043e\u0431\u044b \u043d\u0430\u0441\u0442\u043e\u043b\u044c\u043a\u043e \u0432\u043d\u0438\u043c\u0430\u0442\u0435\u043b\u044c\u043d\u043e \u043e\u0442\u043d\u043e\u0441\u0438\u0442\u044c\u0441\u044f \u043a \u0442\u0435\u0440\u043c\u0438\u043d\u0430\u043c. \u0420\u0435\u0434\u043a\u0438\u0439 \u0447\u0435\u043b\u043e\u0432\u0435\u043a \u0437\u0430\u043f\u043e\u043c\u0438\u043d\u0430\u0435\u0442 \u0438 \u043f\u043e\u043d\u0438\u043c\u0430\u0435\u0442 \u043d\u043e\u0432\u044b\u0435 \u0441\u043b\u043e\u0432\u0430 \u0441 \u043f\u0435\u0440\u0432\u043e\u0433\u043e \u0440\u0430\u0437\u0430. \u041f\u0440\u0430\u043a\u0442\u0438\u0447\u0435\u0441\u043a\u0438 \u043d\u0438\u043a\u0442\u043e \u043d\u0435 \u043c\u043e\u0436\u0435\u0442 \u043f\u0440\u0430\u0432\u0438\u043b\u044c\u043d\u043e \u043f\u0440\u0438\u043c\u0435\u043d\u0438\u0442\u044c \u0442\u0435\u0440\u043c\u0438\u043d, \u043a\u043e\u0442\u043e\u0440\u044b\u0439 \u0442\u043e\u043b\u044c\u043a\u043e \u0447\u0442\u043e \u0443\u0437\u043d\u0430\u043b. \u041f\u043e\u044d\u0442\u043e\u043c\u0443 \u0443 \u0441\u0442\u0443\u0434\u0435\u043d\u0442\u043e\u0432 \u0432\u043e\u0437\u043d\u0438\u043a\u0430\u0435\u0442 \u043a\u0443\u0447\u0430 \u043e\u0448\u0438\u0431\u043e\u043a \u0432\u0440\u043e\u0434\u0435 \u00ab\u043d\u0430\u043f\u0438\u0441\u0430\u043b while(a&lt;7);{, \u0430 \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0430 \u043d\u0435 \u0440\u0430\u0431\u043e\u0442\u0430\u0435\u0442\u00bb.<br \/>\nSipas mendimit tim, n\u00eb fillim \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb jepet sintaksa e konstrukcionit menj\u00ebher\u00eb me kllapa. Variante pa kllapa t\u00eb shpjegohet vet\u00ebm n\u00ebse nx\u00ebn\u00ebsi ka nj\u00eb pyetje specifike \"pse k\u00ebtu pa kllapa dhe punon\".<\/p>\n<p>N\u00eb librin e Okulov \u00abBazat e Programimit\u00bb t\u00eb vitit 2012, njohja me ciklet fillon me modelin for, pastaj jepen rekomandime p\u00ebr p\u00ebrdorimin e tij, dhe m\u00eb pas, menj\u00ebher\u00eb ndodhet seksioni eksperimental i m\u00ebsimit. Un\u00eb e kuptoj se libri \u00ebsht\u00eb shkruar p\u00ebr at\u00eb pakic\u00eb nx\u00ebn\u00ebsish shum\u00eb t\u00eb aft\u00eb, t\u00eb cil\u00ebt rrall\u00eb vijn\u00eb n\u00eb m\u00ebsimet e mia.<\/p>\n<p>N\u00eb librat popullor\u00eb gjithmon\u00eb jepet rezultati i fragmenteve t\u00eb kodit. P\u00ebr shembull, te Shilti \u00abJava 8. Udh\u00ebzues i plot\u00eb\u00bb e botuar n\u00eb vitin 2015. Fillimisht jepet modeli, pastaj nj\u00eb shembull programi dhe menj\u00ebher\u00eb pas tij \u2014 rezultati i ekzekutimit.<\/p>\n<blockquote><p>Si shembull do t\u00eb marrim ciklin while, n\u00eb t\u00eb cilin kryhet num\u00ebrimi mbrapsht, duke filluar nga 10, dhe sakt\u00ebsisht 10 rreshta \u00abtaktesh\u00bb printohen:<br \/>\n\u043e\u0442\u0441\u0447\u0435\u0442, \u043d\u0430\u0447\u0438\u043d\u0430\u044f \u0441 10, \u0438 \u0432\u044b\u0432\u043e\u0434\u0438\u0442\u0441\u044f \u0440\u043e\u0432\u043d\u043e 10 \u0441\u0442\u0440\u043e\u043a \u00ab\u0442\u0430\u043a\u0442\u043e\u0432\u00bb: <\/p>\n<pre><code class=\"java\">\/\/\u041f\u0440\u043e\u0434\u0435\u043c\u043e\u043d\u0441\u0442\u0440\u0438\u0440\u043e\u0432\u0430\u0442\u044c \u043f\u0440\u0438\u043c\u0435\u043d\u0435\u043d\u0438\u0435 \u043e\u043f\u0435\u0440\u0430\u0442\u043e\u0440\u0430 \u0446\u0438\u043a\u043b\u0430 while\nclass While {\n    public static void main(String args []) {\n        int n = 10;\n        while (n &gt; 0) {\n            System.out.println(&quot;\u0442\u0430\u043a\u0442 &quot; + n);\n            n--;\n        }\n    }\n}<\/code><\/pre>\n<p>\nPas nism\u00ebs, ky program printon dhjet\u00eb \u00abtaktesh\u00bb n\u00eb m\u00ebnyr\u00ebn e m\u00ebposhtme:<br \/>\n<code>takti 10<br \/>\ntakti 9<br \/>\ntakti 8<br \/>\ntakti 7<br \/>\ntakti 6<br \/>\ntakti 5<br \/>\ntakti 4<br \/>\ntakti 3<br \/>\ntakti 2<br \/>\ntakti 1<\/code><\/p><\/blockquote>\n<p>\nPranimi me p\u00ebrshkrimin e modelit, shembullit t\u00eb programit dhe rezultatit t\u00eb funksionimit t\u00eb k\u00ebtij programi p\u00ebrdoret gjithashtu n\u00eb librin \u00abJavascript p\u00ebr f\u00ebmij\u00eb\u00bb dhe n\u00eb kursin js n\u00eb w3schools.com. Formati i faqes s\u00eb internetit madje lejon q\u00eb ky shembull t\u00eb b\u00ebhet interaktiv.<\/p>\n<p>N\u00eb librin e Stroustrup \u00abParimet dhe Praktika me C++\u00bb t\u00eb vitit 2016, autori kishte shkuar edhe m\u00eb tej. S\u00eb pari shpjegohet \u00e7far\u00eb rezultati duhej t\u00eb arrihej, dhe vet\u00ebm m\u00eb pas shfaqet teksti i programit. Madje, si shembull, nuk merret nj\u00eb program i rast\u00ebsish\u00ebm, por jepet nj\u00eb ekskurs n\u00eb histori. Kjo ndihmon t\u00eb t\u00ebrhiqet v\u00ebmendja: \u00abShiko, kjo nuk \u00ebsht\u00eb thjesht nj\u00eb tekst i padobish\u00ebm. Ti sheh di\u00e7ka t\u00eb r\u00ebnd\u00ebsishme\u00bb.<\/p>\n<blockquote><p>Si nj\u00eb shembull i iteracionit, do t\u00eb shqyrtojm\u00eb programin e par\u00eb t\u00eb ekzekutuar n\u00eb nj\u00eb makin\u00eb me program t\u00eb ruajtur (EDSAC). Ai u shkrua nga David Wheeler n\u00eb laboratorin kompjuterik t\u00eb Universitetit t\u00eb Kembrixhit (Cambridge University, Angli) m\u00eb 6 maj 1949. Ky program llogarit dhe printon nj\u00eb list\u00eb t\u00eb thjesht\u00eb t\u00eb katror\u00ebve.<br \/>\n<code>0 0<br \/>\n1 1<br \/>\n2 4<br \/>\n3 9<br \/>\n4 16<br \/>\n...