题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might h…

P1360 [USACO07MAR]黄金阵容均衡Gold Balanced L… 题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For ex…

P1360 [USACO07MAR]黄金阵容均衡Gold Balanced L… 题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For ex…

https://www.luogu.org/problem/show?pid=1360 题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For…

题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might h…

传送门 真的骚的一个题,看了半天只会个前缀和+暴力.. 纯考思维.. 良心题解 #include <cstdio> #include <cstring> #include <algorithm> #define M 41 #define N 100011 #define max(x, y) ((x) > (y) ? (x) : (y)) int n, m, ans, last; struct node { int sum[M], id; node() { id =…

题目 不得不说这个题非常毒瘤. 简化题意 这个题的暴力还是非常好想的,完全可以过\(50\%\)的数据.但是\(100\%\)就很难想了. 因为数据很大,所以我们需要用\(O(\sqrt n)\)的时间求出每个时间的前面第一个跟它满足均衡区间的时间. 此时我们会很快速的想到\(Hash\)或\(Map\),他们的时间复杂度是小于\(O(\sqrt n)的\)所以满足要求,因此一般看到寻找相同得值的时候,最好还是想想怎么优化吧 \(Code\) #include <iostream> #incl…

[LuoguP1360][USACP07MAR]黄金阵容均衡(Link) 每天会增加一个数\(A\),将\(A\)二进制分解为\(a[i]\),对于每一个\(i\)都增加\(a[i]\),如果一段时间之内所有的位数上的数都增加了同一个数,那么成这个天数区间为均衡的.现在要求最长的均衡区间. 这道题很显然可以使用前缀和.我们统计每一天的前缀和为一个\(map\)数组\(A\).如果区间\([L, R]\)为均衡区间,那么\(A[R] - A[L - 1]\)就一定满足每一个二进制分解的位数都相等.…

P1360 [USACO07MAR]Gold Balanced Lineup G (前缀和+思维) 前言 题目链接 本题作为一道Stl练习题来说,还是非常不错的,解决的思维比较巧妙 算是一道不错的题 思路分析 第一眼看到这题,我还以为是数据结构题,看来半天没看出来数据结构咋做(我还是太菜了) 我们对\(m\)种能力有\(n\)次操作,需要找到对每种能力提升相同的最大操作区间的长度,求最大 区间,我们考虑维护这\(m\)种技能提升值的前缀和,假设第\(l+1\)次操作到第\(r\)次操作对\(m\…

Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…

1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列 Time Limit: 5 Sec  Memory Limit: 64 MBSubmit: 510  Solved: 196[Submit][Status][Discuss] Description Farmer John's N cows (1 <= N <= 100,000) share many similarities. In fact, FJ has been able to narrow…

Gold Balanced Lineup Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 13540   Accepted: 3941 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared…

L.Map API各种类中的核心部分,用来在页面中创建地图并操纵地图. 使用 example // initialize the map on the "map" div with a given center and zoom var map = L.map('map', { center: [51.505, -0.09], zoom: 13 }); 构造器 构造器 使用 描述 L.Map( <HTMLElement|String> id, <Map options…

Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10924 Accepted: 3244 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…

P3121 [USACO15FEB]审查(黄金)Censoring (Gold) (银的正解是KMP) AC自动机+栈 多字符串匹配--->AC自动机 删除单词的特性--->栈 所以我们先打个AC自动机模板 然后搞2个栈维护: AC自动机目前跑到字典树上的哪个点 已经跑过且没被删除的字符(答案栈) 每次碰到有结尾标记的点,就让2个栈弹出这个点所对应的单词的长度  最后输出第二个栈就行了 attention:输出答案后要换行,否则会蜜汁爆炸7pts P3121 code(P4824要稍作修改):…

题目描述 Farmer John has purchased a subscription to Good Hooveskeeping magazine for his cows, so they have plenty of material to read while waiting around in the barn during milking sessions. Unfortunately, the latest issue contains a rather inappropria…

