{"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\u00ebsoni t\u00eb kap\u00ebrceni v\u00ebshtir\u00ebsit\u00eb dhe po ashtu t\u00eb shkruani cikle","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p>Megjith\u00ebse po flasim p\u00ebr nj\u00eb nga temat themelore, ky artikull \u00ebsht\u00eb shkruar p\u00ebr profesionist\u00eb t\u00eb p\u00ebrvojsh\u00ebm. Q\u00ebllimi \u00ebsht\u00eb t\u00eb tregojm\u00eb se cilat jan\u00eb gabimet e zakonshme q\u00eb b\u00ebjn\u00eb fillestar\u00ebt n\u00eb programim. P\u00ebr zhvilluesit q\u00eb praktikojn\u00eb, k\u00ebto probleme jan\u00eb zgjidhur prej koh\u00ebsh, jan\u00eb harruar ose nuk kan\u00eb qen\u00eb fare t\u00eb v\u00ebrejtura. Artikulli mund t\u00eb jet\u00eb i dobish\u00ebm n\u00ebse ndonj\u00ebher\u00eb duhet t'i ndihmoni dikujt me k\u00ebt\u00eb tem\u00eb. N\u00eb artikull, b\u00ebhen paralele me materialin nga disa libra p\u00ebr programim nga autor\u00ebt Shildt, Stroustrup, Okulov.<\/p>\n<p>Tema p\u00ebr ciklet \u00ebsht\u00eb zgjedhur sepse shum\u00eb njer\u00ebz eliminohen gjat\u00eb m\u00ebsimit t\u00eb programimit.<\/p>\n<p>Kjo metodologji \u00ebsht\u00eb e destinuar p\u00ebr student\u00eb t\u00eb dob\u00ebt. N\u00eb p\u00ebrgjith\u00ebsi, t\u00eb fort\u00ebt nuk mbesin t\u00eb ngat\u00ebrruar n\u00eb k\u00ebt\u00eb tem\u00eb dhe nuk \u00ebsht\u00eb nevoja t\u00eb inventohen metoda t\u00eb ve\u00e7anta p\u00ebr ta. Q\u00ebllimi dyt\u00ebsor i artikullit \u00ebsht\u00eb t\u00eb transferoj\u00eb k\u00ebt\u00eb metodologji nga klasa \"funksionon p\u00ebr t\u00eb gjith\u00eb student\u00ebt, por vet\u00ebm me nj\u00eb m\u00ebsues\" n\u00eb klas\u00ebn \"funksionon p\u00ebr t\u00eb gjith\u00eb student\u00ebt, t\u00eb gjith\u00eb m\u00ebsuesit\". Nuk pretendoj p\u00ebr origjinalitet t\u00eb plot\u00eb. N\u00ebse tashm\u00eb po p\u00ebrdorni nj\u00eb metodologji t\u00eb ngjashme p\u00ebr t\u00eb m\u00ebsuar k\u00ebt\u00eb tem\u00eb, ju lutem na tregoni se si ndryshon varianti juaj. N\u00ebse vendosni ta aplikoni, do t\u00eb isha i interesuar p\u00ebr rezultatet. N\u00ebse nj\u00eb metodologji e ngjashme p\u00ebrshkruhet n\u00eb ndonj\u00eb lib\u00ebr, ju lutem na jepni titullin.<\/p>\n<p><noindex><a rel=\"nofollow\" name=\"habracut\"><\/a><\/noindex><br \/>\nK\u00ebt\u00eb metodologji e kam punuar p\u00ebr 4 vjet, duke u angazhuar 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, student\u00ebt ngeleshin p\u00ebr nj\u00eb koh\u00eb t\u00eb gjat\u00eb n\u00eb k\u00ebt\u00eb tem\u00eb dhe iknin. Pas \u00e7do studenti, metodologjia dhe materialet jan\u00eb korrigjuar. N\u00eb vitin e fundit, student\u00ebt nuk ngel\u00ebn m\u00eb n\u00eb k\u00ebt\u00eb tem\u00eb, prandaj vendosa t\u00eb ndaj p\u00ebrvoj\u00ebn time.<\/p>\n<h4>Pse ka kaq shum\u00eb shkronja? Ciklet jan\u00eb elementare!<\/h4>\n<p>\nSi\u00e7 kam shkruar m\u00eb par\u00eb, p\u00ebr zhvilluesit praktikues dhe student\u00ebt e fort\u00eb, kompleksiteti i konceptit t\u00eb cikleve mund t\u00eb n\u00ebnvler\u00ebsohet. P\u00ebr shembull, mund t\u00eb organizoni nj\u00eb leksion t\u00eb gjat\u00eb, t\u00eb shihni kryeveprat q\u00eb l\u00ebvizin dhe syt\u00eb e men\u00e7ur. Por kur p\u00ebrpiqen t\u00eb zgjidhin ndonj\u00eb problem, fillon nj\u00eb ngecje dhe probleme t\u00eb paqart\u00eb. Pas leksionit, student\u00ebt patjet\u00ebr se kan\u00eb formuar vet\u00ebm nj\u00eb kuptim t\u00eb pjessh\u00ebm. Situata p\u00ebrkeq\u00ebsohet nga fakti q\u00eb student\u00ebt nuk mund t\u00eb shprehin vet\u00eb se ku q\u00ebndron gabimi i tyre.<br \/>\nNj\u00ebher\u00eb e kuptova se student\u00ebt i perceptojn\u00eb shembujt e mi si hieroglif\u00eb. Pra, si copa t\u00eb pandashme teksti, n\u00eb t\u00eb cilat duhet t\u00eb shtosh ndonj\u00eb \"shkronj\u00eb magjike\" dhe do t\u00eb funksionoj\u00eb.<br \/>\nNdonj\u00ebher\u00eb kam v\u00ebn\u00eb re se student\u00ebt mendojn\u00eb se p\u00ebr t\u00eb zgjidhur nj\u00eb problem t\u00eb konkret, nevojitet <i>ndonj\u00eb konstrukcion tjet\u00ebr<\/i> q\u00eb un\u00eb thjesht nuk e kam p\u00ebrmendur akoma. Megjithat\u00eb, p\u00ebr zgjidhjen e nevojshme ishte vet\u00ebm pak modifikim i shembujve.<\/p>\n<p>Prandaj arrita n\u00eb ide q\u00eb v\u00ebmendja kryesore duhet t\u00eb p\u00ebrqendrohet jo n\u00eb sintaks\u00ebn e shprehjeve, por n\u00eb iden\u00eb e ristrukturimit t\u00eb kodit t\u00eb p\u00ebrs\u00ebritur me ndihm\u00ebn e cikleve. Sa her\u00eb q\u00eb nx\u00ebn\u00ebsit e kuptojn\u00eb k\u00ebt\u00eb ide, \u00e7do sintaks\u00eb arrin ta zot\u00ebrojn\u00eb me an\u00eb t\u00eb ushtrimeve t\u00eb vogla.<\/p>\n<h4>K\u00eb t\u00eb m\u00ebsoj dhe p\u00ebrse<\/h4>\n<p>\nDuke qen\u00eb se nuk ka prova hyr\u00ebse, n\u00eb m\u00ebsim mund t\u00eb jen\u00eb si student\u00eb t\u00eb fort\u00eb ashtu dhe 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 mbr\u00ebmjes<\/a><\/noindex><br \/>\nKam synuar q\u00eb programimi t\u00eb m\u00ebsohet nga \u00e7do njeri q\u00eb d\u00ebshiron.