{"id":67,"date":"2015-04-29T12:56:00","date_gmt":"2015-04-29T05:56:00","guid":{"rendered":""},"modified":"2024-04-21T13:25:02","modified_gmt":"2024-04-21T06:25:02","slug":"jawaban-3n1-dengan-java","status":"publish","type":"post","link":"https:\/\/rikiag.com\/blog\/read\/jawaban-3n1-dengan-java\/","title":{"rendered":"Algoritma 3n+1 Dengan Java"},"content":{"rendered":"\n\n
\n

Masalah 3n + 1<\/h1>\n
Deskripsi<\/b><\/div>\n
Dalam masalah 3n+1, Anda diminta menganalisis sifat-sifat statu algoritma untuk suatu input yang diberikan.<\/div>\n
Perhatikan algoritma berikut ini:<\/div>\n
    \n
  • input n<\/li>\n
  • print n<\/li>\n
  • if n = 1 then STOP<\/li>\n
  • if n ganjil then n = 3n+1<\/li>\n
  • else n = n\/2<\/li>\n
  • GOTO 2<\/li>\n<\/ul>\n


    <\/span><\/p>\n

     <\/div>\n
    Diberikan input 22, barisan bilangan-bilangan berikut akan dicetak:<\/div>\n
    22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1<\/div>\n
    Menurut dugaan algoritma di atas akan berhenti ketika angka 1 dicetak untuk sebarang bilangan bulat (integer) yang diinputkan. Meskipun algoritma di atas cukup sederhana, tidak diketahui apakah dugaan ini benar. Telah diverifikasi bahwa untuk semua bilangan bulat n sehingga 0<n<1,000,000 dugaan tersebut benar adanya. Jika diberikan input n, adalah mungkin untuk menentukan jumlah angka tercetak (termasuk 1). Jumlah angka tercetak ini disebut panjang siklus dari n. Dalam contoh di atas, panjang siklus dari 22 adalah 16 . Dalam tugas ini, untuk setiap dua bilangan i dan j, Anda diminta untuk menentukan panjang siklus maksimum dari semua bilangan di antara i dan j.<\/div>\n
     <\/div>\n
    Masukan<\/b><\/div>\n
    Input akan terdiri dari pasangan bilangan bulat (integer) i dan j. Semua bilangan bulat akan kurang dari 1.000.000 dan lebih besar dari 0. Anda harus memproses semua pasangan bilangan bulat dan untuk menentukan panjang siklus maksimum dari semua bilangan bulat antara dan termasuk i dan j.<\/div>\n
     <\/div>\n
     <\/div>\n
    Keluaran<\/b><\/div>\n
    Untuk setiap pasangan integer masukan i dan j, Anda harus mencetak keluaran i, j, dan panjang siklus maksimum untuk integer di antara dan termasuk i dan j. Ketiga angka harus dipisahkan oleh satu spasi dengan kesemua tiga angka tersebut dalam satu baris. Bilangan bulat i dan j harus muncul dalam keluaran dalam urutan yang sama di mana mereka muncul dalam masukan dan harus diikuti dengan panjang siklus maksimum (pada baris yang sama).<\/div>\n

                     Contoh Masukan                                              Contoh Keluaran<\/b>
                     1 10                                                                     1 10 20
                     100 200                                                               100 200 125
                     201 210                                                               201 210 89
                     900 1000                                                             900 1000 174<\/p>\n<\/div>\n\n\n\n\n