题目描述 Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might h…

Description N(1<=N<=100000)头牛,一共K(1<=K<=30)种特色, 每头牛有多种特色,用二进制01表示它的特色ID.比如特色ID为13(1101),则它有第1.3.4种特色.[i,j]段被称为balanced当且仅当K种特色在[i,j]内拥有次数相同.求最大的[i,j]段长度. Input * Line 1: Two space-separated integers, N and K. * Lines 2..N+1: Line i+1 contains…

\(\mathbf{P1360}\) 题解 思路 设\(sum[t][i]\)为截至第t天第i项能力的提升总次数. 由题意可知一个时期为均衡时期\([t_1,t_2]\),当且仅当 \(\forall\;1\leq i \leq m,sum[t_2][i]-sum[t_1-1][i]\)都相等. 由上,对于每个\(t\),可以将序列\(sum[t]\)的每个数减去\(sum[t][1]\),得到一个序列\(f\)(长为\(m\)),它对应值\(t\). 也可以用差分的方法构造序列\(f\),使\…

Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1…

要求一段最大的区间里每个能力的增长值是一样的. 我们首先求一遍前缀和,发现,如果区间内[l,r]每个能力的增长值是一样的话,那么前缀和[r]和[l-1]的差分也应该是一样的. 那么我们把前缀和的差分hash一下,排个序,求出hash值相同的且跨越最远的两个端点,就是答案了. # include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <…

Description N(1<=N<=100000)头牛,一共K(1<=K<=30)种特色,每头牛有多种特色,用二进制01表示它的特色ID.比如特色ID为13(1101),则它有第1.3.4种特色.[i,j]段被称为balanced当且仅当K种特色在[i,j]内拥有次数相同.求最大的[i,j]段长度. Input 第一行给出数字N,K 下面N行每行给出一个数字,代表这头牛的特征值 Output 求出一个区间值,在这个区间中,所有牛的这K种特征值的总和是相等的. Sample In…

Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1 might have s…

题目:http://poj.org/problem?id=3274 #include <iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<stack> #include<queue> #include<iomanip> #include<cmath> #include<map> #include<ve…

Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10711   Accepted: 3182 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list…

题面 传送门 Sol AC自动机+栈,每次匹配到栈顶减去这个单词的长度,回到之前的状态 最后栈中留下的就是答案 # include <bits/stdc++.h> # define IL inline # define RG register # define Fill(a, b) memset(a, b, sizeof(a)) using namespace std; typedef long long ll; const int _(1e5 + 5); int ch[26][_], tot…

[题意]给定n头牛,k个特色,给出每头牛拥有哪些特色的二进制对应数字,[i,j]平衡当且仅当第i~j头牛的所有特色数量都相等,求最长区间长度. [算法]平衡树+数学转化 [题解]统计前缀和sum[i][j]表示前i头牛特色为j的数量,则区间i~j平衡需要满足: sum[j][1]-sum[i-1][1]=sum[j][2]-sum[i-1][2]=sum[j][3]-sum[i-1][3]=... 移项可得,只须 sum[j][1]-sum[j][2]=sum[i-1][1]-sum[i-1][…

Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 ≤ K ≤ 30). For example, cows exhibiting feature #1…

这题,看到别人的解题报告做出来的,分析: 大概意思就是: 数组sum[i][j]表示从第1到第i头cow属性j的出现次数. 所以题目要求等价为: 求满足 sum[i][0]-sum[j][0]=sum[i][1]-sum[j][1]=.....=sum[i][k-1]-sum[j][k-1] (j<i) 中最大的i-j 将上式变换可得到 sum[i][1]-sum[i][0] = sum[j][1]-sum[j][0] sum[i][2]-sum[i][0] = sum[j][2]-sum[j]…

题目链接: https://vjudge.net/problem/POJ-3274 题目大意: 给定多头牛的属性,每头牛的属性由一个非负数表示,该数的二进制表示不会超过K位,它的二进制表示的每一位若为1则表示该牛有对应的第i种属性,若为0则表示没有该属性. 对于给定的牛的顺序,要求输出某一段子序列的长度,这个子序列中的牛的K个属性对应相加以后全部相等. 假设n=3, k = 3 输入的3个数变成的二进制分别为(a1, a2, a3), (b1, b2, b3), (c1, c2, c3) 设su…