{"version":"1.0","author_url":"https://blog.hatena.ne.jp/inamori/","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Finamori.hateblo.jp%2Fentry%2F20110507%2Fp1\" title=\"Scala\u3067Project Euler\uff083\uff09 - inamori\u2019s diary\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","image_url":null,"provider_url":"https://hatena.blog","blog_url":"https://inamori.hateblo.jp/","provider_name":"Hatena Blog","title":"Scala\u3067Project Euler\uff083\uff09","categories":["Scala"],"description":"Problem 1Python\u306a\u3089 from itertools import * N = 1000 print sum(ifilter(lambda n: n % 3 == 0 or n % 5 == 0, xrange(1, N))) Haskell\u306a\u3089 n = 1000 main = print (sum (filter (\\m -> mod m 3 == 0 || mod m 5 == 0) [1..n-1])) \u3053\u3093\u306a\u95a2\u6570\u578b\u3067\u66f8\u304f\u3053\u3068\u3092\u76ee\u6a19\u306b\u3057\u307e\u3057\u3087\u3046\u3002 Range \u307e\u305a\u3001Python\u306exrange\u307f\u305f\u3044\u306a\u306e\u306f\u3053\u3046\u66f8\u304d\u307e\u3059\u3002 println (1 to 10) Range(1, 2,\u2026","width":"100%","type":"rich","height":"190","author_name":"inamori","published":"2011-05-07 00:00:00","blog_title":"inamori\u2019s diary","url":"https://inamori.hateblo.jp/entry/20110507/p1"}