{"html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Ffortran66.hatenablog.com%2Fentry%2F20121230%2F1356881672\" title=\"Problem 14 - fortran66\u306e\u30d6\u30ed\u30b0\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","height":"190","version":"1.0","author_url":"https://blog.hatena.ne.jp/fortran66/","description":"\u5de5\u592b\u305b\u305a\u7d20\u6734\u306b\u6c42\u3081\u307e\u3059\u30024\u30d0\u30a4\u30c8\u6574\u6570\u3067\u306f\u30aa\u30fc\u30d0\u30fc\u30d5\u30ed\u30fc\u3057\u307e\u3057\u305f\u3002 \u30bd\u30fc\u30b9\u30fb\u30d7\u30ed\u30b0\u30e9\u30e0 program PEuler014 implicit none integer :: i, k, kk(1e6) integer(8) :: m do i = 1, 1e6 k = 1 m = i do if (mod(m, 2) == 0) then m = m / 2 else m = 3 * m + 1 end if if (m == 1) exit k = k + 1 end do kk(i) = k end do print *, maxloc(kk), ':', maxval(kk) stop e\u2026","provider_url":"https://hatena.blog","published":"2012-12-30 00:34:32","categories":["ProjectEuler"],"blog_url":"https://fortran66.hatenablog.com/","url":"https://fortran66.hatenablog.com/entry/20121230/1356881672","author_name":"fortran66","image_url":null,"width":"100%","provider_name":"Hatena Blog","blog_title":"fortran66\u306e\u30d6\u30ed\u30b0","type":"rich","title":"Problem 14"}