<br \/>\nM\u00ebsimet e mia jan\u00eb individuale dhe studenti paguan vet\u00eb \u00e7do her\u00eb. Duket se student\u00ebt do t\u00eb optimizojn\u00eb kostot dhe do t\u00eb k\u00ebrkojn\u00eb minimumin. Sidoqoft\u00eb, njer\u00ebzit vin\u00eb n\u00eb m\u00ebsime t\u00eb drejtp\u00ebrdrejta me nj\u00eb m\u00ebsues t\u00eb gjall\u00eb jo vet\u00ebm p\u00ebr diturin\u00eb, por p\u00ebr sigurin\u00eb se ata e kan\u00eb p\u00ebrvet\u00ebsuar, p\u00ebr ndjenj\u00ebn e progresit dhe p\u00ebr miratimin nga eksperti (m\u00ebsuesi). N\u00ebse student\u00ebt nuk do t\u00eb ndjejn\u00eb progres n\u00eb m\u00ebsimin e tyre, ata do t\u00eb largohen. N\u00eb p\u00ebrgjith\u00ebsi, mund t\u00eb organizohen m\u00ebsime n\u00eb m\u00ebnyr\u00eb q\u00eb student\u00ebt t\u00eb ndjejn\u00eb progres n\u00eb rritjen e numrit t\u00eb konstrukcioneve t\u00eb njohura. Pra, fillojm\u00eb duke studiuar n\u00eb detaje while, pastaj m\u00ebsojm\u00eb for, pastaj do while dhe ja, kemi nj\u00eb kurs p\u00ebr nj\u00eb mij\u00eb e nj\u00eb nat\u00eb, ku p\u00ebr dy muaj studiojm\u00eb vet\u00ebm ciklet, dhe n\u00eb fund, nj\u00eb student q\u00eb shkruan bibliotek\u00ebn standarde n\u00eb dikta. Megjithat\u00eb, p\u00ebr t\u00eb zgjidhur probleme praktike, nuk duhet 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 kursin e drejtp\u00ebrdrejt, e konsideroj parimin e duhur \u2014 t\u00eb m\u00ebsoj minimumin dhe t\u00eb inkurajoj studimin e pavarur t\u00eb nuancave dhe temave t\u00eb af\u00ebrta. N\u00eb tem\u00ebn e cikleve, minimumi e shoh si konstrukcionin while. Me t\u00eb mund t\u00eb kuptohet principi. Duke njohur principin, mund t\u00eb zot\u00ebrohet edhe for 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, nuk mjafton t\u00eb p\u00ebrshkruhet sintaksa. Duhet t'u jepet m\u00eb shum\u00eb detyra t\u00eb thjeshta, por t\u00eb larmishme dhe t\u00eb shtjellohet shembujt m\u00eb n\u00eb detaje. N\u00eb fund, shpejt\u00ebsia e p\u00ebrvet\u00ebsimit kufizohet nga aft\u00ebsit\u00eb e studentit p\u00ebr t\u00eb transformuar shprehjet dhe p\u00ebr t\u00eb gjetur rregulla. P\u00ebr student\u00ebt e zgjuar, shumica e detyrave do t\u00eb jen\u00eb t\u00eb m\u00ebrzitshme. Gjat\u00eb m\u00ebsimit 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\">githab.com<\/a><\/noindex>. Megjithat\u00eb, repositori \u00ebsht\u00eb m\u00eb shum\u00eb si nj\u00eb grimuar i nj\u00eb magjistari\u2014askush p\u00ebrve\u00e7 meje nuk do ta kuptoj\u00eb se \u00e7far\u00eb ndodhet kudo, dhe n\u00ebse d\u00ebshton kontrollin, mund t\u00eb \u00e7mendet.<\/p>\n<h4>Metodika \u00ebsht\u00eb e orientuar drejt praktik\u00ebs.<\/h4>\n<p>\nTeoria shpjegohet n\u00eb shembuj zgjidhjeje problemesh. N\u00eb m\u00ebsimet p\u00ebr bazat e programimit, ku studiohen ndarjet dhe ciklet, thjesht nuk mund t\u00eb organizohet nj\u00eb ligj\u00ebrat\u00eb e dobishme p\u00ebr nj\u00eb tem\u00eb p\u00ebr nj\u00eb or\u00eb t\u00eb t\u00ebr\u00eb. 15-20 minuta mjaftojn\u00eb p\u00ebr t\u00eb shpjeguar konceptin. V\u00ebshtir\u00ebsit\u00eb kryesore shfaqen gjat\u00eb realizimit t\u00eb detyrave praktike.<br \/>\nM\u00ebsuesit fillestar\u00eb mund t\u00eb hedhin operator\u00eb, ndarje, cikle dhe masa gjat\u00eb nj\u00eb ligj\u00ebrate. Por student\u00ebt e tyre do t\u00eb hasin v\u00ebshtir\u00ebsi n\u00eb t\u00eb kuptuarit e k\u00ebsaj informacioni.<br \/>\nN\u00eb fakt, nuk mjafton vet\u00ebm t\u00eb tregosh materialin, por edhe t\u00eb sigurohesh se d\u00ebgjuarit e kuptojn\u00eb at\u00eb.<\/p>\n<p>Fakti i p\u00ebrvet\u00ebsimit t\u00eb tem\u00ebs p\u00ebrcaktohet nga m\u00ebnyra se si studenti p\u00ebrballet me pun\u00ebn e pavarur. <br \/>\nN\u00ebse studenti arrin t\u00eb zgjidh\u00eb nj\u00eb problem mbi tem\u00ebn pa ndihm\u00ebn e m\u00ebsuesit, at\u00ebher\u00eb tema \u00ebsht\u00eb p\u00ebrvet\u00ebsuar. P\u00ebr t\u00eb siguruar kontrollin e pavarur, \u00e7do detyre p\u00ebrshkruhet me 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 v\u00ebshtir\u00eb, kalimi n\u00eb t\u00eb ardhshme \u00ebsht\u00eb i pafrytsh\u00ebm. Ajo \u00ebsht\u00eb akoma m\u00eb e v\u00ebshtir\u00eb. P\u00ebr t\u00eb ndihmuar studentin t\u00eb p\u00ebrballoj\u00eb detyr\u00ebn e tanishme t\u00eb v\u00ebshtir\u00eb, i shpjegohet disa teknika duke marr\u00eb shembull nga detyra e par\u00eb. N\u00eb thelb, e gjith\u00eb p\u00ebrmbajtja e tem\u00ebs p\u00ebrqendrohet n\u00eb teknikat e p\u00ebrballimit 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 kryhet \"vet\u00ebm\" menj\u00ebher\u00eb pas t\u00eb par\u00ebs n\u00ebn mbik\u00ebqyrjen e m\u00ebsuesit. T\u00eb gjitha detyrat pasuese jan\u00eb t\u00eb orientuara p\u00ebr t\u00ebrhequr v\u00ebmendjen n\u00eb detaje t\u00eb ndryshme q\u00eb mund t\u00eb shkaktojn\u00eb keqkuptime.<\/p>\n<p>Shpjegimi i shembullit p\u00ebrb\u00ebn nj\u00eb dialog, ku studenti duhet t\u00eb shkaktoj\u00eb back propagation dhe cross-validation p\u00ebr t\u00eb siguruar p\u00ebrvet\u00ebsimin e materialit.<\/p>\n<p>Do t\u00eb jem banal dhe do t\u00eb them se shembulli i par\u00eb mbi tem\u00ebn \u00ebsht\u00eb shum\u00eb i r\u00ebnd\u00ebsish\u00ebm. N\u00ebse ka materiale p\u00ebr pun\u00eb t\u00eb gjer\u00eb t\u00eb pavarur, l\u00ebnia pas e shembullit t\u00eb par\u00eb mund t\u00eb rregullohet. N\u00ebse p\u00ebrve\u00e7 shembullit nuk ka m\u00eb asgj\u00eb, at\u00ebher\u00eb studenti ndoshta nuk do ta p\u00ebrvet\u00ebsoj\u00eb tem\u00ebn.<\/p>\n<h4>While apo for?