    <\/p>\n\n\n

    \n
    \"Algoritma<\/figure>\n<\/div>\n\n\n

    Berikut Code nya dengan menggunakan Java.<\/p>\n\n\n\n

    Java<\/span><\/path><\/path><\/svg><\/span>
    import<\/span> <\/span>java.io.BufferedReader<\/span>;<\/span><\/span>\nimport<\/span> <\/span>java.io.IOException<\/span>;<\/span><\/span>\nimport<\/span> <\/span>java.io.InputStreamReader<\/span>;<\/span><\/span>\n<\/span>\npublic<\/span> <\/span>class<\/span> <\/span>Main<\/span> {<\/span><\/span>\n <\/span>public<\/span> <\/span>static<\/span> <\/span>void<\/span> <\/span>main<\/span>(<\/span>String<\/span>[] <\/span>args<\/span>) <\/span>throws<\/span> <\/span>IOException<\/span> {<\/span><\/span>\n  <\/span>Q100<\/span> q100 <\/span>=<\/span> <\/span>new<\/span> <\/span>Q100<\/span>();<\/span><\/span>\n  <\/span>final<\/span> <\/span>BufferedReader<\/span> reader <\/span>=<\/span> <\/span>new<\/span> <\/span>BufferedReader<\/span>(<\/span>new<\/span> <\/span>InputStreamReader<\/span>(System.in));<\/span><\/span>\n  <\/span>String<\/span> line;<\/span><\/span>\n  <\/span>while<\/span> ((line <\/span>=<\/span> reader.<\/span>readLine<\/span>()) <\/span>!=<\/span> <\/span>null<\/span>) {<\/span><\/span>\n   <\/span>String<\/span> REGEX_WHITESPACE <\/span>=<\/span> <\/span>"<\/span>\\s<\/span>+"<\/span>;<\/span><\/span>\n   <\/span>String<\/span> cleanLine <\/span>=<\/span> line.<\/span>trim<\/span>().<\/span>replaceAll<\/span>(REGEX_WHITESPACE, <\/span>" "<\/span>);<\/span><\/span>\n   <\/span>String<\/span>[] tokens <\/span>=<\/span> cleanLine.<\/span>split<\/span>(REGEX_WHITESPACE);<\/span><\/span>\n   <\/span>if<\/span> (tokens.length <\/span>==<\/span> <\/span>2<\/span>) {<\/span><\/span>\n    <\/span>long<\/span> a <\/span>=<\/span> <\/span>new<\/span> <\/span>Long<\/span>(tokens[<\/span>0<\/span>]).<\/span>longValue<\/span>();<\/span><\/span>\n    <\/span>long<\/span> b <\/span>=<\/span> <\/span>new<\/span> <\/span>Long<\/span>(tokens[<\/span>1<\/span>]).<\/span>longValue<\/span>();<\/span><\/span>\n    System.out.<\/span>println<\/span>(cleanLine <\/span>+<\/span> <\/span>" "<\/span> <\/span>+<\/span> q100.<\/span>countMaxIterationBetweenTwoNumbers<\/span>(a, b));<\/span><\/span>\n   }<\/span><\/span>\n  }<\/span><\/span>\n }<\/span><\/span>\n<\/span>\n <\/span>public<\/span> <\/span>static<\/span> <\/span>class<\/span> <\/span>Q100<\/span> {<\/span><\/span>\n  <\/span>private<\/span> <\/span>final<\/span> <\/span>long<\/span> MAX_CACHE_SIZE <\/span>=<\/span> <\/span>65536<\/span>;<\/span><\/span>\n  <\/span>private<\/span> <\/span>final<\/span> <\/span>long<\/span>[] cache <\/span>=<\/span> <\/span>new<\/span> <\/span>long<\/span>[(<\/span>int<\/span>) MAX_CACHE_SIZE];<\/span><\/span>\n  <\/span>public<\/span> <\/span>long<\/span> <\/span>countIteration<\/span>(<\/span>long<\/span> <\/span>originalNumber<\/span>) {<\/span><\/span>\n   <\/span>long<\/span> n <\/span>=<\/span> originalNumber;<\/span><\/span>\n   <\/span>long<\/span> count <\/span>=<\/span> <\/span>1<\/span>;<\/span><\/span>\n   <\/span>while<\/span> (n <\/span>><\/span> <\/span>1<\/span>) {<\/span><\/span>\n    <\/span>if<\/span> (((n <\/span>-<\/span> <\/span>1<\/span>) <\/span>>><\/span> <\/span>1<\/span>) <\/span>==<\/span> (n <\/span>>><\/span> <\/span>1<\/span>)) {<\/span><\/span>\n     <\/span>long<\/span> newNumber <\/span>=<\/span> (<\/span>3<\/span> <\/span>*<\/span> n) <\/span>+<\/span> <\/span>1<\/span>;<\/span><\/span>\n     <\/span>if<\/span> (newNumber <\/span><<\/span> n) {<\/span><\/span>\n      <\/span>throw<\/span> <\/span>new<\/span> <\/span>RuntimeException<\/span>(<\/span>"overflow :"<\/span> <\/span>+<\/span> originalNumber);<\/span><\/span>\n     } <\/span>else<\/span> {<\/span><\/span>\n      n <\/span>=<\/span> newNumber;<\/span><\/span>\n     }<\/span><\/span>\n    } <\/span>else<\/span> {<\/span><\/span>\n     n <\/span>=<\/span> n <\/span>>><\/span> <\/span>1<\/span>;<\/span><\/span>\n    }<\/span><\/span>\n    <\/span>if<\/span> (n <\/span><<\/span> MAX_CACHE_SIZE <\/span>&&<\/span> cache[(<\/span>int<\/span>) n] <\/span>!