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…

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…

http://poj.org/problem?id=3274 题意 :农夫约翰的n(1 <= N <= 100000)头奶牛,有很多相同之处,约翰已经将每一头奶牛的不同之处,归纳成了K种特性,比如1号特性代表它身上有斑点,2号特性代表它比较喜欢用passcal 写程序而不是C .约翰使用“特性标识符”来描述奶牛的各种特性,例如:一头奶牛的特性标识符是13,将13写成二进制1101,从右向左看,就表示这头奶牛具有1,3,4这三个特性,但没有2号特性,约翰把n头奶牛排成一排,发现在有些连续区间里的…

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…

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…

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…

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\…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 34140   Accepted: 16044 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 39046   Accepted: 18291 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…

  Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 75294   Accepted: 34483 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer J…

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…

http://poj.org/problem?id=3274 ***** #include <stdio.h> #include <iostream> #include <string.h> #include <stdlib.h> using namespace std; ; ; int Maxdis,k; ],sum[N][]; ]; struct node { int row_i; struct node *next; }*hash[N]; bool c…

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…

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…

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…

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…

http://poj.org/problem?id=3264 题意:给出n个数,还有q个询问,询问[l,r]区间里面最大值和最小值的差值. 思路:RMQ模板题,开两个数组维护最大值和最小值就行. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; #define N 50010 #define INF 0x3f3f3f3…

#include<cstdio> #include<cstring> #include<cmath> #include <cstdlib> #define MAXN 100001 using namespace std; ; ]; ]; ]; int n,k,len; struct node { int xx; node *next; }; node *hash[mod]; bool cmp(int a,int b) { ; j<k; j++) if(…

ST表模版 #include<cstdio> #include<cstdlib> #include<algorithm> #include<cstring> #include<cmath> #define MAXN 50000+10 #define LOG 20 #define pii pair<int,int> using namespace std; int dmax[MAXN][LOG]; int dmin[MAXN][LOG]…

[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][…

n<=100000个数表示每头牛在K<=30种物品的选取情况,该数在二进制下某位为0表示不选1表示选,求一个最大的区间使区间内选择每种物品的牛一样多. 数学转化,把不同状态间单变量的关系通过不等式移项转变为单状态的多变量关系. sum[i,j]表示前i头牛有多少选了物品j,那么问题要求即对任意j∈[1,K],sum[p,j]-sum[q,j]相等,使p-q最大.(多状态,单变量) 列出来,sum[p,1]-sum[q,1]=sum[p,2]-sum[q,2]=……,移项,sum[p,2]-su…

我%&&--&()&%????? 双模hashWA,unsigned long longAC,而且必须判断hash出来的数不能为0???? 我可能学了假的hash 这个题求个前缀和,然后目标是找到距离当前位置最远,且能使这两个数组差分后2-k位相同 hash把差分后数组的2到k位压起来即可,用map存这个hash值最早出现的位置 但是我还是不明白为啥hash值不能为0啊?? #include<iostream> #include<cstdio> #i…

\(\mathbf{P1360}\) 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\…

Gold Balanced Lineup Time Limit : 4000/2000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 3   Accepted Submission(s) : 3 Problem Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, F…

题目链接:点击打开链接 Gold Balanced Lineup Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 16978   Accepted: 4796 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 feat…

/******************************************************* 题目: Balanced Lineup(poj 3264) 链接: http://poj.org/problem?id=3264 题意: 给个数列,查询一段区间的最大值与最小值的差 算法: RMQ ********************************************************/ #include<cstdio> #include<cstrin…

Balanced Lineup Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 53703   Accepted: 25237 Case Time Limit: 2000MS Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer Joh…