<\/h4>\n<p>\nNj\u00eb nga pyetjet e diskutueshme \u00ebsht\u00eb zgjedhja e nd\u00ebrtimit p\u00ebr shembullin: while apo for. Nj\u00ebher\u00eb, nj\u00eb mik im, nj\u00eb zhvillues praktik pa p\u00ebrvoj\u00eb m\u00ebsimdh\u00ebnie, m\u00eb bindte p\u00ebr nj\u00eb or\u00eb se cikli for \u00ebsht\u00eb m\u00eb i lehti p\u00ebr t'u kuptuar. Argumentet reduktoheshin n\u00eb \"n\u00eb t\u00eb gjith\u00e7ka \u00ebsht\u00eb e qart\u00eb dhe i organizuar n\u00eb vend\". Megjithat\u00eb, arsyeja kryesore e v\u00ebshtir\u00ebsive p\u00ebr fillestar\u00ebt real\u00eb q\u00ebndron te ideja e ciklit, e jo n\u00eb shkrimin e tij. N\u00ebse nj\u00eb person nuk e kupton k\u00ebt\u00eb ide, at\u00ebher\u00eb do t\u00eb ket\u00eb v\u00ebshtir\u00ebsi me sintaks\u00ebn. Sapo ideja t\u00eb kuptohet, at\u00ebher\u00eb problemet me formatimin e kodit zhduken vetvetiu.<\/p>\n<p>N\u00eb materialet e mia, temat e cikleve pasojn\u00eb ato t\u00eb ndarjeve. Ngjashm\u00ebria e jashtme midis if dhe while lejon nj\u00eb analogji t\u00eb drejtp\u00ebrdrejt\u00eb: \"kur kushti n\u00eb titull \u00ebsht\u00eb i v\u00ebrtet\u00eb, at\u00ebher\u00eb ekzekutohet trupi\". Ve\u00e7antia 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 budallaqe me munges\u00ebn e virgullave dhe kall\u00ebpeve. Fillestar\u00ebt nuk kan\u00eb aq shum\u00eb zhvillim t\u00eb v\u00ebmendjes dhe detajimit, sa p\u00ebr t\u00eb shmangur automatikisht gabimet sintaksore.<br \/>\nArgumenti i tret\u00eb - n\u00eb shum\u00eb libra t\u00eb mira, while shpjegohet i pari.<\/p>\n<p>N\u00ebse studenti arrin t\u00eb transformoj\u00eb leht\u00ebsisht shprehjet, at\u00ebher\u00eb t\u00eb flitet p\u00ebr for p\u00ebrmes kalimit. Studenti pastaj vet\u00eb do t\u00eb zgjedh\u00eb at\u00eb q\u00eb i p\u00eblqen m\u00eb shum\u00eb. N\u00ebse megjithat\u00eb transformimet shkaktojn\u00eb v\u00ebshtir\u00ebsi, at\u00ebher\u00eb \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb mos shp\u00ebrndahen v\u00ebmendjen. Le t\u00eb zgjidh\u00eb studentin s\u00eb pari gjith\u00e7ka me while. Kur tema e cikleve \u00ebsht\u00eb p\u00ebrvet\u00ebsuar, mund t\u00eb rishkruhen zgjidhjet p\u00ebr t\u00eb punuar n\u00eb transformimin e while n\u00eb for.<br \/>\nCiklet me kusht pas jan\u00eb nj\u00eb makth mjaft i rrall\u00eb. P\u00ebr to nuk humbas koh\u00eb fare. N\u00ebse studenti ka p\u00ebrvet\u00ebsuar idet\u00eb e zbuleve t\u00eb rregullave dhe transformimin e shprehjeve, ai do t\u00eb jet\u00eb n\u00eb gjendje t\u00eb kuptoj\u00eb pa ndihm\u00ebn time.<\/p>\n<p>Gjat\u00eb demonstrimit t\u00eb shembullit t\u00eb par\u00eb p\u00ebr student\u00ebt t\u00eb fort\u00eb, theksoj se \u00ebsht\u00eb e r\u00ebnd\u00ebsishme q\u00eb n\u00eb shembullin e par\u00eb t\u00eb regjistrohet jo vet\u00ebm zgjidhja, por gjithashtu e gjith\u00eb zinxhiri i veprimeve q\u00eb \u00e7uan n\u00eb rezultat. Student\u00ebt q\u00eb jan\u00eb dembel\u00eb mund t\u00eb n\u00ebnvler\u00ebsojn\u00eb shkrimin dhe t\u00eb transferojn\u00eb vet\u00ebm algoritmin p\u00ebrfundimtar. Duhet t\u2019i bindim ata se nj\u00ebher\u00eb do t'u dal\u00eb nj\u00eb detyr\u00eb e komplikuar. P\u00ebr ta zgjidhur at\u00eb, do t\u00eb nevojitet t\u00eb ndjekin hapat si n\u00eb k\u00ebt\u00eb shembull. Prandaj, \u00ebsht\u00eb e r\u00ebnd\u00ebsishme t\u00eb regjistrohen t\u00eb gjitha etapat. N\u00eb detyrat e ardhshme mund t\u00eb mbetet vet\u00ebm varianti p\u00ebrfundimtar i zgjidhjes.<\/p>\n<p>Ideja kryesore e automatizimit \u00ebsht\u00eb se i angazhojm\u00eb kompjuterin t\u00eb kryej\u00eb pun\u00eb rutin\u00eb p\u00ebr njeriun. Nj\u00eb nga teknikat baz\u00eb \u00ebsht\u00eb shkrimi i cikleve. Ajo aplikohet kur n\u00eb program shkruhen disa veprime t\u00eb nj\u00ebjta nj\u00eb pas nj\u00eb.<\/p>\n<h4>E qarta \u00ebsht\u00eb m\u00eb e mir\u00eb se e paqart\u00eb<\/h4>\n<p>\nMund t\u00eb duket nj\u00eb ide e mir\u00eb n\u00eb detyr\u00ebn e par\u00eb p\u00ebr ciklet t\u00eb nxjerr\u00ebsh n\u00eb ekran nj\u00eb fraz\u00eb t\u00eb nj\u00ebjt\u00eb disa her\u00eb. P\u00ebr shembull:<\/p>\n<blockquote><p>Hurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!<br \/>\nHurrah, punon!\n<\/p><\/blockquote>\n<p>\nKjo mund\u00ebsi \u00ebsht\u00eb e keqe sepse n\u00eb daljen e rezultateve nuk tregohet vlera e num\u00ebruesit. Kjo \u00ebsht\u00eb nj\u00eb problem p\u00ebr fillestar\u00ebt. Nuk duhet ta n\u00ebnvler\u00ebsojm\u00eb at\u00eb. N\u00eb fillim, kjo detyr\u00eb ishte e para, dhe detyra p\u00ebr t\u00eb nxjerr\u00eb nj\u00eb seri numrash n\u00eb rritje ishte e dyta. Duhej t\u00eb futeshin terma shtes\u00eb si \u201ccikli N her\u00eb\u201d dhe \u201ccikli nga A deri n\u00eb B\u201d, t\u00eb cilat jan\u00eb n\u00eb thelb t\u00eb nj\u00ebjta. P\u00ebr t\u00eb mos krijuar entitete t\u00eb panevojshme, vendosa t\u00eb shfaq vet\u00ebm shembullin me nxjerrjen e nj\u00eb serie numrash. Pak njer\u00ebz e kan\u00eb t\u00eb leht\u00eb t\u00eb m\u00ebsojn\u00eb t\u00eb mbajn\u00eb n\u00eb mend num\u00ebruesin dhe t\u00eb modelojn\u00eb sjelljen e programit n\u00eb mendje. Disa student\u00eb p\u00ebrballen p\u00ebr her\u00eb t\u00eb par\u00eb me modelimin \u201cn\u00eb mendje\u201d pik\u00ebrisht n\u00eb tem\u00ebn e cikleve.<br \/>\nPas disa praktikave, detyr\u00ebn p\u00ebr p\u00ebrs\u00ebritjen e tekstit t\u00eb nj\u00ebjt\u00eb e jap p\u00ebr zgjidhje t\u00eb pavarur. N\u00ebse m\u00eb par\u00eb shfaq nj\u00eb num\u00ebrues t\u00eb duksh\u00ebm dhe pastaj nj\u00eb t\u00eb paduksh\u00ebm, student\u00ebt kan\u00eb m\u00eb pak probleme. N sometimes \u00ebsht\u00eb e mjaftueshme nj\u00eb k\u00ebshill\u00eb \u201cmos e shkruaj num\u00ebruesin n\u00eb ekran\u201d.