<br \/>\n98 9604<br \/>\n99 9801<\/code><br \/>\n\u0417\u0434\u0435\u0441\u044c \u0432 \u043a\u0430\u0436\u0434\u043e\u0439 \u0441\u0442\u0440\u043e\u043a\u0435 \u0441\u043e\u0434\u0435\u0440\u0436\u0438\u0442\u0441\u044f \u0447\u0438\u0441\u043b\u043e, \u0437\u0430 \u043a\u043e\u0442\u043e\u0440\u044b\u043c \u0441\u043b\u0435\u0434\u0443\u044e\u0442 \u0437\u043d\u0430\u043a \u0442\u0430\u0431\u0443\u043b\u044f\u0446\u0438\u0438 (&#8216;t&#8217;) \u0438 \u043a\u0432\u0430\u0434\u0440\u0430\u0442 \u044d\u0442\u043e\u0433\u043e \u0447\u0438\u0441\u043b\u0430. \u0412\u0435\u0440\u0441\u0438\u044f \u044d\u0442\u043e\u0439 \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u044b \u043d\u0430 \u044f\u0437\u044b\u043a\u0435 C++ \u0432\u044b\u0433\u043b\u044f\u0434\u0438\u0442 \u0442\u0430\u043a:<\/p>\n<pre><code class=\"cpp\">\/\/\u0412\u044b\u0447\u0438\u0441\u043b\u044f\u0435\u043c \u0438 \u0440\u0430\u0441\u043f\u0435\u0447\u0430\u0442\u044b\u0432\u0430\u0435\u043c \u0442\u0430\u0431\u043b\u0438\u0446\u0443 \u043a\u0432\u0430\u0434\u0440\u0430\u0442\u043e\u0432 \u0447\u0438\u0441\u0435\u043b 0-99\nint main()\n{\n    int i = 0; \/\/ \u041d\u0430\u0447\u0438\u043d\u0430\u0435\u043c \u0441 \u043d\u0443\u043b\u044f\n    while(i &lt; 100){\n        cout &lt;&lt; i &lt;&lt; 't' &lt;&lt; square(i) &lt;&lt; 'n';\n        ++i;\n    }\n}<\/code><\/pre>\n<\/blockquote>\n<p>\n\u00cbsht\u00eb interesante q\u00eb modeli i sintaks\u00ebs n\u00eb k\u00ebt\u00eb lib\u00ebr nuk p\u00ebrshkruhet. Stroustrup n\u00eb udh\u00ebzuesin e instruktorit (<noindex><a rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/post\/349342\/\">p\u00ebrkthim<\/a><\/noindex>) thekson se respekton inteligjenc\u00ebn e student\u00ebve t\u00eb tij. Ndoshta aft\u00ebsia p\u00ebr t\u00eb identifikuar modelin n\u00eb disa shembuj dhe konsiderohet nj\u00eb shprehje e till\u00eb e inteligjenc\u00ebs.<\/p>\n<h4>Si\u00e7 e shpjegoj un\u00eb vet\u00eb<\/h4>\n<p>\nQasja e Stroustrup: p\u00ebrshkrimi i rezultatit, pastaj zgjidhja e problemit, dhe m\u00eb pas analiza e pavarur nga studenti \u2014 duket si m\u00eb e menduar. Prandaj vendosa ta marr si baz\u00eb at\u00eb, por do t\u00eb flas p\u00ebr nj\u00eb shembull m\u00eb pak historik \u2014 problemin e shfaqjes s\u00eb \"p\u00ebrmbajtjes\". Kjo formon nj\u00eb pik\u00eb referimi t\u00eb njohur, p\u00ebr t'u th\u00ebn\u00eb m\u00eb pas \"kujto problemin p\u00ebr p\u00ebrmbajtjen\" dhe p\u00ebr student\u00ebt t\u00eb kujtojn\u00eb sakt\u00ebsisht at\u00eb. N\u00eb shembullin tim, p\u00ebrpiqesha t\u00eb parandaloj dy nga keqkuptimet m\u00eb t\u00eb zakonshme. M\u00eb pas do t\u00eb shkruaj p\u00ebr ta m\u00eb holl\u00ebsisht.<\/p>\n<blockquote><p>N\u00eb k\u00ebt\u00eb problem njohim teknikat e zgjidhjes s\u00eb problemeve t\u00eb nd\u00ebrlikuara. Zgjidhja fillestare duhet t\u00eb jet\u00eb primitiv\u00eb dhe e thjesht\u00eb. Pastaj mund t\u00eb mendojm\u00eb p\u00ebr p\u00ebrmir\u00ebsimin e k\u00ebsaj zgjidhjeje.<br \/>\n<code>Hyrje<br \/>\nKapitulli 1<br \/>\nKapitulli 2<br \/>\nKapitulli 3<br \/>\nKapitulli 4<br \/>\nKapitulli 5<br \/>\nKapitulli 6<br \/>\nKapitulli 7<br \/>\nP\u00ebrfundim<\/code><\/p><\/blockquote>\n<p>\n\u041f\u043e \u043c\u043e\u0438\u043c \u043d\u0430\u0431\u043b\u044e\u0434\u0435\u043d\u0438\u044f\u043c, \u043f\u043e\u0434\u0445\u043e\u0434 \u00ab\u0448\u0430\u0431\u043b\u043e\u043d-\u043f\u0440\u0438\u043c\u0435\u0440-\u0440\u0435\u0437\u0443\u043b\u044c\u0442\u0430\u0442\u00bb \u0432 \u0440\u0430\u0437\u043d\u044b\u0445 \u043a\u043e\u043c\u0431\u0438\u043d\u0430\u0446\u0438\u044f\u0445 \u0432\u0441\u0435 \u0440\u0430\u0432\u043d\u043e \u043f\u0440\u0438\u0432\u043e\u0434\u0438\u0442 \u043a \u0442\u043e\u043c\u0443, \u0447\u0442\u043e \u0441\u0442\u0443\u0434\u0435\u043d\u0442\u044b \u0432\u043e\u0441\u043f\u0440\u0438\u043d\u0438\u043c\u0430\u044e\u0442 \u0446\u0438\u043a\u043b \u043a\u0430\u043a \u0438\u0435\u0440\u043e\u0433\u043b\u0438\u0444. \u042d\u0442\u043e \u043f\u0440\u043e\u044f\u0432\u043b\u044f\u043b\u043e\u0441\u044c \u0432 \u0442\u043e\u043c, \u0447\u0442\u043e \u043e\u043d\u0438 \u043d\u0435 \u043f\u043e\u043d\u0438\u043c\u0430\u043b\u0438 \u0437\u0430\u0447\u0435\u043c \u0442\u0430\u043c \u043f\u0438\u0441\u0430\u0442\u044c \u0443\u0441\u043b\u043e\u0432\u0438\u0435, \u043a\u0430\u043a \u0432\u044b\u0431\u0438\u0440\u0430\u0442\u044c \u043c\u0435\u0436\u0434\u0443 i++ \u0438 i&#8212; \u0438 \u043f\u0440\u043e\u0447\u0438\u0435 \u0432\u0440\u043e\u0434\u0435 \u0431\u044b \u043e\u0447\u0435\u0432\u0438\u0434\u043d\u044b\u0435 \u0432\u0435\u0449\u0438. \u0414\u043b\u044f \u0438\u0437\u0431\u0435\u0436\u0430\u043d\u0438\u044f \u044d\u0442\u0438\u0445 \u0437\u0430\u0431\u043b\u0443\u0436\u0434\u0435\u043d\u0438\u0439, \u043f\u043e\u0434\u0445\u043e\u0434 \u043a \u0440\u0430\u0441\u0441\u043a\u0430\u0437\u0443 \u043e \u0446\u0438\u043a\u043b\u0430\u0445 \u0434\u043e\u043b\u0436\u0435\u043d \u043f\u043e\u0434\u0447\u0451\u0440\u043a\u0438\u0432\u0430\u0442\u044c \u0441\u043c\u044b\u0441\u043b \u043f\u043e\u0432\u0442\u043e\u0440\u0435\u043d\u0438\u044f \u043e\u0434\u0438\u043d\u0430\u043a\u043e\u0432\u044b\u0445 \u0434\u0435\u0439\u0441\u0442\u0432\u0438\u0439 \u0438 \u0442\u043e\u043b\u044c\u043a\u043e \u043f\u043e\u0442\u043e\u043c \u2014 \u043e\u0444\u043e\u0440\u043c\u043b\u0435\u043d\u0438\u0435 \u0438\u0445 \u0441 \u043f\u043e\u043c\u043e\u0449\u044c\u044e \u043a\u043e\u043d\u0441\u0442\u0440\u0443\u043a\u0446\u0438\u0438. \u041f\u043e\u044d\u0442\u043e\u043c\u0443 \u043f\u0440\u0435\u0436\u0434\u0435 \u0447\u0435\u043c \u0434\u0430\u0432\u0430\u0442\u044c \u0441\u0438\u043d\u0442\u0430\u043a\u0441\u0438\u0441 \u0446\u0438\u043a\u043b\u0430, \u043d\u0443\u0436\u043d\u043e \u0440\u0435\u0448\u0438\u0442\u044c \u0437\u0430\u0434\u0430\u0447\u0443 \u00ab\u0432 \u043b\u043e\u0431\u00bb. \u041f\u0440\u0438\u043c\u0438\u0442\u0438\u0432\u043d\u043e\u0435 \u0440\u0435\u0448\u0435\u043d\u0438\u0435 \u0437\u0430\u0434\u0430\u0447\u0438 \u043f\u0440\u043e \u043e\u0433\u043b\u0430\u0432\u043b\u0435\u043d\u0438\u0435 \u0432\u044b\u0433\u043b\u044f\u0434\u0438\u0442 \u0442\u0430\u043a:<\/p>\n<pre><code class=\"cs\">Console.WriteLine(&quot;\u0412\u0432\u0435\u0434\u0435\u043d\u0438\u0435&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 1&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 2&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 3&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 4&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 5&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 6&quot;);\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 7&quot;);\nConsole.WriteLine(&quot;\u0417\u0430\u043a\u043b\u044e\u0447\u0435\u043d\u0438\u0435&quot;);\n<\/code><\/pre>\n<p>\nSi mund t\u00eb p\u00ebrmir\u00ebsohet?<br \/>\nZ\u00ebvend\u00ebsojeni veprimet e nj\u00ebtrajtshme me nj\u00eb cik\u00ebl.<br \/>\nCilat veprime p\u00ebrs\u00ebriten k\u00ebtu nj\u00ebra pas tjetr\u00ebs pa ndryshime?<br \/>\nN\u00eb k\u00ebt\u00eb fragment nuk ka t\u00eb till\u00eb. Megjithat\u00eb, komandat p\u00ebr t\u00eb printuar fjal\u00ebn \"Kapitulli\" me num\u00ebr jan\u00eb shum\u00eb t\u00eb ngjashme me nj\u00ebra-tjetr\u00ebn.<br \/>\nPrandaj, hapi tjet\u00ebr \u00ebsht\u00eb t\u00eb k\u00ebrkoni dallimin midis fragmenteve. Kjo vet\u00ebm n\u00eb k\u00ebt\u00eb problem \u00ebsht\u00eb e qart\u00eb, m\u00eb von\u00eb do t\u00eb p\u00ebrs\u00ebriten jo komandat e ve\u00e7anta, por blloqet e kodit nga 5 rreshta e m\u00eb shum\u00eb. Do t\u00eb duhet t\u00eb k\u00ebrkoni jo vet\u00ebm n\u00eb list\u00ebn e komandave, por n\u00eb konstruktet e deklasifikimit ose ciklit.<br \/>\nN\u00eb shembullin, dallimi midis komandave \u00ebsht\u00eb n\u00eb numrin pas fjal\u00ebs \"Kapitulli\".<br \/>\nSapo t\u00eb gjendet dallimi, duhet t\u00eb kuptohet ligj\u00ebsia e ndryshimit. A \u00ebsht\u00eb fragmeti q\u00eb ndryshon nj\u00eb num\u00ebr? A rritet apo zvog\u00eblohet vazhdimisht? Si ndryshon vlera e numrit midis dy komandave ngjitur?<br \/>\nN\u00eb shembullin e dh\u00ebn\u00eb, numri pas fjal\u00ebs \u201cKapitulli\u201d rritet me hapin 1. Diferenca \u00ebsht\u00eb gjetur, dhe nj\u00eb model \u00ebsht\u00eb zbuluar. Tani mund t\u00eb z\u00ebvend\u00ebsojm\u00eb fragmentin q\u00eb ndryshon me nj\u00eb variab\u00ebl.<br \/>\n\u00cbsht\u00eb e nevojshme t\u00eb shpallim k\u00ebt\u00eb variab\u00ebl para fragmentit t\u00eb par\u00eb q\u00eb p\u00ebrs\u00ebritet. Kjo variab\u00ebl zakonisht quhet I ose j ose ndonj\u00eb em\u00ebr m\u00eb t\u00eb gjat\u00eb. Vlera e saj fillestare duhet t\u00eb jet\u00eb e barabart\u00eb me vler\u00ebn e par\u00eb q\u00eb shfaqet n\u00eb ekran. N\u00eb shembullin, vlera e par\u00eb \u00ebsht\u00eb 1. <br \/>\nCila \u00ebsht\u00eb vlere fillestare q\u00eb duhet t\u00eb merret p\u00ebr t\u00eb shfaqur serin\u00eb e numrave \u00ab100, 101, 102, 103, 104, 105\u00bb?<br \/>\nN\u00eb k\u00ebt\u00eb seri, numri i par\u00eb \u00ebsht\u00eb 100.<br \/>\nPas \u00e7do komande p\u00ebr shfaqjen e numrave, duhet t\u00eb rritet vlera e k\u00ebsaj variabli me 1. Ky nj\u00eb \u00ebsht\u00eb hapi i ndryshimit.<br \/>\nCili do t\u00eb jet\u00eb hapi n\u00eb serin\u00eb e numrave \u00ab100, 102, 104, 106\u00bb?<br \/>\nN\u00eb k\u00ebt\u00eb seri hapi \u00ebsht\u00eb 2.<br \/>\nPas z\u00ebvend\u00ebsimit t\u00eb fragmentit q\u00eb ndryshon me nj\u00eb variab\u00ebl, kodi do t\u00eb duket k\u00ebshtu:<\/p>\n<pre><code class=\"cs\">Console.WriteLine(&quot;\u0412\u0432\u0435\u0434\u0435\u043d\u0438\u0435&quot;);\nint i;\ni = 0;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\ni = i + 1;\nConsole.WriteLine(&quot;\u0417\u0430\u043a\u043b\u044e\u0447\u0435\u043d\u0438\u0435&quot;);\n<\/code><\/pre>\n<p>\nPas aplikimit t\u00eb teknik\u00ebs \"shprehja e modelit me variab\u00ebl\", n\u00eb kod rezulton disa grupe veprimesh t\u00eb nj\u00ebjta, t\u00eb cilat vijn\u00eb nj\u00ebra pas tjetr\u00ebs. Tani veprimet e p\u00ebrs\u00ebritura mund t\u00eb z\u00ebvend\u00ebsohen me nj\u00eb cik\u00ebl.<\/p>\n<p>Seqeuca e zgjidhjes s\u00eb problemit, ku do t\u00eb nevojiten ciklet, p\u00ebrb\u00ebhet nga etapet:<\/p>\n<ol>\n<li>Ta zgjidhim \"n\u00eb m\u00ebnyr\u00eb direkte\" me shum\u00eb komanda t\u00eb ve\u00e7anta<\/li>\n<li>T\u00eb gjejm\u00eb nj\u00eb model<\/li>\n<li>T\u00eb shprehni modelin me nj\u00eb variab\u00ebl<\/li>\n<li>T\u00eb formohet si nj\u00eb cik\u00ebl<\/li>\n<\/ol>\n<p>M\u00eb pas futen terma t\u00eb rinj, p\u00ebr t\u00eb mos l\u00ebn\u00eb studentin n\u00eb situat\u00ebn \"t\u00eb gjith\u00eb e kuptoj, por nuk mund t\u00eb flas\":<br \/>\n \u2014 num\u00ebrues \u2014 gjithmon\u00eb nj\u00eb variab\u00ebl, e cila \u00ebsht\u00eb e nevojshme p\u00ebr t\u00eb ndjekur numrin e hapave t\u00eb ciklit. Zakonisht nj\u00eb num\u00ebr t\u00ebr\u00eb, q\u00eb krahasohet me kufirin.<br \/>\n \u2014 hapi i num\u00ebruesit \u2014 p\u00ebrshkrimi i modelit t\u00eb ndryshimit t\u00eb num\u00ebruesit.