=<\/span> <\/span>0<\/span>) {<\/span><\/span>\n     <\/span>return<\/span> count <\/span>+<\/span> cache[(<\/span>int<\/span>) n];<\/span><\/span>\n    }<\/span><\/span>\n    count<\/span>++<\/span>;<\/span><\/span>\n   }<\/span><\/span>\n   <\/span>return<\/span> count;<\/span><\/span>\n  }<\/span><\/span>\n<\/span>\n  <\/span>public<\/span> <\/span>long<\/span> <\/span>countMaxIterationBetweenTwoNumbers<\/span>(<\/span>long<\/span> <\/span>a<\/span>, <\/span>long<\/span> <\/span>b<\/span>) {<\/span><\/span>\n   <\/span>final<\/span> <\/span>long<\/span> big;<\/span><\/span>\n   <\/span>final<\/span> <\/span>long<\/span> small;<\/span><\/span>\n   <\/span>if<\/span> (a <\/span>><\/span> b) {<\/span><\/span>\n    big <\/span>=<\/span> a;<\/span><\/span>\n    small <\/span>=<\/span> b;<\/span><\/span>\n   } <\/span>else<\/span> {<\/span><\/span>\n    small <\/span>=<\/span> a;<\/span><\/span>\n    big <\/span>=<\/span> b;<\/span><\/span>\n   }<\/span><\/span>\n   <\/span>long<\/span> winner <\/span>=<\/span> <\/span>-<\/span>1<\/span>;<\/span><\/span>\n   <\/span>for<\/span> (<\/span>long<\/span> i <\/span>=<\/span> small; i <\/span><=<\/span> big; i<\/span>++<\/span>) {<\/span><\/span>\n    <\/span>long<\/span> count <\/span>=<\/span> <\/span>countIteration<\/span>(i);<\/span><\/span>\n    <\/span>if<\/span> (i <\/span><<\/span> MAX_CACHE_SIZE) {<\/span><\/span>\n     cache[(<\/span>int<\/span>) i] <\/span>=<\/span> count;<\/span><\/span>\n    }<\/span><\/span>\n    <\/span>if<\/span> (count <\/span>><\/span> winner) {<\/span><\/span>\n     winner <\/span>=<\/span> count;<\/span><\/span>\n    }<\/span><\/span>\n   }<\/span><\/span>\n   <\/span>return<\/span> winner;<\/span><\/span>\n  }<\/span><\/span>\n }<\/span><\/span>\n}<\/span><\/span><\/code><\/pre><\/div>\n\n\n\n
    Berikut ini Hasilnya :<\/p>\n\n\n
    \n
    \"\"\/<\/figure>\n<\/div>","protected":false},"excerpt":{"rendered":"
    Masalah 3n + 1 Deskripsi Dalam masalah 3n+1, Anda diminta menganalisis sifat-sifat statu algoritma untuk suatu input yang diberikan. Perhatikan algoritma berikut ini: input n print n if n = 1 then STOP if n ganjil then n = 3n+1 else n = n\/2 GOTO 2   Diberikan input 22, barisan bilangan-bilangan berikut akan dicetak: […]<\/p>\n","protected":false},"author":1,"featured_media":325,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[7],"tags":[],"class_list":["post-67","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-programing"],"yoast_head":"\nAlgoritma 3n+1 Dengan Java - Blog Riki<\/title>\n<meta name=\"description\" content=\"Dalam masalah 3n+1, Anda diminta menganalisis sifat-sifat statu algoritma untuk suatu input yang diberikan.Dalam masalah 3n+1.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/rikiag.com\/blog\/read\/jawaban-3n1-dengan-java\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Algoritma 3n+1 Dengan Java - Blog Riki\" \/>\n<meta property=\"og:description\" content=\"Dalam masalah 3n+1, Anda diminta menganalisis sifat-sifat statu algoritma untuk suatu input yang diberikan.Dalam masalah 3n+1.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/rikiag.com\/blog\/read\/jawaban-3n1-dengan-java\/\" \/>\n<meta property=\"og:site_name\" content=\"Blog Riki\" \/>\n<meta property=\"article:published_time\" content=\"2015-04-29T05:56:00+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-04-21T06:25:02+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/rikiag.com\/blog\/wp-content\/uploads\/2015\/04\/java-3n1-algoritm.webp\" \/>\n\t<meta property=\"og:image:width\" content=\"1110\" \/>\n\t<meta property=\"og:image:height\" content=\"570\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/webp\" \/>\n<meta name=\"author\" content=\"rikiag\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"rikiag\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"3 minutes\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\\\/\\\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/#article\",\"isPartOf\":{\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/\"},\"author\":{\"name\":\"rikiag\",\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/#\\\/schema\\\/person\\\/7b5d3bb9b4a9a4d87a3dcb1a45799979\"},\"headline\":\"Algoritma 3n+1 Dengan Java\",\"datePublished\":\"2015-04-29T05:56:00+00:00\",\"dateModified\":\"2024-04-21T06:25:02+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/\"},\"wordCount\":479,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/#\\\/schema\\\/person\\\/7b5d3bb9b4a9a4d87a3dcb1a45799979\"},\"image\":{\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/#primaryimage\"},\"thumbnailUrl\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/wp-content\\\/uploads\\\/2015\\\/04\\\/java-3n1-algoritm.webp\",\"articleSection\":[\"Programing\"],\"inLanguage\":\"en-US\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/\",\"url\":\"https:\\\/\\\/rikiag.com\\\/blog\\\/read\\\/jawaban-3n1-dengan-java\\\/\",\"name\":\"Algoritma 3n+1 Dengan Java - 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