<\/p>\n<h4>Si shpjegohet te t\u00eb tjer\u00ebt?<\/h4>\n<p>\nN\u00eb shumic\u00ebn e materialeve m\u00ebsimore n\u00eb internet, sintaksa e ciklit jepet si pjes\u00eb e \"lekture\". P\u00ebr shembull, n\u00eb developer.mozilla.org (n\u00eb k\u00ebt\u00eb moment) s\u00eb bashku me ciklin while p\u00ebrmenden edhe disa konstrukte t\u00eb tjera. N\u00eb k\u00ebt\u00eb rast jepen ekskluzivisht vet\u00eb konstruktionet n\u00eb form\u00eb shablloni. Rezultati i ekzekutimit t\u00eb tyre p\u00ebrshkruhet me fjal\u00eb, nd\u00ebrsa ilustruese mungon. N\u00eb mendimin tim, nj\u00eb paraqitje e till\u00eb e tem\u00ebs shum\u00ebfishohet me zero dobishm\u00ebrin\u00eb e k\u00ebtyre materialeve. Nx\u00ebn\u00ebsi mund ta kopjoj\u00eb kodin dhe ta ekzekutoj\u00eb vet\u00eb, por nj\u00eb referenc\u00eb p\u00ebr krahasim akoma \u00ebsht\u00eb e nevojshme. Si t\u00eb kuptohet q\u00eb shembulli \u00ebsht\u00eb kopjuar n\u00eb m\u00ebnyr\u00eb t\u00eb duhur n\u00ebse nuk ka \u00e7far\u00eb t\u00eb krahasosh rezultatin?<br \/>\nKur jepet vet\u00ebm shablloni, pa shembuj, studentit i b\u00ebhet akoma m\u00eb e v\u00ebshtir\u00eb. Si t\u00eb kuptohet q\u00eb fragmentet e kodit jan\u00eb vendosur n\u00eb shabllon n\u00eb m\u00ebnyr\u00eb t\u00eb duhur? Mund t\u00eb provohet t\u00eb shkruhet ndonj\u00ebher\u00eb, dhe pastaj t\u00eb ekzekutohet. Por n\u00ebse nuk ka nj\u00eb standard p\u00ebr krahasim t\u00eb rezultatit, at\u00ebher\u00eb ekzekutimi gjithashtu nuk ndihmon. <i>ndonj\u00ebher\u00eb<\/i>, dhe pastaj t\u00eb ekzekutohet. Por n\u00ebse nuk ka nj\u00eb standard p\u00ebr krahasim t\u00eb rezultatit, at\u00ebher\u00eb 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 \/>\nN\u00eb mendimin tim, n\u00eb fillim \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb jepet sintaksa e konstruktionit menj\u00ebher\u00eb me kllapa. Varianti pa kllapa t\u00eb shpjegohet vet\u00ebm n\u00ebse nx\u00ebn\u00ebsi ngre nj\u00eb pyetje t\u00eb ve\u00e7ant\u00eb \"pse k\u00ebtu pa kllapa dhe funksionon\".<\/p>\n<p>N\u00eb librin e Okulov \"Themel\u00ebt e programimit\" 2012, njohja me ciklet fillon me shabllonin for, pastaj jepen rekomandime p\u00ebr p\u00ebrdorimin e tij, dhe pastaj menj\u00ebher\u00eb shkon n\u00eb nj\u00eb seksion eksperimental t\u00eb m\u00ebsimit. Un\u00eb e kuptoj q\u00eb 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 gjithmon\u00eb shkruhet rezultati i fragmenteve t\u00eb kodit. P\u00ebr shembull, te Shildt \"Java 8. Udh\u00ebzimi i Plot\u00eb\" i botuar n\u00eb vitin 2015. N\u00eb fillim jepet shablloni, pastaj shembulli i programit dhe menj\u00ebher\u00eb pas tij \u2014 rezultati i ekzekutimit.<\/p>\n<blockquote><p>Si nj\u00eb shembull, le t\u00eb shohim ciklin while, ku b\u00ebhet num\u00ebrimi pasues duke filluar nga 10 dhe printohen sakt\u00ebsisht 10 rreshta \"taktesh\":<br \/>\nPas nisjes, ky program printon dhjet\u00eb \"takte\" n\u00eb k\u00ebt\u00eb m\u00ebnyr\u00eb: <\/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>\nQasja me p\u00ebrshkrimin e modelit, shembullin e programit dhe rezultatin e pun\u00ebs s\u00eb k\u00ebtij programi p\u00ebrdoret gjithashtu n\u00eb librin \"JavaScript p\u00ebr f\u00ebmij\u00eb\" 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.<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>\nN\u00eb librin e Stroustrupit \"Principet dhe praktika me C++\" 2016, autori e \u00e7on m\u00eb tej. Fillimisht shpjegohet cilat rezultate duhen arritur, dhe pastaj tregohet teksti i programit. P\u00ebr m\u00eb tep\u00ebr, si shembull nuk merret thjesht nj\u00eb program rast\u00ebsor, por jepet nj\u00eb udh\u00ebtim n\u00eb histori. Kjo ndihmon n\u00eb fokusimin mbi t\u00eb: \"Shiko, kjo nuk \u00ebsht\u00eb thjesht ndonj\u00eb tekst t\u00eb padobish\u00ebm. Ti sheh di\u00e7ka t\u00eb r\u00ebnd\u00ebsishme\".<\/p>\n<p>Si nj\u00eb shembull t\u00eb iteracionit, le t\u00eb shqyrtojm\u00eb programin e par\u00eb t\u00eb ekzekutuar n\u00eb makin\u00ebn me program t\u00eb ruajtur (EDSAC). Ai u shkrua nga David Wheeler n\u00eb laboratorin e kompjuter\u00ebve t\u00eb Universitetit t\u00eb Kembrixhit (Cambridge University, Angli) m\u00eb 6 maj 1949. Ky program kalkulon dhe printon nj\u00eb list\u00eb t\u00eb thjesht\u00eb t\u00eb katror\u00ebve.<\/p>\n<blockquote><p>K\u00ebtu, n\u00eb \u00e7do rresht p\u00ebrfshihet nj\u00eb num\u00ebr, i cili pason nga shenja e tabulimit (\u2018t\u2019) dhe katrori i k\u00ebtij numri. Versioni i k\u00ebtij programi n\u00eb gjuh\u00ebn C++ dukej k\u00ebshtu:<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) thekson se respekton inteligjenc\u00ebn e student\u00ebve t\u00eb tij. Ndoshta aft\u00ebsia p\u00ebr t\u00eb zbuluar nj\u00eb model n\u00eb disa shembuj konsiderohet nj\u00eb shenj\u00eb e till\u00eb inteligjence.