20150812 常州培训 扫坑
1. 给出一个数的$absurd$值为：
[1]去掉前导后导0；[2]设去掉后的数为$x$，最后一位数为$p$，$x$的位数为$m$；[3]$absurd$值为：当$m=5$时，$absurd=2\times p-1$；否则，$absurd=2\times p$。
定义一个数$c$是$absurd$的，当且仅当在$[0.95c,1.05c]$内，存在一个数$d$，使得$d$的$absurd$值小于$c$的$absurd$值，那么就是$absurd$的。
5000组数据，每组数据给出$c$，求是否为$absurd$的。
法1:打表，总共281个数是$not absurd$的，存储即可。复杂度$O(281 \times T)$
法2:200以内的我们可以暴力处理；对于一个数c，我们先去零，然后找他前面和后面的3个数即可。
2. 一个傻逼$DynamicProgramming$= =
求最长连续段，若相同最大化段数。
3. kmp，我竟然没看出来
问两个时钟旋转后能否重合。
求差后，复制一遍，kmp匹配，如果匹配=n，那么就是possible，否则impossible。
我的情况：
t1t2 AC无压力
t3 哈希炸了，没判定先后，发现了好多问题
后来得知kmp太神了。orzorz
2024年1月10日 13:26
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