{"id":16421,"date":"2022-10-16T16:05:48","date_gmt":"2022-10-16T10:35:48","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-modulus-of-a-very-large-number\/"},"modified":"2022-10-16T16:05:48","modified_gmt":"2022-10-16T10:35:48","slug":"solved-modulus-of-a-very-large-number","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-modulus-of-a-very-large-number\/","title":{"rendered":"[Solved] Modulus of a very large number"},"content":{"rendered":"<p> [ad_1]<br \/>\n<\/p>\n<div id=\"answer-33223058\" class=\"answer js-answer accepted-answer js-accepted-answer\" data-answerid=\"33223058\" data-parentid=\"33223005\" data-score=\"4\" data-position-on-page=\"1\" data-highest-scored=\"1\" data-question-has-accepted-highest-score=\"1\" itemprop=\"acceptedAnswer\" itemscope itemtype=\"https:\/\/schema.org\/Answer\">\n<div class=\"post-layout\">\n<div class=\"votecell post-layout--left\"><\/div>\n<div class=\"answercell post-layout--right\">\n<div class=\"s-prose js-post-body\" itemprop=\"text\">\n<p><a rel=\"nofollow noopener\" target=\"_blank\" href=\"http:\/\/docs.oracle.com\/javase\/8\/docs\/api\/java\/math\/BigInteger.html\">BigInteger<\/a> has the <a rel=\"nofollow noopener\" target=\"_blank\" href=\"http:\/\/docs.oracle.com\/javase\/8\/docs\/api\/java\/math\/BigInteger.html#divideAndRemainder-java.math.BigInteger-\">divideAndRemainder(&#8230;)<\/a> method, one that returns a BigInteger array, the first item the division result, and the second, the remainder (which is what mod really does in Java).<\/p>\n<p><strong>Update<\/strong><\/p>\n<p>Including comment by Mark Dickinson to other answer:<\/p>\n<blockquote>\n<p>There&#8217;s a much simpler linear-time algorithm: set acc to 0, then for each digit d in x in turn (left to right), convert d to an integer and set acc = (acc * 10 + d) % y. Once the digits are consumed, acc is your result.<\/p>\n<\/blockquote>\n<p>Implemented all 3 algorithms as described:<\/p>\n<pre><code>private static int testMarkDickinson(String x, int y) {\n    int acc = 0;\n    for (int i = 0; i &lt; x.length(); i++)\n        acc = (acc * 10 + x.charAt(i) - '0') % y;\n    return acc;\n}\n\nprivate static int testHovercraftFullOfEels(String x, int y) {\n    return new BigInteger(x).divideAndRemainder(BigInteger.valueOf(y))[1].intValue();\n}\n\nprivate static int testMrM(String x, int y) {\n    String s = x;\n    while (s.length() &gt;= 7) {\n        int len = Math.min(9, s.length());\n        s = Integer.parseInt(s.substring(0, len)) % y + s.substring(len);\n    }\n    return Integer.parseInt(s) % y;\n}\n<\/code><\/pre>\n<p>Testing with seeded random number for verifiable test (didn&#8217;t want to hardcode a 98765 digit number):<\/p>\n<pre><code>public static void main(String[] args) {\n    Random r = new Random(98765);\n    char[] buf = new char[98765];\n    for (int i = 0; i &lt; buf.length; i++)\n        buf[i] = (char)('0' + r.nextInt(10));\n\n    String x = new String(buf);\n    int y = 98765;\n\n    System.out.println(testMarkDickinson(x, y));\n    System.out.println(testHovercraftFullOfEels(x, y));\n    System.out.println(testMrM(x, y));\n\n    long test1nano = 0, test2nano = 0, test3nano = 0;\n    for (int i = 0; i &lt; 10; i++) {\n        long nano1 = System.nanoTime();\n        testMarkDickinson(x, y);\n        long nano2 = System.nanoTime();\n        testHovercraftFullOfEels(x, y);\n        long nano3 = System.nanoTime();\n        testMrM(x, y);\n        long nano4 = System.nanoTime();\n        test1nano += nano2 - nano1;\n        test2nano += nano3 - nano2;\n        test3nano += nano4 - nano3;\n    }\n    System.out.printf(\"%11d%n%11d%n%11d%n\", test1nano, test2nano, test3nano);\n}\n<\/code><\/pre>\n<p>Output:<\/p>\n<pre><code>23134\n23134\n23134\n    8765773\n 1514329736\n 7563954071\n<\/code><\/pre>\n<p>All 3 produced same result, but there&#8217;s a clear difference in performance, and Mr. M&#8217;s &#8220;better solution&#8221; is the worst of them all.<\/p>\n<\/p><\/div>\n<div class=\"mt24\"><\/div>\n<\/div>\n<p>            <span class=\"d-none\" itemprop=\"commentCount\">12<\/span> <\/p><\/div>\n<\/div>\n<p>[ad_2]<\/p>\n<p>solved Modulus of a very large number <\/p>\n","protected":false},"excerpt":{"rendered":"<p>[ad_1] BigInteger has the divideAndRemainder(&#8230;) method, one that returns a BigInteger array, the first item the division result, and the second, the remainder (which is what mod really does in Java). Update Including comment by Mark Dickinson to other answer: There&#8217;s a much simpler linear-time algorithm: set acc to 0, then for each digit d &#8230; <a title=\"[Solved] Modulus of a very large number\" class=\"read-more\" href=\"https:\/\/jassweb.com\/solved\/solved-modulus-of-a-very-large-number\/\" aria-label=\"More on [Solved] Modulus of a very large number\">Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[457,323,4170,1369],"class_list":["post-16421","post","type-post","status-publish","format-standard","hentry","category-solved","tag-algorithm","tag-java","tag-numeric","tag-optimization"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.5 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>[Solved] Modulus of a very large number - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-modulus-of-a-very-large-number\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Modulus of a very large number - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] BigInteger has the divideAndRemainder(&#8230;) method, one that returns a BigInteger array, the first item the division result, and the second, the remainder (which is what mod really does in Java). 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