<noindex><a rel=\"nofollow\" href=\"https:\/\/habr.com\/ru\/post\/349342\/\">p\u00ebrkthim<\/a><\/noindex>Si\u00e7 shpjegoj un\u00eb vet\u00eb<\/p>\n<h4>Qasja e Stroustrupit: p\u00ebrshkrimi i rezultatit, pastaj zgjidhja e problemit, dhe m\u00eb pas analiza e pavarur nga studenti \u2014 duket si m\u00eb e arsyeshmja. Prandaj, vendosa ta marr si baz\u00eb pik\u00ebrisht k\u00ebt\u00eb qasje, por p\u00ebr ta treguar n\u00eb nj\u00eb shembull m\u00eb pak historik \u2014 nj\u00eb problem lidhur me printimin e \"p\u00ebrmbajtjes\". Kjo formon nj\u00eb ankor\u00eb t\u00eb njohur, q\u00eb m\u00eb pas t\u00eb themi \"kujto problemin mbi p\u00ebrmbajtjen\" dhe q\u00eb student\u00ebt ta kujtojn\u00eb at\u00eb. N\u00eb shembullin tim, p\u00ebrpiqem t\u00eb paralajm\u00ebroj dy nga e keqja m\u00eb t\u00eb zakonshme. M\u00eb pas do t\u00eb shkruaj m\u00eb shum\u00eb rreth tyre.<\/h4>\n<p>\nN\u00eb k\u00ebt\u00eb problem kemi njohuri me teknikat p\u00ebr zgjidhjen e problemeve komplekse. Zgjidhja fillestare duhet t\u00eb jet\u00eb primitive dhe e thjesht\u00eb. M\u00eb pas mund t\u00eb mendojm\u00eb se si ta p\u00ebrmir\u00ebsojm\u00eb k\u00ebt\u00eb zgjidhje.<\/p>\n<blockquote><p>Nga v\u00ebzhgimet e mia, qasja \"model-shembull-rezultati\" n\u00eb kombinime t\u00eb ndryshme ende sjell n\u00eb m\u00ebnyr\u00eb q\u00eb student\u00ebt ta perceptojn\u00eb ciklin si nj\u00eb hieroglif. Kjo u shfaq n\u00eb faktin se ata nuk kuptonin pse duhet t\u00eb shkruanin kushte, si t\u00eb zgjidhnin mes i++ dhe i-- dhe gj\u00ebra t\u00eb tjera q\u00eb dukeshin t\u00eb qarta. P\u00ebr t\u00eb shmangur k\u00ebto keqkuptime, qasja p\u00ebr t\u00eb folur mbi ciklet duhet t\u00eb theksoj\u00eb kuptimin e p\u00ebrs\u00ebritjes s\u00eb veprimeve t\u00eb nj\u00ebjta dhe vet\u00ebm pastaj \u2014 formatizimin e tyre p\u00ebrmes konstrukcioneve. Prandaj, para se t\u00eb jepet sintaksa e ciklit, duhen zgjidhur problemet \"drejt\". Nj\u00eb zgjidhje primitive p\u00ebr problemin e p\u00ebrmbajtjes \u00ebsht\u00eb si m\u00eb posht\u00eb:<br \/>\n<code>Hyrje<br \/>\nKreu 1<br \/>\nKreu 2<br \/>\nKreu 3<br \/>\nKreu 4<br \/>\nKapitulli 5<br \/>\nKreu 6<br \/>\nKreu 7<br \/>\nP\u00ebrfundimi<\/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>\nT\u00eb z\u00ebvend\u00ebsojm\u00eb veprimet uniforme me nj\u00eb cik\u00ebl.<br \/>\nCilat veprime p\u00ebrs\u00ebriten k\u00ebtu radhazi pa ndryshime?<br \/>\nN\u00eb k\u00ebt\u00eb fragment nuk ka asnj\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 i ardhsh\u00ebm \u00ebsht\u00eb t\u00eb gjejm\u00eb diferenc\u00ebn nd\u00ebrmjet fragmenteve. Kjo vet\u00ebm n\u00eb k\u00ebt\u00eb problem \u00ebsht\u00eb gjith\u00e7ka e qart\u00eb, m\u00eb von\u00eb do t\u00eb p\u00ebrs\u00ebriten jo vet\u00ebm komandat individuale, por blloqe kodi prej 5 rreshtash e m\u00eb shum\u00eb. Do t\u00eb duhet t\u00eb k\u00ebrkohet jo thjesht n\u00eb list\u00ebn e komandave, por n\u00eb struktura t\u00eb ndarjes ose ciklit.<br \/>\nN\u00eb shembullin, diferenca midis komandave \u00ebsht\u00eb numri pas fjal\u00ebs \"Kapitulli\".<br \/>\nSapoh q\u00eb ndodhet diferenca, duhen kuptuar rregullat e ndryshimit. A \u00ebsht\u00eb frangenti ndryshe numri? A rritet ose zvog\u00eblohet vazhdimisht? Si ndryshon vlera e numrit nd\u00ebrmjet dy komandave af\u00ebr nj\u00ebra-tjetr\u00ebs?<br \/>\nN\u00eb shembull, numri pas fjal\u00ebs \"Kapitulli\" rritet me hap 1. Diferenca \u00ebsht\u00eb gjetur, rregulli \u00ebsht\u00eb zbuluar. Tani mund t\u00eb z\u00ebvend\u00ebsojm\u00eb fragmentin e ndryshuar me nj\u00eb variab\u00ebl.<br \/>\nDeklarimi i till\u00eb i variabl\u00ebs duhet t\u00eb b\u00ebhet p\u00ebrpara t\u00eb par\u00ebs nga fragmentet q\u00eb p\u00ebrs\u00ebriten. Kjo variab\u00ebl zakonisht quhet I ose j ose ndonj\u00eb em\u00ebr m\u00eb t\u00eb avancuar. Vlera e saj fillestare duhet t\u00eb jet\u00eb e barabart\u00eb me vler\u00ebn e par\u00eb q\u00eb printohet n\u00eb ekran. N\u00eb shembull vlera e par\u00eb \u00ebsht\u00eb 1.<br \/>\nCila vler\u00eb fillestare duhet t\u00eb merret p\u00ebr printimin e radh\u00ebs s\u00eb numrave \"100, 101, 102, 103, 104, 105\"? <br \/>\nN\u00eb k\u00ebt\u00eb radh\u00eb, numri i par\u00eb \u00ebsht\u00eb 100.<br \/>\nPas \u00e7do komande printimi, duhet t\u00eb rritet vlera e k\u00ebsaj variabile me 1. Ky hap \u00ebsht\u00eb ndryshimi.<br \/>\nPas \u00e7do komand\u00eb t\u00eb daljes, \u00ebsht\u00eb e nevojshme t\u00eb rritet vlera e k\u00ebsaj variabli me 1. Ky nj\u00ebsi \u00ebsht\u00eb hapi i ndryshimit.<br \/>\nCili \u00ebsht\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 t\u00eb ndrysh\u00ebm 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 \u00abshprehi rregullin me nj\u00eb variab\u00ebl\u00bb n\u00eb kod, rezultojn\u00eb disa grupe t\u00eb nj\u00ebjtave veprime, 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>Sezoni i zgjidhjes s\u00eb detyr\u00ebs, ku duhet t\u00eb p\u00ebrdoren ciklet, p\u00ebrb\u00ebhet nga etapet:<\/p>\n<ol>\n<li>Zgjidhni \u00abdrejt\u00bb me nj\u00eb s\u00ebr\u00eb komandash t\u00eb ve\u00e7anta<\/li>\n<li>Gjeni rregullin<\/li>\n<li>Shprehni rregullin me nj\u00eb variab\u00ebl<\/li>\n<li>Formuloni n\u00eb form\u00ebn e nj\u00eb cikli<\/li>\n<\/ol>\n<p>M\u00eb pas prezantohen terma t\u00eb rinj, p\u00ebr t\u00eb mos e l\u00ebn\u00eb studentin n\u00eb situat\u00ebn \u00abpo kuptoj gjith\u00e7ka, por nuk mund t\u00eb them\u00bb:<br \/>\n \u2014 num\u00ebrues \u2014 gjithmon\u00eb \u00ebsht\u00eb nj\u00eb variab\u00ebl q\u00eb \u00ebsht\u00eb e nevojshme p\u00ebr t\u00eb ndjekur numrin e hapat t\u00eb ciklit. Zakonisht \u00ebsht\u00eb nj\u00eb num\u00ebr i t\u00ebr\u00eb, i cili krahasohet me kufirin.