<br \/>\n \u2014 kufiri \u2014 numri ose variabla, me t\u00eb cilin krahasohet num\u00ebruesi, n\u00eb m\u00ebnyr\u00eb q\u00eb algoritmi t\u00eb jet\u00eb p\u00ebrfundimtar. Vlera e num\u00ebruesit ndryshon n\u00eb m\u00ebnyr\u00eb q\u00eb t\u00eb afrohet me kufirin.<br \/>\n \u2014 trupi i ciklit \u2014 grupi i komandave, q\u00eb do t\u00eb p\u00ebrs\u00ebriten. Kur thuhet \"komanda \u00ebsht\u00eb shkruar brenda ciklit\", n\u00ebnkuptojm\u00eb sakt\u00ebsisht trupin.<br \/>\n \u2014 iterimi i ciklit \u2014 nj\u00eb her\u00eb kryerja e trupit t\u00eb ciklit.<br \/>\n \u2014 kushti i ciklit \u2014 shprehje logjike, nga e cila varet n\u00ebse do t\u00eb realizohet nj\u00eb iteracion tjet\u00ebr. (K\u00ebtu mund t\u00eb ket\u00eb ngat\u00ebrrim me strukturat e deg\u00ebzimit)<br \/>\nDuhet t\u00eb jeni t\u00eb gatsh\u00ebm q\u00eb n\u00eb fillim student\u00ebt do t\u00eb p\u00ebrdorin terminologjin\u00eb ndryshe nga ajo q\u00eb \u00ebsht\u00eb e duhura. Kjo i p\u00ebrket si atyre t\u00eb fort\u00eb, ashtu edhe atyre t\u00eb dob\u00ebt. Krijimi i nj\u00eb gjuhe t\u00eb p\u00ebrbashk\u00ebt \u00ebsht\u00eb nj\u00eb art i t\u00ebr\u00eb. Tani do ta shkruaj shkurt: duhet t\u00eb vendosni detyr\u00ebn \"ndara fragmentin e kodit me &lt;termin&gt;\" dhe t\u00eb p\u00ebrdorni k\u00ebto terma sakt\u00eb n\u00eb bised\u00eb.<br \/>\nPas transformimit me ciklin, rezulton ky fragment:<\/p>\n<pre><code class=\"cs\">Console.WriteLine(&quot;\u0412\u0432\u0435\u0434\u0435\u043d\u0438\u0435&quot;);\nint i = 0;\nwhile (i &lt; 7) {\n    Console.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\n    i = i + 1;\n}\nConsole.WriteLine(&quot;\u0417\u0430\u043a\u043b\u044e\u0447\u0435\u043d\u0438\u0435&quot;);<\/code><\/pre>\n<h4>Mendimi kryesor<\/h4>\n<p>\nNj\u00eb ngat\u00ebrres\u00eb e njohur te student\u00ebt \u00ebsht\u00eb se ata vendosin brenda struktur\u00ebs s\u00eb ciklit veprime q\u00eb duhet t\u00eb b\u00ebhen vet\u00ebm nj\u00eb her\u00eb. P\u00ebr shembull, k\u00ebshtu:<\/p>\n<pre><code class=\"cs\">;\nint i = 0;\nwhile (i &lt; 7) {\n    Console.WriteLine(&quot;\u0412\u0432\u0435\u0434\u0435\u043d\u0438\u0435&quot;)\n    Console.WriteLine(&quot;\u0413\u043b\u0430\u0432\u0430 &quot; + i);\n    i = i + 1;\n    Console.WriteLine(&quot;\u0417\u0430\u043a\u043b\u044e\u0447\u0435\u043d\u0438\u0435&quot;);\n}\n<\/code><\/pre>\n<p>\nNx\u00ebn\u00ebsit vazhdimisht hasin k\u00ebt\u00eb problem, si n\u00eb fillim ashtu edhe n\u00eb detyrat m\u00eb t\u00eb komplikuara.<br \/>\nK\u00ebshilla kryesore n\u00eb k\u00ebt\u00eb rast:<\/p>\n<blockquote><p>Sa her\u00eb duhet t\u00eb p\u00ebrs\u00ebritet ekzekutimi i komand\u00ebs: nj\u00eb her\u00eb apo shum\u00eb?\n<\/p><\/blockquote>\n<p>\nKomandat p\u00ebr t\u00eb shfaqur fjal\u00ebt \"Hyrja\" dhe \"P\u00ebrfundimi\", si dhe deklarimi dhe inicializimi i ndryshores i, nuk jan\u00eb t\u00eb ngjashme me veprimet e tjera q\u00eb p\u00ebrs\u00ebriten. Ato kryhen vet\u00ebm nj\u00eb her\u00eb, prandaj duhet t'i shkruajm\u00eb jasht\u00eb trupit t\u00eb ciklit.<\/p>\n<p>N\u00eb kod duhet t\u00eb mbeten t\u00eb tri fazat e zgjidhjes, n\u00eb m\u00ebnyr\u00eb q\u00eb m\u00eb pas t\u00eb referohemi te to n\u00eb rast t\u00eb ndonj\u00eb v\u00ebshtir\u00ebsie. Dy variantet e para mjafton t'i komentoni, q\u00eb t\u00eb mos shqet\u00ebsojn\u00eb.<br \/>\nV\u00ebmendja e studentit duhet t\u00eb fokusohet n\u00eb faktet e m\u00ebposhtme:<br \/>\n \u2014 N\u00eb kushtin e ciklit zakonisht krahasohet num\u00ebruesi dhe kufiri. Num\u00ebruesi mund t\u00eb ndryshoj\u00eb n\u00eb trupin e ciklit, nd\u00ebrsa kufiri \u2014 jo. P\u00ebr ta shkelur k\u00ebt\u00eb rregull, jan\u00eb t\u00eb nevojshme arsye t\u00eb forta.<br \/>\n \u2014 Komandat p\u00ebr t\u00eb shfaqur fjal\u00ebt \"Hyrja\" dhe \"P\u00ebrfundimi\" ndodhen jasht\u00eb trupit t\u00eb ciklit. Na nevojitet t'i ekzekutojm\u00eb 1 her\u00eb. \"Hyrja\" \u2014 para p\u00ebrs\u00ebritjes s\u00eb veprimeve, \"P\u00ebrfundimi\" \u2014 pas.<br \/>\nN\u00eb procesin e konsolidimit t\u00eb k\u00ebsaj teme, mastering t\u00eb m\u00ebvonsh\u00ebm, si dhe p\u00ebr t\u00eb sqaruar nd\u00ebrlikimet, edhe student\u00ebt e fort\u00eb \u00ebsht\u00eb e dobishme t'i b\u00ebjn\u00eb pyetje: \"Dhe ky veprim sa her\u00eb duhet t\u00eb kryhet? Nj\u00eb apo shum\u00eb?\".<\/p>\n<h4>Zhvillimi i aft\u00ebsive t\u00eb tjera<\/h4>\n<p>\nGjat\u00eb procesit t\u00eb studimit t\u00eb cikleve, student\u00ebt ende ushtrojn\u00eb aft\u00ebsin\u00eb p\u00ebr diagnoz\u00ebn dhe zgjidhjen e problemeve. P\u00ebr t\u00eb kryer diagnostikimin, studenti duhet t\u00eb paraqes\u00eb rezultatin e d\u00ebshiruar dhe ta krahasoj\u00eb at\u00eb me rezultatin aktual. Veprimet p\u00ebr t\u00eb korrigjuar varen nga dallimi midis tyre.<br \/>\nPasi student\u00ebt n\u00eb k\u00ebt\u00eb faz\u00eb ende nuk e kuptojn\u00eb mir\u00eb rezultatin e \"d\u00ebshiruar\", ata mund t\u00eb orientojn\u00eb n\u00eb t\u00eb dh\u00ebnat testuese. N\u00eb p\u00ebrgjith\u00ebsi, askush n\u00eb k\u00ebt\u00eb faz\u00eb nuk e kupton ende se \u00e7far\u00eb mund t\u00eb shkoj\u00eb keq dhe si t\u00eb merret me k\u00ebt\u00eb. Prandaj, un\u00eb jap p\u00ebr sh\u00ebnim n\u00eb nj\u00eb ditar p\u00ebrshkrimin e problemeve tipike dhe disa m\u00ebnyra p\u00ebr t'i zgjidhur ato. Zgjedhja e asaj m\u00eb t\u00eb p\u00ebrshtatshme \u00ebsht\u00eb detyr\u00eb e vet\u00eb studentit.<br \/>\nSh\u00ebnimi \u00ebsht\u00eb i nevojsh\u00ebm p\u00ebr t\u00eb pyetur \"u arrit ajo q\u00eb pritej?\", \"Cila nga k\u00ebto situata ndodhi tani?\", \"A ndihmoi zgjidhja e aplikuar?