USACO 6.1 Cow XOR

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Cow XORAdrian Vladu -- 2005 Farmer John is stuck with another problem while feeding his cows. All of his N (1 ≤ N ≤ 100,000) cows (numbered 1..N) are lined up in front of the barn, sorted by their rank in their social hierarchy. Cow #1 has the highes…

NBUT 1525 Cow Xor（01字典树+前缀思想）

[1525] Cow Xor 时间限制: 2000 ms 内存限制: 65535 K 问题描述农民约翰在喂奶牛的时候被另一个问题卡住了.他的所有N(1 <= N <= 100,000)个奶牛在他面前排成一行(按序号1..N的顺序),按照它们的社会等级排序.奶牛#1有最高的社会等级,奶牛#N最低.每个奶牛同时被指定了一个不唯一的附加值,这个数在0..2^21 - 1的范围内. 帮助农民约翰找出应该从哪一头奶牛开始喂,使得从这头奶牛开始的一个连续的子序列上,奶牛的附加值的异或最大. 如果有多个这…

[USACO]6.1.3 cow xor（二进制+Trie）

题意:给你一个序列(n<=100000),求出一个连续的子序列[i,j]使得ai xor ai+1 xor…… xor aj最大,求出这个最大值(其中每个数<=2^21) 分析:题目和求一段子序列的和或者积很像,只是运算法则改变,所以可以往求子段和.积的方法上面考虑首先如果设sum[i]=a1 xor a2 xor a3……xor ai,则子序列[i,j]的xor值可以表达为sum[j] xor sum[i-1],这样是O(n^2)的是不行的有以往的处理经验我们知道,后面那个j一定要枚举的…

USACO 2.3 Cow Pedigrees

Cow Pedigrees Silviu Ganceanu -- 2003 Farmer John is considering purchasing a new herd of cows. In this new herd, each mother cow gives birth to two children. The relationships among the cows can easily be represented by one or more binary trees with…

USACO 2012 Feb Cow Coupons

2590: [Usaco2012 Feb]Cow Coupons Time Limit: 10 Sec Memory Limit: 128 MB Submit: 349 Solved: 181 [Submit][Status][Discuss] Description Farmer John needs new cows! There are N cows for sale (1 <= N <= 50,000), and FJ has to spend no more than his bud…

USACO 2.4 Cow Tours

Cow Tours Farmer John has a number of pastures on his farm. Cow paths connect some pastures with certain other pastures, forming a field. But, at the present time, you can find at least two pastures that cannot be connected by any sequence of cow pat…

【USACO 6.1 Cow XOR】的更多相关文章

USACO 6.1 Cow XOR

NBUT 1525 Cow Xor（01字典树+前缀思想）

[USACO]6.1.3 cow xor（二进制+Trie）

USACO 2.3 Cow Pedigrees

USACO 2012 Feb Cow Coupons

USACO 2.4 Cow Tours

[1525] Cow Xor

[Usaco 2012 Feb]Cow coupons牛券：反悔型贪心

USACO 2011 November Cow Lineup /// map set 尺取法 oj25279

USACO 2009 Open Cow Line /// 队列 oj26220