https://blog.youkuaiyun.com/2301_80353190zh-cnhttps://blog.youkuaiyun.com/https://blog.youkuaiyun.com/2301_80353190/article/details/145793111https://blog.youkuaiyun.com/2301_80353190/article/details/1457931112301_80353190Sun, 23 Feb 2025 14:13:17 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145590200https://blog.youkuaiyun.com/2301_80353190/article/details/1455902002301_80353190Sat, 15 Feb 2025 18:20:56 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145532356https://blog.youkuaiyun.com/2301_80353190/article/details/1455323562301_80353190Sun, 09 Feb 2025 15:40:52 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145531631https://blog.youkuaiyun.com/2301_80353190/article/details/1455316312301_80353190Sun, 09 Feb 2025 14:44:02 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145512984https://blog.youkuaiyun.com/2301_80353190/article/details/1455129842301_80353190Sat, 08 Feb 2025 20:42:32 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145498487https://blog.youkuaiyun.com/2301_80353190/article/details/1454984872301_80353190Fri, 07 Feb 2025 18:12:05 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145365564https://blog.youkuaiyun.com/2301_80353190/article/details/1453655642301_80353190Sun, 26 Jan 2025 11:24:23 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145357626https://blog.youkuaiyun.com/2301_80353190/article/details/1453576262301_80353190Sat, 25 Jan 2025 22:06:19 +0800m时构造会不一样，这种情况的具体构造可以变成回文串，前n-m个和后n-m个对称，中间n-2m个相等的就好了。首先我们找k的下界，通过D题我们可以知道，k与两个相同字母之间的间隔距离有关，那么想要最小化相隔距离，我们就要尽量地填充不一样的字母，但是每26个字母之后必定会出现相同的字母，那么不难推出来下界就是max(0,n-26)，上界很容易知道是n-1。]]>https://blog.youkuaiyun.com/2301_80353190/article/details/145357078https://blog.youkuaiyun.com/2301_80353190/article/details/1453570782301_80353190Sat, 25 Jan 2025 17:38:03 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/145340828https://blog.youkuaiyun.com/2301_80353190/article/details/1453408282301_80353190Fri, 24 Jan 2025 16:03:55 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/141094995https://blog.youkuaiyun.com/2301_80353190/article/details/1410949952301_80353190Sat, 10 Aug 2024 19:30:10 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140813755https://blog.youkuaiyun.com/2301_80353190/article/details/1408137552301_80353190Sat, 10 Aug 2024 19:08:42 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140957050https://blog.youkuaiyun.com/2301_80353190/article/details/1409570502301_80353190Tue, 06 Aug 2024 16:37:09 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140895232https://blog.youkuaiyun.com/2301_80353190/article/details/1408952322301_80353190Sun, 04 Aug 2024 12:52:09 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140890157https://blog.youkuaiyun.com/2301_80353190/article/details/1408901572301_80353190Sat, 03 Aug 2024 16:15:38 +0800=2）进行操作是最有价值的（因为i=1时能对任意值进行加减），所以我们考虑能不能把所有数都加到a1上。我们发现ai是严格大于等于1的，那么我们从左往右遍历，总和必然大于等于i，也就是说，我们可以通过先把a1借给ai，再将ai全部还给a1，这样一次操作数就是2，最后我们只需要将平均数分给每个ai（i>=2）就可以了，操作数为3*(n-1)。反之A就要占前面放最小的。换位思考会发现，到A操作时，如果B剩余的字母全小于等于A的，那么A就要占最后放手中最大的。]]>https://blog.youkuaiyun.com/2301_80353190/article/details/140830943https://blog.youkuaiyun.com/2301_80353190/article/details/1408309432301_80353190Thu, 01 Aug 2024 13:28:25 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140774922https://blog.youkuaiyun.com/2301_80353190/article/details/1407749222301_80353190Mon, 29 Jul 2024 17:01:17 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140739236https://blog.youkuaiyun.com/2301_80353190/article/details/1407392362301_80353190Sat, 27 Jul 2024 18:57:00 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140722818https://blog.youkuaiyun.com/2301_80353190/article/details/1407228182301_80353190Sat, 27 Jul 2024 15:23:09 +0800https://blog.youkuaiyun.com/2301_80353190/article/details/140733380https://blog.youkuaiyun.com/2301_80353190/article/details/1407333802301_80353190Sat, 27 Jul 2024 13:33:13 +0800