\".<\/p>\n<ol>\n<li>Numri i veprimeve \u00ebsht\u00eb 1 m\u00eb pak ose m\u00eb shum\u00eb se sa pritej. M\u00ebnyrat e zgjidhjes:<br \/>\n \u2014 rritni vler\u00ebn fillestare t\u00eb num\u00ebruesit me 1.<br \/>\n \u2014 z\u00ebvend\u00ebsoni operatorin e rrept\u00eb t\u00eb krahasimit (&lt; ose &gt;) me at\u00eb t\u00eb pa rrept\u00eb (&lt;= ose &gt;=).<br \/>\n \u2014 ndryshoni vler\u00ebn e kufirit me 1.<\/li>\n<li>Veprimet n\u00eb cik\u00ebl ekzekutohen pa ndales\u00eb, n\u00eb m\u00ebnyr\u00eb t\u00eb pafund. M\u00ebnyrat e zgjidhjes:<br \/>\n \u2014 shtoni komand\u00ebn p\u00ebr ndryshimin e num\u00ebruesit n\u00ebse ajo mungon.<br \/>\n \u2014 rregulloni komand\u00ebn p\u00ebr ndryshimin e num\u00ebruesit n\u00eb m\u00ebnyr\u00eb q\u00eb vlera e tij t\u00eb b\u00ebhet m\u00eb e af\u00ebrt me kufirin.<br \/>\n \u2014 hiqni komand\u00ebn p\u00ebr ndryshimin e kufirit, n\u00ebse ajo ndodhet n\u00eb trupin e ciklit.<\/li>\n<li>Numri i veprimeve n\u00eb cik\u00ebl \u00ebsht\u00eb m\u00eb shum\u00eb se 1 m\u00eb pak ose m\u00eb shum\u00eb se sa pritej. Nj\u00eb veprim n\u00eb cik\u00ebl nuk \u00ebsht\u00eb ekzekutuar asnj\u00ebher\u00eb. Fillimisht, duhet t\u00eb zbulohet vlerat e v\u00ebrteta t\u00eb variablave direkt para fillimit t\u00eb ciklit. M\u00ebnyrat e zgjidhjes:<br \/>\n \u2014 ndryshoni vler\u00ebn fillestare t\u00eb kufirit<br \/>\n \u2014 ndryshoni vler\u00ebn fillestare t\u00eb num\u00ebruesit<\/li>\n<\/ol>\n<p>\nN\u00eb p\u00ebrgjith\u00ebsi, problemi 3 lidhet me p\u00ebrdorimin e variabl\u00ebs s\u00eb gabuar ose me mos-anulimin e num\u00ebruesit.<\/p>\n<p>Pas k\u00ebtij shpjegimi, studenti ende mund t\u00eb ket\u00eb keqkuptime t\u00eb ndryshme n\u00eb lidhje me funksionimin e cikleve.<br \/>\nP\u00ebr t\u00eb shuar k\u00ebto m\u00eb t\u00eb zakonshmet, jap detyra:<\/p>\n<ol>\n<li>N\u00eb t\u00eb cil\u00ebn kufiri, vlera fillestare e num\u00ebruesit ose hapi i num\u00ebruesit e fut p\u00ebrdoruesi.<\/li>\n<li>N\u00eb t\u00eb cil\u00ebn vlera e num\u00ebruesit duhet t\u00eb p\u00ebrdoret n\u00eb nj\u00eb shprehje aritmetike. Preferohet q\u00eb t\u00eb jet\u00eb me num\u00ebrues n\u00eb shprehjen n\u00ebnrr\u00ebnj\u00eb ose n\u00eb em\u00ebrues, n\u00eb m\u00ebnyr\u00eb q\u00eb dallimi t\u00eb jet\u00eb jo-linear.<\/li>\n<li>Ku nj\u00eb q\u00eb vlera e num\u00ebruesit nuk shfaqet n\u00eb ekran gjat\u00eb pun\u00ebs s\u00eb ciklit. P\u00ebr shembuj, shfaqja e numrit t\u00eb nevojsh\u00ebm t\u00eb fragmenteve t\u00eb nj\u00ebjta t\u00eb tekstit ose t\u00eb vizatoni nj\u00eb figur\u00eb me grafik\u00ebn e breshk\u00ebs.<\/li>\n<li>Ku duhet t\u00eb kryhen fillimisht disa veprime t\u00eb p\u00ebrs\u00ebritura, dhe pastaj t\u00eb tjera.<\/li>\n<li>Ku duhet t\u00eb kryhen veprime t\u00eb tjera para dhe pas p\u00ebrs\u00ebritjeve.<\/li>\n<\/ol>\n<p>\nP\u00ebr \u00e7do detyr\u00eb \u00ebsht\u00eb e nevojshme t\u00eb jepen t\u00eb dh\u00ebnat testuese dhe rezultati i pritur.<\/p>\n<p>P\u00ebr t\u00eb kuptuar sa shpejt mund t\u00eb l\u00ebvize, \u00ebsht\u00eb e nevojshme t\u00eb lejohet t\u00eb lexohen kushtet e k\u00ebtyre detyrave dhe t\u00eb pyeten: \"\u00e7far\u00eb i ndryshon nga shembulli?\", \"\u00c7far\u00eb duhet t\u00eb ndryshoj\u00eb n\u00eb shembull p\u00ebr t'i zgjidhur ato?\". N\u00ebse studenti p\u00ebrgjigjet me kuptim, at\u00ebher\u00eb le t\u00eb zgjidh\u00eb t\u00eb pakt\u00ebn nj\u00eb n\u00eb m\u00ebsim, nd\u00ebrsa t\u00eb tjerat - vet\u00eb n\u00eb sht\u00ebpi. N\u00ebse zgjidhja do t\u00eb jet\u00eb e suksesshme, mund t\u00eb filloni t\u00eb shpjegoni rreth kushteve brenda cikleve.<br \/>\nN\u00ebse ka v\u00ebshtir\u00ebsi me zgjidhjen e pavarur, \u00ebsht\u00eb e nevojshme t\u00eb punoni gjith\u00e7ka n\u00eb m\u00ebsim. P\u00ebr t\u00eb mos e b\u00ebr\u00eb zgjidhjen e detyr\u00ebs t\u00eb duket si vizatimi i nj\u00eb sorkadhe, rekomandoj fillimisht t\u00eb zgjidhni detyr\u00ebn jo n\u00eb m\u00ebnyr\u00eb universale. Pra, p\u00ebr ta zgjidhur at\u00eb n\u00eb m\u00ebnyr\u00eb q\u00eb zgjidhja t\u00eb kaloj\u00eb testin e par\u00eb dhe t\u00eb mos p\u00ebrdor\u00eb nd\u00ebrtimin e ciklit. Pastaj mund t\u00eb aplikoni transformime p\u00ebr t\u00eb arritur universialitetin e zgjidhjes.<\/p>\n<h4>Ciklet dhe deget<\/h4>\n<p>\nSipas mendimit tim, \u00ebsht\u00eb e dobishme t\u00eb jepet tem\u00eb \"ciklet brenda deg\u00ebve\" ve\u00e7mas. K\u00ebshtu q\u00eb m\u00eb pas t\u00eb duket ndarja midis kontrollit t\u00eb shum\u00ebfisht\u00eb t\u00eb kushteve dhe atij nj\u00eb her\u00eb.<br \/>\nDetyrat p\u00ebr konsolidim do t\u00eb p\u00ebrmendin shfaqjen e numrave nga A n\u00eb B, t\u00eb cilat futen nga p\u00ebrdoruesi:<br \/>\n \u2014 gjithmon\u00eb n\u00eb rritje.<br \/>\n \u2014 n\u00eb rritje ose n\u00eb r\u00ebnien e varur nga vlerat e A dhe B.<\/p>\n<p>\u041a \u0442\u0435\u043c\u0435 \u00ab\u0432\u0435\u0442\u0432\u043b\u0435\u043d\u0438\u044f \u0432\u043d\u0443\u0442\u0440\u0438 \u0446\u0438\u043a\u043b\u043e\u0432\u00bb \u043d\u0443\u0436\u043d\u043e \u043f\u0435\u0440\u0435\u0445\u043e\u0434\u0438\u0442\u044c \u0442\u043e\u043b\u044c\u043a\u043e \u043f\u043e\u0441\u043b\u0435 \u0442\u043e\u0433\u043e, \u043a\u0430\u043a \u0441\u0442\u0443\u0434\u0435\u043d\u0442 \u043e\u0441\u0432\u043e\u0438\u043b \u043f\u0440\u0438\u0435\u043c\u044b: \u00ab\u0437\u0430\u043c\u0435\u043d\u0430 \u0437\u0430\u043a\u043e\u043d\u043e\u043c\u0435\u0440\u043d\u043e\u0441\u0442\u0438 \u043d\u0430 \u043f\u0435\u0440\u0435\u043c\u0435\u043d\u043d\u0443\u044e\u00bb \u0438 &quot; \u0437\u0430\u043c\u0435\u043d\u0430 \u043f\u043e\u0432\u0442\u043e\u0440\u044f\u044e\u0449\u0438\u0445\u0441\u044f \u0434\u0435\u0439\u0441\u0442\u0432\u0438\u0439 \u043d\u0430 \u0446\u0438\u043a\u043b&quot;.