<br \/>\n \u2014 hapi i num\u00ebruesit \u2014 p\u00ebrshkrimi i rregullit t\u00eb ndryshimit t\u00eb num\u00ebruesit.<br \/>\n \u2014 kufiri \u2014 numri ose variabla me t\u00eb cil\u00ebn krahasohet num\u00ebruesi, q\u00eb algoritmi t\u00eb jet\u00eb i p\u00ebrfunduar. 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 \u00abkomanda \u00ebsht\u00eb shkruar brenda ciklit\u00bb, n\u00ebnkuptohet pik\u00ebrisht trupi.<br \/>\n \u2014 iteracioni i ciklit \u2014 ekzekutimi i nj\u00eb her\u00eb i trupit t\u00eb ciklit.<br \/>\n \u2014 kushti i ciklit \u2014 nj\u00eb shprehje logjike, nga e cila varet n\u00ebse do t\u00eb ekzekutohet nj\u00eb iteracion tjet\u00ebr. (K\u00ebtu mund t\u00eb ket\u00eb konfuzion me ndarjet e deg\u00ebve)<br \/>\nDuhet t\u00eb jeni t\u00eb p\u00ebrgatitur p\u00ebr faktin se p\u00ebr her\u00eb t\u00eb par\u00eb student\u00ebt do t\u00eb aplikojn\u00eb terma jo sipas q\u00ebllimit. Kjo p\u00ebrfshin si t\u00eb fort\u00ebt ashtu edhe t\u00eb dob\u00ebtit. 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'i v\u00ebn\u00eb detyr\u00ebn \u00abizolo fragmentin e kodit me &lt;termin&gt;\u00bb dhe vet\u00eb t\u00eb p\u00ebrdorin k\u00ebto terma sakt\u00ebsisht n\u00eb bised\u00eb.<br \/>\nPas transformimit me ciklin, rezulton fragmenti:<\/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>Keqkuptimi kryesor<\/h4>\n<p>\nNj\u00eb keqkuptim i njohur nga 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 hasin vazhdimisht n\u00eb 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 shkruarja e komand\u00ebs: nj\u00eb her\u00eb apo shum\u00eb?\n<\/p><\/blockquote>\n<p>\nKomandat p\u00ebr daljen e fjal\u00ebve \u00abHyrje\u00bb dhe \u00abP\u00ebrfundim\u00bb, si dhe shpallja dhe inicializimi i variabl\u00ebs i nuk jan\u00eb t\u00eb ngjashme me veprimet e tjera t\u00eb p\u00ebrs\u00ebritura. Ato ekzekutohen vet\u00ebm nj\u00eb her\u00eb, prandaj duhen shkruar jasht\u00eb trupit t\u00eb ciklit.<\/p>\n<p>N\u00eb kod duhet t\u00eb mbesin t\u00eb tre etapet e zgjidhjes, q\u00eb m\u00eb pas t\u00eb referohen p\u00ebr t'u ndihmuar n\u00eb rast v\u00ebshtir\u00ebsish. Dy variantet e para mjafton t\u00eb komentohet, q\u00eb t\u00eb mos pengojn\u00eb.<br \/>\nV\u00ebmendja e student\u00ebve duhet t\u00eb p\u00ebrqendrohet 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 brenda trupit t\u00eb ciklit, nd\u00ebrsa kufiri \u2014 jo. P\u00ebr ta shkelur k\u00ebt\u00eb rregull, duhet t\u00eb formulohet nj\u00eb arsye t\u00eb fort\u00eb.<br \/>\n \u2014 Komandat p\u00ebr t\u00eb dal\u00eb me fjal\u00ebt \u00abHyrje\u00bb dhe \u00abP\u00ebrfundim\u00bb ndodhen jasht\u00eb trupit t\u00eb ciklit. Ne duam t'i ekzekutojm\u00eb ato 1 her\u00eb. \u00abHyrja\u00bb \u2014 para p\u00ebrs\u00ebritjes s\u00eb veprimeve, \u00abP\u00ebrfundimi\u00bb \u2014 pas saj.<br \/>\nGjat\u00eb procesit t\u00eb konsolidimit t\u00eb k\u00ebsaj teme, p\u00ebrvet\u00ebsimit t\u00eb t\u00eb tjerave, si dhe zgjidhjes s\u00eb v\u00ebshtir\u00ebsive, madje edhe student\u00ebve m\u00eb t\u00eb fort\u00eb u ndihmon t\u00eb p\u00ebrshkruajn\u00eb pyetjen: \u00abSa her\u00eb duhet t\u00eb ekzekutohet ky veprim? Nj\u00eb apo shum\u00eb?\u00bb.<\/p>\n<h4>Zhvillimi i aft\u00ebsive shtes\u00eb<\/h4>\n<p>\nGjat\u00eb procesit t\u00eb studimit t\u00eb cikleve, student\u00ebt gjithashtu st\u00ebrviten n\u00eb diagnostikimin dhe zgjidhjen e problemeve. P\u00ebr t\u00eb kryer nj\u00eb diagnostikim, studentit duhet t\u00eb paraqes\u00eb rezultate t\u00eb d\u00ebshiruara dhe t'i krahasoj\u00eb ato me rezultatet reale. Nga ndryshimi midis tyre varen veprimet p\u00ebr t'u korrigjuar.<br \/>\nDuke qen\u00eb se student\u00ebt n\u00eb k\u00ebt\u00eb faz\u00eb ende nuk e kan\u00eb iden\u00eb e mir\u00eb p\u00ebr rezultatin \u00abt\u00eb d\u00ebshiruar\u00bb, ata mund t\u00eb orientojn\u00eb n\u00eb t\u00eb dh\u00ebnat testuese. Zakonisht askush n\u00eb k\u00ebt\u00eb faz\u00eb nuk e kupton se \u00e7far\u00eb mund t\u00eb shkoj\u00eb keq dhe si t\u00eb b\u00ebjn\u00eb p\u00ebrball\u00eb saj. Prandaj un\u00eb ofroj n\u00eb regjistrim n\u00eb ditar p\u00ebrshkrimin e problemeve tipike dhe disa m\u00ebnyra p\u00ebr t'i zgjidhur ato. Zgjedhja e m\u00eb t\u00eb p\u00ebrshtatshmes \u00ebsht\u00eb detyra e vet\u00eb studentit.<br \/>\nRegjistrimi \u00ebsht\u00eb i nevojsh\u00ebm p\u00ebr t\u00eb pyetur \"a doli si\u00e7 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 jo- t\u00eb rrept\u00eb (&lt;= ose &gt;=).<br \/>\n \u2014 ndryshoni vler\u00ebn e kufizimit me 1.<\/li>\n<li>Veprimet n\u00eb cik\u00ebl ekzekutohen pa ndales\u00eb, pa fund. M\u00ebnyrat e zgjidhjes:<br \/>\n \u2014 shtoni komand\u00ebn p\u00ebr ndryshimin e num\u00ebruesit, n\u00ebse ajo mungon.<br \/>\n \u2014 korrigjoni komand\u00ebn p\u00ebr ndryshimin e num\u00ebruesit n\u00eb m\u00ebnyr\u00eb q\u00eb vlera e tij t\u00eb afrohet kufizimit.<br \/>\n \u2014 hiqni komand\u00ebn p\u00ebr ndryshimin e kufizimit, n\u00ebse ajo ndodhet brenda 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. Veprimi n\u00eb cik\u00ebl nuk u ekzekutua asnj\u00ebher\u00eb. S\u00eb pari, duhet t\u00eb zbulojm\u00eb vlerat aktuale t\u00eb variablave menj\u00ebher\u00eb para fillimit t\u00eb ciklit. M\u00ebnyrat e zgjidhjes:<br \/>\n \u2014 ndryshoni vler\u00ebn fillestare t\u00eb kufizimit<br \/>\n \u2014 ndryshoni vler\u00ebn fillestare t\u00eb num\u00ebruesit<\/li>\n<\/ol>\n<p>\nZakonisht problemi 3 lidhet me p\u00ebrdorimin e variabl\u00ebs s\u00eb gabuar ose pa zeroimin e num\u00ebruesit.