{"type":"rich","image_url":null,"url":"https://inamori.hateblo.jp/entry/20070420/p1","provider_url":"https://hatena.blog","author_url":"https://blog.hatena.ne.jp/inamori/","published":"2007-04-20 00:00:00","height":"190","title":"\u30de\u30eb\u30c1\u30b9\u30ec\u30c3\u30c9\u306e\u52b9\u7387\uff083\uff09","categories":["\u30d7\u30ed\u30b0\u30e9\u30df\u30f3\u30b0"],"width":"100%","version":"1.0","author_name":"inamori","description":"\u524d\u56de\u306eE\u306e\u8a08\u7b97\u306f\u3001 \u3053\u308c\u3092\u4f7f\u3048\u3070\u7c21\u5358\u306a\u3053\u3068\u306b\u6628\u65e5\u306e\u591c\u6c17\u304c\u3064\u3044\u305f\u3002 (n - k)nCk = nn-1Ck \u307e\u305a\u3001 N = 2l + 1 \u3068\u8868\u305b\u308b\u5834\u5408\u3092\u8003\u3048\u308b\u3002 I = NNC0 + ... + (N - l)NCl + (N - l)NCl+1 + ... + NNCN \u3068\u304a\u304f\u3068\u3001 I / 2 = 2N-1E = NNC0 + ... + (N - l)NCl = NN-1C0 + ... + NN-1Cl = N2N-2 + NN-1Cl / 2 \u5e73\u5747\u6642\u9593\u306f\u3001 E = N / 2 + NN-1Cl / 2N \u52b9\u7387\u306f\u3001 \u3068\u306a\u308b\u3002 N\u304c\u5076\u6570\u306e\u5834\u5408\u3082\u540c\u69d8\u306b\u6c42\u3081\u3089\u308c\u3066\u3001 l = [ N / 2 ] \u2026","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Finamori.hateblo.jp%2Fentry%2F20070420%2Fp1\" title=\"\u30de\u30eb\u30c1\u30b9\u30ec\u30c3\u30c9\u306e\u52b9\u7387\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>","provider_name":"Hatena Blog","blog_url":"https://inamori.hateblo.jp/","blog_title":"inamori\u2019s diary"}