{"author_url":"https://blog.hatena.ne.jp/parukii/","description":"dp[i][j]:=\u5de6\u306f\u6587\u5b57i\u307e\u3067\u3001\u53f3\u5074\u306fj\u307e\u3067\u898b\u305f\u3068\u304d\u306e\u5076\u6570\u9577(\u9577\u30550\u4ee5\u4e0a)\u306e\u56de\u5206\u306e\u500b\u6570\u306e\u6570(\u305f\u3060\u30571<=i<=j<=N) \u305f\u3068\u3048\u3070\u3001\u6587\u5b57\u5217s\u304c\u4ee5\u4e0b\u306e\u3088\u3046\u306a\u5f62\u306e\u5834\u5408(1-indexed) abcxycba abc...ba dp[3][7] = 2^2 = 4 1<=j-i<=n\u3092 n\u304b\u3089\u9806\u306b\u78ba\u5b9a\u3057\u3066\u3044\u3051\u3070\u3044\u3044\u3002\u3064\u307e\u308a dp[i][j] = dp[i-1][j] + dp[i][j+1] - dp[i-1][j+1] + (s[i]==s[j]?dp[i-1][j+1]:0) = dp[i-1][j] + dp[i][j+1] - (s[i]==s[j]?0:dp[i-1][j+1])","blog_title":"paruki\u306e\u30d6\u30ed\u30b0","categories":[],"url":"https://par.hateblo.jp/entry/2017/02/10/135714","provider_url":"https://hatena.blog","blog_url":"https://par.hateblo.jp/","version":"1.0","width":"100%","image_url":"https://cdn-ak.f.st-hatena.com/images/fotolife/p/parukii/20170210/20170210135639.jpg","type":"rich","author_name":"parukii","published":"2017-02-10 13:57:14","title":"SRM 708 DIV2 Hard: PalindromicSubseq2","provider_name":"Hatena Blog","height":"190","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fpar.hateblo.jp%2Fentry%2F2017%2F02%2F10%2F135714\" title=\"SRM 708 DIV2 Hard: PalindromicSubseq2 - paruki\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>"}