{"id":6067,"date":"2021-02-16T06:00:31","date_gmt":"2021-02-16T04:00:31","guid":{"rendered":"http:\/\/www.glc.us.es\/~jalonso\/exercitium\/?p=6067"},"modified":"2021-02-23T08:42:29","modified_gmt":"2021-02-23T06:42:29","slug":"cantidad-de-numeros-oblongos-en-un-intervalo","status":"publish","type":"post","link":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/cantidad-de-numeros-oblongos-en-un-intervalo\/","title":{"rendered":"Cantidad de n\u00fameros oblongos en un intervalo"},"content":{"rendered":"

Un n\u00famero oblongo<\/a> es un n\u00famero que es el producto de dos n\u00fameros naturales consecutivos; es decir, n es un n\u00famero oblongo si existe un n\u00famero natural x tal que n = x(x+1). Por ejemplo, 42 es un n\u00famero oblongo porque 42 = 6 x 7.<\/p>\n

La sucesi\u00f3n de los n\u00fameros oblongos es<\/p>\n

\n   0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, ...\n<\/pre>\n
En el intervalo [10,30] hay 3 n\u00fameros oblongos (el 12, el 20 y el 30).<\/p>\n
Definir las funciones<\/p>\n
\n   oblongos            :: [Integer]\n   oblongosEnIntervalo :: Integer -> Integer -> Int\n<\/pre>\n
tales que<\/p>\n
    \n
  • oblongos es la sucesi\u00f3n de los n\u00fameros oblongos. Por ejemplo,<\/li>\n<\/ul>\n
    \n     take 15 oblongos   == [0,2,6,12,20,30,42,56,72,90,110,132,156,182,210]\n     oblongos !! 50     == 2550\n     oblongos !! (10^5) == 10000100000\n     oblongos !! (10^6) == 1000001000000\n     oblongos !! (10^7) == 100000010000000\n<\/pre>\n
      \n
    • (oblongosEnIntervalo a b) es la cantidad de n\u00fameros oblongos en el intervalo [a,b]. Por ejemplo,<\/li>\n<\/ul>\n
      \n     oblongosEnIntervalo 10 30           ==  3\n     oblongosEnIntervalo (10^3) (10^10)  ==  99968\n     oblongosEnIntervalo (10^3) (10^12)  ==  999968\n     oblongosEnIntervalo (10^3) (10^14)  ==  9999968\n<\/pre>\n
      Soluciones<\/h4>\n
      \n-- 1\u00aa definici\u00f3n de oblongos\noblongos1 :: [Integer]\noblongos1 = [n*(n+1) | n <- [0..]]\n\n-- 2\u00aa definici\u00f3n de oblongos\noblongos2 :: [Integer]\noblongos2 = zipWith (*) [0..] [1..]\n\n-- 3\u00aa definici\u00f3n de oblongos\noblongos3 :: [Integer]\noblongos3 = scanl1 (+) [0,2..]\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n--    \u03bb> oblongos1 !! (10^7)\n--    100000010000000\n--    (3.05 secs, 1,840,112,008 bytes)\n--    \u03bb> oblongos2 !! (10^7)\n--    100000010000000\n--    (0.90 secs, 2,480,112,304 bytes)\n--    \u03bb> oblongos3 !! (10^7)\n--    100000010000000\n--    (3.38 secs, 2,252,411,640 bytes)\n\n-- Definici\u00f3n de oblongos\n-- ======================\n\n-- En lo que sigue, usaremos la 2\u00aa.\noblongos :: [Integer]\noblongos = oblongos2\n\n-- 1\u00aa definici\u00f3n de oblongosEnIntervalo\n-- ====================================\n\noblongosEnIntervalo1 :: Integer -> Integer -> Int\noblongosEnIntervalo1 a b =\n  length [x | x <- [a..b]\n            , x `elem` takeWhile (<=b) oblongos]\n\n-- 2\u00aa definici\u00f3n de oblongosEnIntervalo\n-- ====================================\n\noblongosEnIntervalo2 :: Integer -> Integer -> Int\noblongosEnIntervalo2 a b =\n  length (takeWhile (<=b) (dropWhile (< a) oblongos))\n\n-- Comparaci\u00f3n de eficiencia\n-- =========================\n\n-- La comparaci\u00f3n es\n--    \u03bb> oblongosEnIntervalo1 100 (10^5)\n--    306\n--    (2.43 secs, 1,784,537,632 bytes)\n--    \u03bb> oblongosEnIntervalo2 100 (10^5)\n--    306\n--    (0.01 secs, 119,112 bytes)\n<\/pre>\n
      Nuevas soluciones<\/h4>\n
        \n
      • En los comentarios se pueden escribir nuevas soluciones.\n
      • El c\u00f3digo se debe escribir entre una l\u00ednea con <pre lang="haskell"> y otra con <\/pre>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"
        Un n\u00famero oblongo es un n\u00famero que es el producto de dos n\u00fameros naturales consecutivos; es decir, n es un n\u00famero oblongo si existe un n\u00famero natural x tal que n = x(x+1). Por ejemplo, 42 es un n\u00famero oblongo porque 42 = 6 x 7. La sucesi\u00f3n de los n\u00fameros oblongos es 0, 2,…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"jetpack_post_was_ever_published":false,"_kad_post_transparent":"","_kad_post_title":"","_kad_post_layout":"","_kad_post_sidebar_id":"","_kad_post_content_style":"","_kad_post_vertical_padding":"","_kad_post_feature":"","_kad_post_feature_position":"","_kad_post_header":false,"_kad_post_footer":false,"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"footnotes":"","_jetpack_memberships_contains_paid_content":false},"categories":[5],"tags":[],"jetpack_featured_media_url":"","jetpack_sharing_enabled":true,"_links":{"self":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6067"}],"collection":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/comments?post=6067"}],"version-history":[{"count":2,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6067\/revisions"}],"predecessor-version":[{"id":6117,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/posts\/6067\/revisions\/6117"}],"wp:attachment":[{"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/media?parent=6067"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/categories?post=6067"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.glc.us.es\/~jalonso\/exercitium\/wp-json\/wp\/v2\/tags?post=6067"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}