<\/p>\n<p>Pas k\u00ebtij shpjegimi, studenti ende mund t\u00eb ket\u00eb keqkuptime t\u00eb ndryshme rreth funksionimit t\u00eb cikleve.<br \/>\nP\u00ebr t\u00eb shuar keqkuptimet m\u00eb t\u00eb zakonshme, po jap detyra:<\/p>\n<ol>\n<li>N\u00eb t\u00eb cil\u00ebn kufizimi, vlera fillestare e num\u00ebruesit ose hapi i num\u00ebruesit hyhet nga p\u00ebrdoruesi.<\/li>\n<li>N\u00eb t\u00eb cil\u00ebn vlera e num\u00ebruesit duhet t\u00eb p\u00ebrdoret n\u00eb ndonj\u00eb shprehje aritmetike. Preferohet me num\u00ebruesin n\u00eb shprehjen n\u00ebnrr\u00ebnjore ose n\u00eb em\u00ebrues, p\u00ebr t\u00eb siguruar q\u00eb diferenca t\u00eb jet\u00eb jo lineare.<\/li>\n<li>N\u00eb t\u00eb cil\u00ebn vlera e num\u00ebruesit nuk shfaqet n\u00eb ekran gjat\u00eb pun\u00ebs s\u00eb ciklit. P\u00ebr shembull, shfaqja e numrit t\u00eb k\u00ebrkuar t\u00eb fragmenteve t\u00eb nj\u00ebjta t\u00eb tekstit ose shkrimi i nj\u00eb forme me grafik\u00ebn e blet\u00ebve.<\/li>\n<li>N\u00eb t\u00eb cil\u00ebn \u00ebsht\u00eb e nevojshme t\u00eb kryhen fillimisht disa veprime t\u00eb p\u00ebrs\u00ebritura, dhe m\u00eb pas disa t\u00eb tjera.<\/li>\n<li>N\u00eb t\u00eb cil\u00ebn duhet t\u00eb kryhen veprime t\u00eb tjera para dhe pas atyre t\u00eb p\u00ebrs\u00ebritura.<\/li>\n<\/ol>\n<p>\nP\u00ebr secil\u00ebn detyr\u00eb duhet t\u00eb jepni t\u00eb dh\u00ebna testuese dhe rezultatin e pritur.<\/p>\n<p>P\u00ebr t\u00eb kuptuar se sa shpejt mund t\u00eb ecet, duhet t\u00eb lejohet t\u00eb lexoj\u00eb kushtet e k\u00ebtyre detyrave dhe t\u00eb pyesim: \"si ndryshojn\u00eb ato nga shembulli?\", \"\u00c7far\u00eb duhet t\u00eb ndryshohet n\u00eb shembull p\u00ebr t'i zgjidhur ato?\". N\u00ebse studenti p\u00ebrgjigjet n\u00eb m\u00ebnyr\u00eb t\u00eb kuptueshme, at\u00ebher\u00eb le t'i zgjidh\u00eb t\u00eb pakt\u00ebn nj\u00eb n\u00eb klas\u00eb, dhe t\u00eb tjerat n\u00eb sht\u00ebpi. N\u00ebse zgjidhja \u00ebsht\u00eb e suksesshme, mund t\u00eb fillojm\u00eb shpjegimin mbi kushtet brenda cikleve.<br \/>\nN\u00ebse ka v\u00ebshtir\u00ebsi me zgjidhjen e vetme, at\u00ebher\u00eb duhet t\u00eb punojm\u00eb gjith\u00e7ka n\u00eb klas\u00eb. P\u00ebr t\u00eb mos i ngjallur detyrat, un\u00eb rekomandoj q\u00eb fillimisht t\u00eb zgjidhni detyr\u00ebn n\u00eb nj\u00eb m\u00ebnyr\u00eb jo universale. Pra, n\u00eb m\u00ebnyr\u00eb q\u00eb zgjidhja t\u00eb kaloj\u00eb testin e par\u00eb dhe t\u00eb mos p\u00ebrdor\u00eb struktur\u00ebn e ciklit. Pastaj, t\u00eb aplikohet transformimi p\u00ebr t\u00eb arritur universalisht.<\/p>\n<h4>Ciklet dhe deg\u00ebzimet<\/h4>\n<p>\nSipas mendimit tim, \u00ebsht\u00eb e dobishme t\u00eb jepet tema \"ciklet brenda deg\u00ebzimeve\" ndaras. K\u00ebshtu, n\u00eb m\u00ebnyr\u00eb q\u00eb m\u00eb pas t\u00eb shihet dallimi midis kontrollit t\u00eb shum\u00ebfisht\u00eb t\u00eb kushteve dhe atij nj\u00ebfisht\u00eb.<br \/>\nDetyrat p\u00ebr konsolidim do t\u00eb jen\u00eb p\u00ebr shfaqjen e numrave nga A n\u00eb B, t\u00eb cilat hyjn\u00eb nga p\u00ebrdoruesi:<br \/>\n \u2014 gjithmon\u00eb n\u00eb rritje.<br \/>\n \u2014 n\u00eb rritje ose n\u00eb r\u00ebnie n\u00eb var\u00ebsi t\u00eb vlerave 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 \/>\nShkaku kryesor p\u00ebr aplikimin e deg\u00ebzimeve brenda cikleve \u2014 anomalit\u00eb n\u00eb rregull. Ajo prishet n\u00eb mes n\u00eb var\u00ebsi t\u00eb t\u00eb dh\u00ebnave hyr\u00ebse.<br \/>\nStudent\u00ebve q\u00eb jan\u00eb n\u00eb gjendje t\u00eb k\u00ebrkojn\u00eb zgjidhje duke kombinuar metoda t\u00eb thjeshta, mjafton t\u00eb thuhet \"deg\u00ebzimet mund t\u00eb shkruhen brenda cikleve\" dhe t'u jepet detyra \"p\u00ebr shembull\" plot\u00ebsisht p\u00ebr zgjidhje t\u00eb pavarur.<br \/>\nDetyr\u00eb 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 u fut 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 u fut 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 u fut 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 u fut 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 shpjegimi i shkurt\u00ebr mjafton p\u00ebr t\u00eb shkruar me cik\u00ebl, at\u00ebher\u00eb duhet t\u00eb arrijm\u00eb nj\u00eb zgjidhje universale t\u00eb k\u00ebsaj detyre pa cik\u00ebl.<br \/>\nDo t\u00eb rezultoj\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 mundur<\/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 jap m\u00eb her\u00ebt, gjat\u00eb studimit t\u00eb tem\u00ebs p\u00ebr deg\u00ebzimin.<br \/>\nN\u00ebse studentit i doli nj\u00eb variant \"i mundsh\u00ebm\", duhet t\u00eb tregoni se zgjidhjet e s\u00eb nj\u00ebjt\u00ebs detyr\u00eb mund t\u00eb jen\u00eb t\u00eb shumta. Megjithat\u00eb, ato ndryshojn\u00eb n\u00eb q\u00ebndrueshm\u00ebrin\u00eb ndaj ndryshimeve t\u00eb k\u00ebrkesave. B\u00ebni pyetjen: \"Sa vende n\u00eb kod do t\u00eb duhej t\u00eb rregulloheshin, n\u00ebse do t\u00eb duhet t\u00eb shtoni nj\u00eb num\u00ebr t\u00eb ri?\" N\u00eb variantin \"e mundsh\u00ebm\" do t\u00eb duhej t\u00eb shtonit nj\u00eb deg\u00ebzim t\u00eb ri dhe t\u00eb shkruani n\u00eb 10 vende t\u00eb tjera numrin e ri. N\u00eb \"t\u00eb d\u00ebshiruara\" \u00ebsht\u00eb mjaft t\u00eb shtoni vet\u00ebm nj\u00eb deg\u00ebzim.