<br \/>\nArsyeja kryesore p\u00ebr aplikimin e deg\u00ebve brenda cikleve \u00ebsht\u00eb anomali n\u00eb model. N\u00eb mes t\u00eb saj, ajo shkelet n\u00eb var\u00ebsi t\u00eb t\u00eb dh\u00ebnave fillestare.<br \/>\nAtyre student\u00ebve q\u00eb jan\u00eb n\u00eb gjendje t\u00eb k\u00ebrkojn\u00eb zgjidhje duke kombinuar teknik\u00eb t\u00eb thjeshta, \u00ebsht\u00eb mjaft t\u00eb thuhet \"deget mund t\u00eb shkruhen brenda cikleve\" dhe t\u00eb jepet nj\u00eb detyr\u00eb \"p\u00ebr shembull\" plot\u00ebsisht p\u00ebr zgjidhje t\u00eb pavarur.<br \/>\nDetyra p\u00ebr shembull:<\/p>\n<blockquote><p>\u041f\u043e\u043b\u044c\u0437\u043e\u0432\u0430\u0442\u0435\u043b\u044c \u0432\u0432\u043e\u0434\u0438\u0442 \u0447\u0438\u0441\u043b\u043e \u0425. \u0412\u044b\u0432\u0435\u0441\u0442\u0438 \u0432 \u0441\u0442\u043e\u043b\u0431\u0438\u043a \u0447\u0438\u0441\u043b\u0430 \u043e\u0442 0 \u0434\u043e 9 \u0438 \u043f\u043e\u0441\u0442\u0430\u0432\u0438\u0442\u044c \u0437\u043d\u0430\u043a &#8216;+&#8217; \u043d\u0430\u043f\u0440\u043e\u0442\u0438\u0432 \u0442\u043e\u0433\u043e \u0447\u0438\u0441\u043b\u0430, \u043a\u043e\u0442\u043e\u0440\u043e\u0435 \u0440\u0430\u0432\u043d\u043e \u0425.<\/p><\/blockquote>\n<p>\n<b class=\"spoiler_title\">N\u00ebse \u00ebsht\u00eb futur 0<\/b>0+<br \/>\n1<br \/>\n2<br \/>\n3<br \/>\n4<br \/>\n5<br \/>\n6<br \/>\n7<br \/>\n8<br \/>\n9<\/p>\n<p><b class=\"spoiler_title\">N\u00ebse \u00ebsht\u00eb futur 6<\/b>0<br \/>\n1<br \/>\n2<br \/>\n3<br \/>\n4<br \/>\n5<br \/>\n6+<br \/>\n7<br \/>\n8<br \/>\n9<\/p>\n<p><b class=\"spoiler_title\">N\u00ebse \u00ebsht\u00eb futur 9<\/b>0<br \/>\n1<br \/>\n2<br \/>\n3<br \/>\n4<br \/>\n5<br \/>\n6<br \/>\n7<br \/>\n8<br \/>\n9+<\/p>\n<p><b class=\"spoiler_title\">N\u00ebse \u00ebsht\u00eb futur 777<\/b>0<br \/>\n1<br \/>\n2<br \/>\n3<br \/>\n4<br \/>\n5<br \/>\n6<br \/>\n7<br \/>\n8<br \/>\n9<\/p>\n<p>N\u00ebse mungon nj\u00eb shpjegim t\u00eb shkurt\u00ebr p\u00ebr t\u00eb shkruar me cik\u00ebl, at\u00ebher\u00eb duhet t\u00eb arrijm\u00eb nj\u00eb zgjidhje universale p\u00ebr k\u00ebt\u00eb detyr\u00eb pa cik\u00ebl.<br \/>\nDo t\u00eb dal\u00eb nj\u00eb nga dy opsionet:<br \/>\n<b class=\"spoiler_title\">E d\u00ebshiruara<\/b><\/p>\n<pre><code class=\"cs\">string temp;\ntemp = Console.ReadLine();\nint x;\nx = int.Parse(temp);\nif (x==0) {\n    Console.WriteLine(0 + &quot;+&quot;);\n} else {\n    Console.WriteLine(0);\n}\nif (x==1) {\n    Console.WriteLine(1 + &quot;+&quot;);\n} else {\n    Console.WriteLine(1);\n}\nif (x==2) {\n    Console.WriteLine(2 + &quot;+&quot;);\n} else {\n    Console.WriteLine(2);\n}\nif (x==3) {\n    Console.WriteLine(3 + &quot;+&quot;);\n} else {\n    Console.WriteLine(3);\n}\nif (x==4) {\n    Console.WriteLine(4 + &quot;+&quot;);\n} else {\n    Console.WriteLine(4);\n}\nif (x==5) {\n    Console.WriteLine(5 + &quot;+&quot;);\n} else {\n    Console.WriteLine(5);\n}\nif (x==6) {\n    Console.WriteLine(6 + &quot;+&quot;);\n} else {\n    Console.WriteLine(6);\n}\nif (x==7) {\n    Console.WriteLine(7 + &quot;+&quot;);\n} else {\n    Console.WriteLine(7);\n}\nif (x==8) {\n    Console.WriteLine(8 + &quot;+&quot;);\n} else {\n    Console.WriteLine(8);\n}\nif (x==9) {\n    Console.WriteLine(9 + &quot;+&quot;);\n} else {\n    Console.WriteLine(9);\n}\n<\/code><\/pre>\n<p><b class=\"spoiler_title\">E mundshme<\/b><\/p>\n<pre><code class=\"cs\">string temp;\ntemp = Console.ReadLine();\nint x;\nx = int.Parse(temp);\nif (x==0) {\n    Console.WriteLine(&quot;0+n1n2n3n4n5n6n7n8n9&quot;);\n}\nif (x==1) {\n    Console.WriteLine(&quot;0n1+n2n3n4n5n6n7n8n9&quot;);\n}\nif (x==2) {\n    Console.WriteLine(&quot;0n1n2+n3n4n5n6n7n8n9&quot;);\n}\nif (x==3) {\n    Console.WriteLine(&quot;0n1n2n3+n4n5n6n7n8n9&quot;);\n}\nif (x==4) {\n    Console.WriteLine(&quot;0n1n2n3n4+n5n6n7n8n9&quot;);\n}\nif (x==5) {\n    Console.WriteLine(&quot;0n1n2n3n4n5+n6n7n8n9&quot;);\n}\nif (x==6) {\n    Console.WriteLine(&quot;0n1n2n3n4n5n6+n7n8n9&quot;);\n}\nif (x==7) {\n    Console.WriteLine(&quot;0n1n2n3n4n5n6n7+n8n9&quot;);\n}\nif (x==8) {\n    Console.WriteLine(&quot;0n1n2n3n4n5n6n7n8+n9&quot;);\n}\nif (x==9) {\n    Console.WriteLine(&quot;0n1n2n3n4n5n6n7n8n9+&quot;);\n}\n<\/code><\/pre>\n<p>Nj\u00eb detyr\u00eb t\u00eb ngjashme e jap m\u00eb p\u00ebrpara, gjat\u00eb m\u00ebsimit p\u00ebr tem\u00ebn e deg\u00ebve.<br \/>\nN\u00ebse studenti ka arritur nj\u00eb variant \"t\u00eb mundsh\u00ebm\", duhet t\u00eb tregoni se zgjidhje t\u00eb nj\u00ebjt\u00ebs detyr\u00eb mund t\u00eb ket\u00eb shum\u00eb. Megjithat\u00eb, ato ndryshojn\u00eb n\u00eb q\u00ebndrueshm\u00ebrin\u00eb ndaj ndryshimeve t\u00eb k\u00ebrkesave. B\u00ebni pyetje: \"Sa vende n\u00eb kod do t\u00eb duhet t\u00eb rregullohen, n\u00ebse duhet t\u00eb shtojm\u00eb nj\u00eb num\u00ebr tjet\u00ebr?\" N\u00eb variantin \"t\u00eb mundsh\u00ebm\", do t\u00eb duhet t\u00eb shtohet nj\u00eb deg\u00ebzim tjet\u00ebr dhe t\u00eb shkruhen n\u00eb 10 vende t\u00eb tjera numri i ri. N\u00eb \"variantin e d\u00ebshiruar\", mjafton t\u00eb shtohet vet\u00ebm nj\u00eb deg\u00ebzim.<br \/>\nVendosni detyr\u00ebn p\u00ebr t\u00eb riprodhuar \"variantin e d\u00ebshiruar\", pastaj gjeni n\u00eb kod nj\u00eb rregull dhe b\u00ebni z\u00ebvend\u00ebsimin e variablit dhe shkruani ciklin.