https://blog.youkuaiyun.com/ACMer_ZPzh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/ACMer_ZP/article/details/80448951https://blog.youkuaiyun.com/ACMer_ZP/article/details/80448951ACMer_ZPFri, 25 May 2018 11:04:52 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/79499312https://blog.youkuaiyun.com/ACMer_ZP/article/details/79499312ACMer_ZPFri, 09 Mar 2018 16:58:59 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/79269599https://blog.youkuaiyun.com/ACMer_ZP/article/details/79269599ACMer_ZPTue, 06 Feb 2018 13:16:57 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78759213https://blog.youkuaiyun.com/ACMer_ZP/article/details/78759213ACMer_ZPSat, 09 Dec 2017 14:50:58 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78570926https://blog.youkuaiyun.com/ACMer_ZP/article/details/78570926ACMer_ZPSat, 18 Nov 2017 20:17:37 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78219257https://blog.youkuaiyun.com/ACMer_ZP/article/details/78219257ACMer_ZPThu, 12 Oct 2017 20:17:46 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78150261https://blog.youkuaiyun.com/ACMer_ZP/article/details/78150261ACMer_ZPMon, 02 Oct 2017 11:08:32 +0800 #includehttps://blog.youkuaiyun.com/ACMer_ZP/article/details/78144840https://blog.youkuaiyun.com/ACMer_ZP/article/details/78144840ACMer_ZPSat, 30 Sep 2017 18:59:42 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78128162https://blog.youkuaiyun.com/ACMer_ZP/article/details/78128162ACMer_ZPThu, 28 Sep 2017 21:08:12 +0800 #include #include #include #include #include #define LL long long using namespace std; const int maxn=4e6+10; const int inf=0x]]>https://blog.youkuaiyun.com/ACMer_ZP/article/details/78116938https://blog.youkuaiyun.com/ACMer_ZP/article/details/78116938ACMer_ZPWed, 27 Sep 2017 20:36:16 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/78055005https://blog.youkuaiyun.com/ACMer_ZP/article/details/78055005ACMer_ZPThu, 21 Sep 2017 17:51:55 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/77895859https://blog.youkuaiyun.com/ACMer_ZP/article/details/77895859ACMer_ZPFri, 08 Sep 2017 16:01:04 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/77506580https://blog.youkuaiyun.com/ACMer_ZP/article/details/77506580ACMer_ZPWed, 23 Aug 2017 15:27:39 +0800 #include #include #include]]>https://blog.youkuaiyun.com/ACMer_ZP/article/details/77367258https://blog.youkuaiyun.com/ACMer_ZP/article/details/77367258ACMer_ZPFri, 18 Aug 2017 11:08:36 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/77184590https://blog.youkuaiyun.com/ACMer_ZP/article/details/77184590ACMer_ZPTue, 15 Aug 2017 10:29:46 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/77113147https://blog.youkuaiyun.com/ACMer_ZP/article/details/77113147ACMer_ZPSat, 12 Aug 2017 11:21:59 +0800 #include #include #includehttps://blog.youkuaiyun.com/ACMer_ZP/article/details/77100030https://blog.youkuaiyun.com/ACMer_ZP/article/details/77100030ACMer_ZPFri, 11 Aug 2017 16:44:17 +0800k时才会有结果所以可以预处理处所有的答案为k的结果最后O(1)输出就好了。枚举k然后从大到小枚举bi每次都将bi的倍数统计出来 然后每次就可以算出(ai-k)%bi==0的有多少个了，具体的看代码和注释。 这里有一个关于bits]]>https://blog.youkuaiyun.com/ACMer_ZP/article/details/76864671https://blog.youkuaiyun.com/ACMer_ZP/article/details/76864671ACMer_ZPTue, 08 Aug 2017 08:42:31 +0800L)，且他们之间两两互素，然后求出这些L的和最大值。 思路:首先可以想到L肯定是n的因子]]>https://blog.youkuaiyun.com/ACMer_ZP/article/details/76855030https://blog.youkuaiyun.com/ACMer_ZP/article/details/76855030ACMer_ZPMon, 07 Aug 2017 18:01:30 +0800https://blog.youkuaiyun.com/ACMer_ZP/article/details/76854293https://blog.youkuaiyun.com/ACMer_ZP/article/details/76854293ACMer_ZPMon, 07 Aug 2017 17:40:44 +0800 #include #include #include #include #define LL long long using namespace std; const int maxn=1]]>