<br \/>\nT\u00eb vendosni detyr\u00ebn p\u00ebr t\u00eb riprodhuar variantin \"e d\u00ebshiruar\", pastaj t\u00eb gjeni n\u00eb kod nj\u00eb rregull, t\u00eb b\u00ebni z\u00ebvend\u00ebsimin e variablit dhe t\u00eb shkruani nj\u00eb cik\u00ebl.<br \/>\nN\u00ebse keni nj\u00eb ide se si ta zgjidhni k\u00ebt\u00eb detyr\u00eb pa cik\u00ebl p\u00ebrmes nj\u00eb metode tjet\u00ebr, ju lutemi 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 num\u00ebruesit p\u00ebr ciklin e brendsh\u00ebm dhe t\u00eb jasht\u00ebm duhet t\u00eb jen\u00eb variabla t\u00eb ndrysh\u00ebm.<br \/>\n \u2014 num\u00ebruesi p\u00ebr ciklin e brendsh\u00ebm duhet t\u00eb riparoj\u00eb shum\u00eb her\u00eb (pra brenda trupit t\u00eb ciklit t\u00eb jasht\u00ebm).<br \/>\n \u2014 n\u00eb detyrat p\u00ebr nxjerrjen e tekstit, nuk mund t\u00eb shkruani fillimisht nj\u00eb shkronj\u00eb n\u00eb disa rreshta, dhe pastaj nj\u00eb tjet\u00ebr. Duhet fillimisht t\u00eb nxirrni t\u00eb gjitha shkronjat e rreshtit t\u00eb par\u00eb, pastaj t\u00eb gjitha shkronjat e rreshtit t\u00eb dyt\u00eb e k\u00ebshtu me radh\u00eb.<\/p>\n<p>Shpjegimi i tem\u00ebs p\u00ebr ciklet brenda cikleve \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb fillohet me shpjegimin e r\u00ebnd\u00ebsis\u00eb s\u00eb riparimit t\u00eb num\u00ebruesit.<br \/>\nDetyr\u00eb 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 R duhet t\u00eb jet\u00eb jo negative<\/p>\n<p><b class=\"spoiler_title\">R=6, T=-2<\/b>Vlera T duhet t\u00eb jet\u00eb jo negative<\/p>\n<p>Sigurisht, q\u00eb kjo detyr\u00eb ka t\u00eb pakt\u00ebn dy variante 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\">I mundsh\u00ebm \u21161<\/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 \u00ebsht\u00eb se n\u00eb zgjidhjen \"e mundsh\u00ebm\" p\u00ebr nxjerrjen e rreshtit t\u00eb dyt\u00eb u p\u00ebrdor nj\u00eb variabl\u00eb e dyt\u00eb. Duhet t\u00eb insistoni n\u00eb p\u00ebrdorimin e t\u00eb nj\u00ebjtit variabl\u00eb p\u00ebr t\u00eb dy ciklet. Mund t\u00eb argumentoni k\u00ebt\u00eb kufizim duke th\u00ebn\u00eb se zgjidhja me nj\u00eb num\u00ebrues p\u00ebr dy cikle do t\u00eb ilustronte termin \"riparimi i num\u00ebruesit\". Kuptimi i k\u00ebtij termini \u00ebsht\u00eb i nevojsh\u00ebm p\u00ebr zgjidhjen e detyrave t\u00eb ardhshme. Si nj\u00eb kompromis, mund t\u00eb mbani t\u00eb dy zgjidhjet e detyr\u00ebs.<\/p>\n<p>Nj\u00eb problem tipik me p\u00ebrdorimin e nj\u00eb variabli num\u00ebrues p\u00ebr dy cikle shfaqet k\u00ebshtu:<br \/>\n<b class=\"spoiler_title\">R=5, T=11<\/b>#####<br \/>\n######<\/p>\n<p>Numri i simboleve n\u00eb rreshtin e dyt\u00eb nuk korrespondon 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 p\u00ebrmbledhjen mbi problemet tipike me ciklet. Ky \u00ebsht\u00eb simptoma \u21163. Diagnostikohet n\u00ebse shtoni nxjerrjen e vler\u00ebs num\u00ebruesit menj\u00ebher\u00eb para ciklit t\u00eb dyt\u00eb. Rregullohet me riparimin. Por \u00ebsht\u00eb m\u00eb mir\u00eb t\u00eb mos e tregoni menj\u00ebher\u00eb. Studenti duhet t\u00eb p\u00ebrpiqet t\u00eb formulloj\u00eb t\u00eb pakt\u00ebn nj\u00eb hipotez\u00eb.<\/p>\n<p>Sigurisht, ka edhe nj\u00eb variant tjet\u00ebr zgjidhjeje. Por un\u00eb nuk e kam par\u00eb at\u00eb tek student\u00ebt. N\u00eb faz\u00ebn e studimit t\u00eb cikleve, tregimi i tij do t\u00eb shp\u00ebrqendronte v\u00ebmendjen. Mund t\u00eb kthehemi te ai m\u00eb von\u00eb, gjat\u00eb studimit t\u00eb funksioneve q\u00eb lidhen me shenjat.<br \/>\n<b class=\"spoiler_title\">I mundsh\u00ebm \u21162<\/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>\nDetyra tjet\u00ebr e obligueshme:<\/p>\n<blockquote><p>Nxirrni n\u00eb ekran numrat nga 0 n\u00eb 9. \u00c7do num\u00ebr duhet t\u00eb jet\u00eb n\u00eb rreshtin e vet. Numri i numrave n\u00eb rresht (W) jepet p\u00ebrmes tastier\u00ebs.<\/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 zot\u00ebruar teknik\u00ebn e z\u00ebvend\u00ebsimit t\u00eb variables, at\u00ebher\u00eb ai do ta p\u00ebrballoj\u00eb mjaft shpejt. Problemi mund t\u00eb jet\u00eb p\u00ebrs\u00ebri n\u00eb zerimin e variables. N\u00ebse nuk arrin t\u00eb kryej\u00eb konvertimin, do t\u00eb thot\u00eb se jeni nxituar dhe duhet t\u00eb zgjidhni detyra m\u00eb t\u00eb lehta.<\/p>\n<p>Faleminderit p\u00ebr v\u00ebmendjen. Jepni p\u00eblqime dhe regjistrohuni n\u00eb kanal.<\/p>\n<p>P.S. N\u00ebse keni gjetur gabime ose shkrime t\u00eb gabuara n\u00eb tekst, ju lutem m\u00eb njoftoni. Kjo mund t\u00eb b\u00ebhet duke theksuar nj\u00eb pjes\u00eb t\u00eb tekstit dhe duke shtypur n\u00eb Mac \u00ab\u2318 + Enter\u00bb, ose n\u00eb tastierat klasike \u00abCtrl \/ Enter\u00bb, ose p\u00ebrmes mesazheve private. 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 lutemi.<\/p>\n<h2 class=\"default-block__polling-title\">Anket\u00eb 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 profesionist, +12<\/p>\n<\/li>\n<li class=\"poll-result__item\">\n<p>                <strong class=\"poll-result__data-percent\">10,0%<\/strong>M\u00ebsoj profesionist, -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>    10 p\u00ebrdorues votuan. 5 p\u00ebrdorues abstenuan.<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 5.0.0.1 - 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