<br \/>\nN\u00ebse keni nj\u00eb ide, si ta zgjidhni k\u00ebt\u00eb detyr\u00eb pa cik\u00ebl n\u00eb ndonj\u00eb m\u00ebnyr\u00eb tjet\u00ebr, ju lutem, shkruani n\u00eb komentet.<\/p>\n<h4>Ciklet brenda cikleve<\/h4>\n<p>\nN\u00eb k\u00ebt\u00eb tem\u00eb duhet t\u00eb kushtoni v\u00ebmendje se:<br \/>\n \u2014 numruesit p\u00ebr ciklin e brendsh\u00ebm dhe at\u00eb t\u00eb jasht\u00ebm duhet t\u00eb jen\u00eb variabla t\u00eb ndrysh\u00ebm.<br \/>\n \u2014 numruesi p\u00ebr ciklin e brendsh\u00ebm duhet t\u00eb ri-inizializohet shum\u00eb her\u00eb (dmth. n\u00eb trupin e ciklit t\u00eb jasht\u00ebm).<br \/>\n N\u00eb problemet e shkarkimit t\u00eb tekstit nuk mund t\u00eb shkruani m\u00eb par\u00eb nj\u00eb shkronj\u00eb n\u00eb disa rreshta dhe pastaj nj\u00eb tjet\u00ebr. Duhet t\u00eb shkruani m\u00eb par\u00eb t\u00eb gjitha shkronjat e rreshtit t\u00eb par\u00eb, pastaj t\u00eb gjitha ato t\u00eb rreshtit t\u00eb dyt\u00eb dhe k\u00ebshtu me radh\u00eb.<\/p>\n<p>Shpjegimi i tem\u00ebs p\u00ebr ciklet brenda cikleve fillon m\u00eb s\u00eb miri me shpjegimin e r\u00ebnd\u00ebsis\u00eb s\u00eb rrying t\u00eb numruesit.<br \/>\nDetyra p\u00ebr shembull:<\/p>\n<blockquote><p>\u041f\u043e\u043b\u044c\u0437\u043e\u0432\u0430\u0442\u0435\u043b\u044c \u0432\u0432\u043e\u0434\u0438\u0442 \u0434\u0432\u0430 \u0447\u0438\u0441\u043b\u0430: R \u0438 T. \u0412\u044b\u0432\u0435\u0441\u0442\u0438 \u0434\u0432\u0435 \u0441\u0442\u0440\u043e\u043a\u0438 \u0441\u0438\u043c\u0432\u043e\u043b\u043e\u0432 &quot;#&quot;. \u0412 \u043f\u0435\u0440\u0432\u043e\u0439 \u0441\u0442\u0440\u043e\u043a\u0435 \u0434\u043e\u043b\u0436\u043d\u043e \u0431\u044b\u0442\u044c R \u0448\u0442\u0443\u043a \u0441\u0438\u043c\u0432\u043e\u043b\u043e\u0432. \u0412\u043e \u0432\u0442\u043e\u0440\u043e\u0439 \u0441\u0442\u0440\u043e\u043a\u0435 T \u0448\u0442\u0443\u043a. \u0415\u0441\u043b\u0438 \u043a\u0430\u043a\u043e\u0435-\u043b\u0438\u0431\u043e \u0447\u0438\u0441\u043b\u043e \u0431\u0443\u0434\u0435\u0442 \u043e\u0442\u0440\u0438\u0446\u0430\u0442\u0435\u043b\u044c\u043d\u043e, \u0432\u044b\u0432\u0435\u0441\u0442\u0438 \u0441\u043e\u043e\u0431\u0449\u0435\u043d\u0438\u0435 \u043e\u0431 \u043e\u0448\u0438\u0431\u043a\u0435.<\/p><\/blockquote>\n<p>\n<b class=\"spoiler_title\">R=5, T=11<\/b>#####<br \/>\n###########<\/p>\n<p><b class=\"spoiler_title\">R=20, T=3<\/b>####################<br \/>\n###<\/p>\n<p><b class=\"spoiler_title\">R=-1, T=6<\/b>Vlera e R duhet t\u00eb jet\u00eb e papajtueshme<\/p>\n<p><b class=\"spoiler_title\">R=6, T=-2<\/b>Vlera e T duhet t\u00eb jet\u00eb e papajtueshme<\/p>\n<p>E qart\u00eb, q\u00eb kjo problematik\u00eb ka t\u00eb pakt\u00ebn dy mund\u00ebsi zgjidhjeje.<br \/>\n<b class=\"spoiler_title\">E d\u00ebshiruara<\/b><\/p>\n<pre><code class=\"cs\">string temp;\nint R;\nint T;\ntemp = Console.ReadLine();\nR = int.Parse(temp);\ntemp = Console.ReadLine();\nT = int.Parse(temp);\nint i = 0;\nwhile (i &lt; R)\n{\n    Console.Write(&quot;#&quot;);\n    i = i + 1;\n}\nConsole.WriteLine();\ni = 0;\nwhile (i &lt; T)\n{\n    Console.Write(&quot;#&quot;);\n    i = i + 1;\n}\n<\/code><\/pre>\n<p><b class=\"spoiler_title\">Mund\u00ebsia nr. 1<\/b><\/p>\n<pre><code class=\"cs\">string temp;\nint R;\nint T;\ntemp = Console.ReadLine();\nR = int.Parse(temp);\ntemp = Console.ReadLine();\nT = int.Parse(temp);\nint i = 0;\nwhile (i &lt; R)\n{\n    Console.Write(&quot;#&quot;);\n    i = i + 1;\n}\nConsole.WriteLine();\nint j = 0;\nj = 0;\nwhile (j &lt; T)\n{\n    Console.Write(&quot;#&quot;);\n    j = j + 1;\n}\n<\/code><\/pre>\n<p>Dallimi q\u00ebndron n\u00eb faktin se n\u00eb zgjidhjen \"e mundshme\" p\u00ebr shfaqjen e rreshtit t\u00eb dyt\u00eb \u00ebsht\u00eb p\u00ebrdorur nj\u00eb variab\u00ebl e dyt\u00eb. Duhet insistuar n\u00eb p\u00ebrdorimin e t\u00eb nj\u00ebjt\u00ebs variab\u00ebl p\u00ebr t\u00eb dy ciklet. Argumentimi i k\u00ebtij kufizimi mund t\u00eb b\u00ebhet p\u00ebr faktin se zgjidhja me nj\u00eb numrues p\u00ebr dy ciklet do t\u00eb ilustronte termin \"rryshje numruesi\". Kuptimi i k\u00ebtij termi \u00ebsht\u00eb i nevojsh\u00ebm p\u00ebr zgjidhjen e problemeve t\u00eb tjera. Si nj\u00eb kompromis, mund t\u00eb mbahen t\u00eb dy zgjidhjet e problemit.<\/p>\n<p>Nj\u00eb problem typik me p\u00ebrdorimin e nj\u00eb varianti numrues p\u00ebr dy ciklet shfaqet k\u00ebshtu:<br \/>\n<b class=\"spoiler_title\">R=5, T=11<\/b>#####<br \/>\n######<\/p>\n<p>Numri i karaktereve n\u00eb rreshtin e dyt\u00eb nuk p\u00ebrputhet me vler\u00ebn T. N\u00ebse keni nevoj\u00eb p\u00ebr ndihm\u00eb me k\u00ebt\u00eb problem, duhet t\u00eb \"tregoni me gishta\" n\u00eb sh\u00ebnimet p\u00ebr problemet tipike me ciklet. Ky \u00ebsht\u00eb simptoma nr. 3. Diagnozohet n\u00ebse shtoni shfaqjen e vler\u00ebs s\u00eb numruesit menj\u00ebher\u00eb p\u00ebrpara ciklit t\u00eb dyt\u00eb. Korrigjohet me nj\u00eb rryshje. Por nuk \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb flisni menj\u00ebher\u00eb p\u00ebr k\u00ebt\u00eb. Studenti duhet t\u00eb p\u00ebrpiqet t\u00eb formuloj\u00eb t\u00eb pakt\u00ebn nj\u00eb hipotez\u00eb.<\/p>\n<p>Natyrisht, ka edhe nj\u00eb variant tjet\u00ebr zgjidhjeje. Por un\u00eb nuk e kam par\u00eb kurr\u00eb at\u00eb tek student\u00ebt. N\u00eb faz\u00ebn e studimit t\u00eb cikleve, tregimi p\u00ebr t\u00eb do t\u00eb shp\u00ebrndante v\u00ebmendjen. Mund t\u00eb kthehemi te ai m\u00eb von\u00eb, gjat\u00eb studimit t\u00eb funksioneve t\u00eb pun\u00ebs me vargje.