Java’da While D\u00f6ng\u00fcs\u00fc ile Say\u0131 \u00d6zellikleri Programlama (9-15. G\u00fcn)<\/p>\n
Bu makalede, Java programlama dilinde Bir say\u0131n\u0131n asal olup olmad\u0131\u011f\u0131n\u0131 kontrol etmek i\u00e7in, say\u0131y\u0131 2’den say\u0131n\u0131n karek\u00f6k\u00fcne kadar olan t\u00fcm say\u0131lara b\u00f6lmeyi deneyebiliriz. E\u011fer say\u0131 bu aral\u0131ktaki herhangi bir say\u0131ya tam b\u00f6l\u00fcn\u00fcrse, asal de\u011fildir. Aktif olarak bu i\u015flemi ger\u00e7ekle\u015ftirdi\u011fimizde, asal olup olmad\u0131\u011f\u0131n\u0131 belirleriz. A\u015fa\u011f\u0131da, Java’da Bu kod, kullan\u0131c\u0131dan bir say\u0131 al\u0131r ve Bir m\u00fckemmel say\u0131, kendisini b\u00f6lenlerin toplam\u0131na e\u015fit olan bir say\u0131d\u0131r (kendisi hari\u00e7). \u00d6rne\u011fin, 6 m\u00fckemmel bir say\u0131d\u0131r \u00e7\u00fcnk\u00fc b\u00f6lenleri 1, 2 ve 3't\u00fcr ve toplamlar\u0131 6'd\u0131r. A\u015fa\u011f\u0131da, Java'da Bu kod, benzer \u015fekilde bir Bir Armstrong say\u0131s\u0131, basamaklar\u0131n\u0131n her birinin, basamak say\u0131s\u0131n\u0131n kuvvetine y\u00fckseltilmesinin toplam\u0131na e\u015fit olan bir say\u0131d\u0131r. \u00d6rne\u011fin, 153 bir Armstrong say\u0131s\u0131d\u0131r \u00e7\u00fcnk\u00fc 1\u00b3 + 5\u00b3 + 3\u00b3 = 153't\u00fcr. Son olarak, Java'da Bu kod, say\u0131n\u0131n basamaklar\u0131n\u0131 ay\u0131rmak ve i\u015flemleri ger\u00e7ekle\u015ftirmek i\u00e7in Bu \u00f6rnekler, Java'da Umar\u0131m bu makale, Java programlama becerilerinizi geli\u015ftirmenize yard\u0131mc\u0131 olur! Herhangi bir sorunuz varsa, l\u00fctfen yorumlarda belirtmekten \u00e7ekinmeyin.<\/p>\n #Etiketler: Java, while d\u00f6ng\u00fcs\u00fc, say\u0131 \u00f6zellikleri, asal say\u0131, m\u00fckemmel say\u0131, Armstrong say\u0131s\u0131, programlama, kod \u00f6rnekleri<\/p>\n <\/body>while<\/code> d\u00f6ng\u00fclerini kullanarak \u00e7e\u015fitli say\u0131 \u00f6zelliklerini hesaplayan program \u00f6rneklerini ele alaca\u011f\u0131z. \u00d6rne\u011fin, asal say\u0131, m\u00fckemmel say\u0131 ve Armstrong say\u0131s\u0131 gibi konulara de\u011finece\u011fiz. Bu \u00f6rnekler, d\u00f6ng\u00fc kullan\u0131m\u0131n\u0131 ve say\u0131sal i\u015flemleri peki\u015ftirmenize yard\u0131mc\u0131 olacakt\u0131r. Ayr\u0131ca, kod okunabilirli\u011fi ve verimlili\u011fi i\u00e7in baz\u0131 ipu\u00e7lar\u0131n\u0131 payla\u015faca\u011f\u0131m.<\/p>\nAsal Say\u0131 Kontrol\u00fc<\/h2>\n
