USACO07MAR Face The Right Way G 差分

【USACO07MAR Face The Right Way G 差分】的更多相关文章

USACO07MAR Face The Right Way G 差分

题目链接 https://www.luogu.com.cn/problem/P2882 分析这个题来看的话好像有点难下手,不如再去读一遍题 N遍,发现一句话很重要Each time the machine is used, it reverses the facing direction of a contiguous group of K cows in the line,就是说只能翻转固定的长度区间,那这样是不是就可以枚举区间了?枚举一层区间,再枚举每次起点,最后加上区间修改,时间复杂度\…

[USACO07MAR]Face The Right Way G

发现选定一个长度后,怎么翻转是固定的. 那我们直接选定一个长度去操作就行. 优化操作过程类似于堆里打持久化标记一样的感觉. [USACO07MAR]Face The Right Way G // Problem: P2882 [USACO07MAR]Face The Right Way G // Contest: Luogu // URL: https://www.luogu.com.cn/problem/P2882 // Memory Limit: 125 MB // Time Limit:…

洛谷 P2882 [USACO07MAR]Face The Right Way G

题目传送门题目描述 Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect. Fortunately, FJ rec…

luogu P2882 [USACO07MAR]Face The Right Way G

题目描述 Farmer John has arranged his N (1 ≤ N ≤ 5,000) cows in a row and many of them are facing forward, like good cows. Some of them are facing backward, though, and he needs them all to face forward to make his life perfect. Fortunately, FJ recently…

BZOJ3453: tyvj 1858 XLkxc(拉格朗日插值)

题意题目链接 Sol 把式子拆开,就是求这个东西 \[\sum_{i = 0} ^n \sum_{j = 1}^{a + id} \sum_{x =1}^j x^k \pmod P\] 那么设$f(x) = \sum_{i = 1}^n i^k$,这是个经典的$k + 1$多项式,直接差值式子就可以化成 \[\sum_{i = 0} ^n \sum_{j = 1}^{a + id} f(j) \pmod P\] 设$g(x) = \sum_{i = 1}^n f(x)$ 对\(g…

BZOJ4650：[NOI2016]优秀的拆分——题解

https://www.lydsy.com/JudgeOnline/problem.php?id=4650 https://www.luogu.org/problemnew/show/P1117 如果一个字符串可以被拆分为 AABB 的形式,其中 A和 B是任意非空字符串,则我们称该字符串的这种拆分是优秀的. 例如,对于字符串 aabaabaa,如果令 A=aab,B=a,我们就找到了这个字符串拆分成 AABB的一种方式. 一个字符串可能没有优秀的拆分,也可能存在不止一种优秀的拆分.比如我们令…

『题解』[NOI2016]优秀的拆分

如果一个字符串可以被拆分为$AABB$的形式,其中$A和 B是任意非空字符串,则我们称该字符串的这种拆分是优秀的. 例如,对于字符串$aabaabaa$,如果令$A=aab$,$B=a$,我们就找到了这个字符串拆分成$AABB$的一种方式. 一个字符串可能没有优秀的拆分,也可能存在不止一种优秀的拆分.比如我们令$A=a$,$B=baa$,也可以用 AABB表示出上述字符串:但是,字符串$abaabaa$就没有优秀的拆分. 现在给出一个长度为$n$的字符串\(S…

CodeChef DGCD Dynamic GCD

CodeChef题面 Time limit 210 ms Code length Limit //内存限制也不说一下,真是的-- 50000 B OS Linux Language limit C, CPP14, JAVA, PYTH, PYTH 3.6, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON…

Storyboards Tutorial 03

这一节主要介绍segues,static table view cells 和 Add Player screen 以及 a game picker screen. Introducing Segues 开启 Main.storyboard拖拉一个Bar Button Item到Players窗口导航栏的右侧.在Attributes inspector中改变Identifier为Add,就会出现一个+号button.…

文件图标SVG