<br \/>\n<b class=\"spoiler_title\">Mund\u00ebsia nr. 2<\/b><\/p>\n<pre><code class=\"cs\">string temp;\nint R;\nint T;\ntemp = Console.ReadLine();\nR = int.Parse(temp);\ntemp = Console.ReadLine();\nT = int.Parse(temp);\nConsole.WriteLine(new String('#', R));\nConsole.WriteLine(new String('#', T));\n<\/code><\/pre>\n<p>\nDetyr\u00eb e ardhshme e obligueshme:<\/p>\n<blockquote><p>Shfaqni numrat nga 0 n\u00eb 9. \u00c7do num\u00ebr duhet t\u00eb jet\u00eb n\u00eb vij\u00ebn e vet. Numri i numrave n\u00eb rresht (W) futet nga tastiera.<\/p><\/blockquote>\n<p>\n<b class=\"spoiler_title\">W=1<\/b>0<br \/>\n1<br \/>\n2<br \/>\n3<br \/>\n4<br \/>\n5<br \/>\n6<br \/>\n7<br \/>\n8<br \/>\n9<\/p>\n<p><b class=\"spoiler_title\">W=10<\/b><code>0000000000<br \/>\n1111111111<br \/>\n2222222222<br \/>\n3333333333<br \/>\n4444444444<br \/>\n5555555555<br \/>\n6666666666<br \/>\n7777777777<br \/>\n8888888888<br \/>\n9999999999<\/code><\/p>\n<p>N\u00ebse studenti ka m\u00ebsuar teknik\u00ebn e z\u00ebvend\u00ebsimit t\u00eb variablave, ai do ta zgjidh\u00eb mjaft shpejt. Problemi i mundsh\u00ebm do t\u00eb jet\u00eb p\u00ebrs\u00ebri n\u00eb zerozimin e variablave. N\u00ebse nuk arrin t\u00eb kryej\u00eb transformimin, do t\u00eb thot\u00eb se keni nxituar dhe duhet t\u00eb zgjidhni probleme m\u00eb t\u00eb thjeshta.<\/p>\n<p>Faleminderit p\u00ebr v\u00ebmendjen. Jepni p\u00eblqime, abonohuni n\u00eb kanal.<\/p>\n<p>P.S. N\u00ebse keni gjetur ndonj\u00eb gabim ose shkrim t\u00eb pasakt\u00eb n\u00eb tekst, ju lutem m\u00eb njoftoni. K\u00ebt\u00eb mund ta b\u00ebni duke theksuar nj\u00eb pjes\u00eb t\u00eb tekstit dhe duke shtypur n\u00eb Mac \u00ab\u2318 + Enter\u00bb, ndersa n\u00eb tastierat klasike \u00abCtrl \/ Enter\u00bb, ose p\u00ebrmes mesazheve t\u00eb drejtp\u00ebrdrejta. N\u00ebse k\u00ebto mund\u00ebsi nuk jan\u00eb t\u00eb disponueshme, shkruani p\u00ebr gabimet n\u00eb komentet. Faleminderit!<\/p>\n<p class=\"for_users_only_msg\">Vet\u00ebm p\u00ebrdoruesit e regjistruar mund t\u00eb marrin pjes\u00eb n\u00eb anket\u00eb. <noindex><a rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/auth\/login\/\">Hyni<\/a><\/noindex>, ju lutem.<\/p>\n<h2 class=\"default-block__polling-title\">Anketa p\u00ebr lexuesit pa karma<\/h2>\n<ul class=\"poll-result\">\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent\">20,0%<\/strong>M\u00ebsoj profesionalisht, +12<\/p>\n<\/li>\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent\">10,0%<\/strong>M\u00ebsoj profesionalisht, -11<\/p>\n<\/li>\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent  poll-result__data-percent_winner\">70,0%<\/strong>Nuk m\u00ebsoj, +17<\/p>\n<\/li>\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent\">0,0%<\/strong>Nuk m\u00ebsoj, -10<\/p>\n<\/li>\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent\">0,0%<\/strong>Tjet\u00ebr0<\/p>\n<\/li>\n<\/ul>\n<p>    Kan\u00eb votuar 10 p\u00ebrdorues. U p\u00ebrmbajt\u00ebn 5 p\u00ebrdorues.<br \/>\n<br \/>Burimi: <a content=\"nofollow\" rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/post\/456500\/\">habr.com<\/a> <\/p>","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>\u041d\u0435\u0441\u043c\u043e\u0442\u0440\u044f \u043d\u0430 \u0442\u043e, \u0447\u0442\u043e \u0440\u0435\u0447\u044c \u043f\u043e\u0439\u0434\u0435\u0442 \u043e\u0431 \u043e\u0434\u043d\u043e\u0439 \u0438\u0437 \u0431\u0430\u0437\u043e\u0432\u044b\u0445 \u0442\u0435\u043c, \u0434\u0430\u043d\u043d\u0430\u044f \u0441\u0442\u0430\u0442\u044c\u044f \u043d\u0430\u043f\u0438\u0441\u0430\u043d\u0430 \u0434\u043b\u044f \u043e\u043f\u044b\u0442\u043d\u044b\u0445 \u0441\u043f\u0435\u0446\u0438\u0430\u043b\u0438\u0441\u0442\u043e\u0432. \u0426\u0435\u043b\u044c \u2014 \u043f\u043e\u043a\u0430\u0437\u0430\u0442\u044c \u043a\u0430\u043a\u0438\u0435 \u0437\u0430\u0431\u043b\u0443\u0436\u0434\u0435\u043d\u0438\u044f \u0431\u044b\u0432\u0430\u044e\u0442 \u0443 \u043d\u043e\u0432\u0438\u0447\u043a\u043e\u0432 \u0432 \u043f\u0440\u043e\u0433\u0440\u0430\u043c\u043c\u0438\u0440\u043e\u0432\u0430\u043d\u0438\u0438. \u0414\u043b\u044f \u043f\u0440\u0430\u043a\u0442\u0438\u043a\u0443\u044e\u0449\u0438\u0445 \u0440\u0430\u0437\u0440\u0430\u0431\u043e\u0442\u0447\u0438\u043a\u043e\u0432 \u044d\u0442\u0438 \u043f\u0440\u043e\u0431\u043b\u0435\u043c\u044b \u0443\u0436\u0435 \u0434\u0430\u0432\u043d\u043e \u0440\u0435\u0448\u0435\u043d\u044b, \u043f\u043e\u0437\u0430\u0431\u044b\u0442\u044b \u0438\u043b\u0438 \u0432\u043e\u043e\u0431\u0449\u0435 \u043d\u0435 \u0437\u0430\u043c\u0435\u0447\u0435\u043d\u044b. \u0421\u0442\u0430\u0442\u044c\u044f \u043c\u043e\u0436\u0435\u0442 \u043f\u0440\u0438\u0433\u043e\u0434\u0438\u0442\u044c\u0441\u044f, \u0435\u0441\u043b\u0438 \u0432\u0434\u0440\u0443\u0433 \u0432\u0430\u043c \u043f\u0440\u0438\u0434\u0435\u0442\u0441\u044f \u043f\u043e\u043c\u043e\u0433\u0430\u0442\u044c \u0441 \u044d\u0442\u043e\u0439 \u0442\u0435\u043c\u043e\u0439 \u043a\u043e\u043c\u0443-\u043d\u0438\u0431\u0443\u0434\u044c. \u0412 \u0441\u0442\u0430\u0442\u044c\u0435 \u043f\u0440\u043e\u0432\u043e\u0434\u044f\u0442\u0441\u044f [&hellip;]<\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-40208","post","type-post","status-publish","format-standard","hentry"],"aioseo_notices":[],"aioseo_head":"\n\t\t<!-- All in One SEO 5.0.1.1 - 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