while<\/code> d\u00f6ng\u00fcs\u00fc kullanarak asal say\u0131 kontrol\u00fc yapan bir kod \u00f6rne\u011fi bulunmaktad\u0131r:<\/p>\n\nimport java.util.Scanner;\n\npublic class AsalSayi {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n System.out.print(\"Bir say\u0131 girin: \");\n int sayi = scanner.nextInt();\n boolean asal = true;\n int i = 2;\n while (i <= Math.sqrt(sayi)) {\n if (sayi % i == 0) {\n asal = false;\n break;\n }\n i++;\n }\n if (asal) {\n System.out.println(sayi + \" asal bir say\u0131d\u0131r.\");\n } else {\n System.out.println(sayi + \" asal bir say\u0131 de\u011fildir.\");\n }\n scanner.close();\n }\n}\n<\/code><\/pre>\nwhile<\/code> d\u00f6ng\u00fcs\u00fc kullanarak asal olup olmad\u0131\u011f\u0131n\u0131 kontrol eder. \u00d6ncelikle asal<\/code> de\u011fi\u015fkeni true<\/code> olarak ba\u015flat\u0131l\u0131r. Daha sonra, i<\/code> de\u011fi\u015fkeni 2'den ba\u015flayarak say\u0131n\u0131n karek\u00f6k\u00fcne kadar art\u0131r\u0131l\u0131r. E\u011fer say\u0131, i<\/code>'ye tam b\u00f6l\u00fcn\u00fcrse, asal<\/code> de\u011fi\u015fkeni false<\/code> olarak de\u011fi\u015ftirilir ve d\u00f6ng\u00fc break<\/code> ifadesiyle sonland\u0131r\u0131l\u0131r. D\u00f6ng\u00fc tamamland\u0131ktan sonra, asal<\/code> de\u011fi\u015fkeninin de\u011feri ekrana yazd\u0131r\u0131l\u0131r.<\/p>\nM\u00fckemmel Say\u0131 Kontrol\u00fc<\/h2>\n
while<\/code> d\u00f6ng\u00fcs\u00fc kullanarak m\u00fckemmel say\u0131 kontrol\u00fc yapan bir kod \u00f6rne\u011fi bulunmaktad\u0131r:<\/p>\n\n\/\/M\u00fckemmel say\u0131 kontrol\u00fc kodu buraya eklenecek.\n<\/code><\/pre>\nwhile<\/code> d\u00f6ng\u00fcs\u00fc kullanarak say\u0131n\u0131n b\u00f6lenlerini bulur ve toplam\u0131n\u0131 hesaplar. Toplam, say\u0131ya e\u015fitse, say\u0131 m\u00fckemmel bir say\u0131d\u0131r.<\/p>\nArmstrong Say\u0131s\u0131 Kontrol\u00fc<\/h2>\n
while<\/code> d\u00f6ng\u00fcs\u00fc kullanarak Armstrong say\u0131s\u0131 kontrol\u00fc yapan bir kod \u00f6rne\u011fi bulunmaktad\u0131r:<\/p>\n\n\/\/Armstrong say\u0131s\u0131 kontrol\u00fc kodu buraya eklenecek.\n<\/code><\/pre>\nwhile<\/code> d\u00f6ng\u00fcs\u00fcn\u00fc kullan\u0131r. Sonu\u00e7 say\u0131ya e\u015fitse, say\u0131 bir Armstrong say\u0131s\u0131d\u0131r.<\/p>\nwhile<\/code> d\u00f6ng\u00fclerinin say\u0131sal i\u015flemlerde nas\u0131l kullan\u0131labilece\u011fini g\u00f6stermektedir. Daha karma\u015f\u0131k say\u0131sal problemler i\u00e7in, bu temel \u00f6rnekleri geni\u015fletebilir ve \u00f6zelle\u015ftirebilirsiniz. Daha fazla \u00f6rnek ve detayl\u0131 bilgi i\u00e7in fatihsoysal.com<\/a> adresini ziyaret edebilirsiniz. Ayr\u0131ca, konu ile ilgili daha fazla bilgi i\u00e7in bu ba\u011flant\u0131ya<\/a> g\u00f6z atabilirsiniz.<\/p>\n
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