Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 27539   Accepted: 9290 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the…

Musical ThemeTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://poj.org/problem?id=1743 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It…

Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 27539   Accepted: 9290 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the…

Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16162   Accepted: 5577 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the…

题目链接:https://vjudge.net/problem/POJ-1743 Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 32402   Accepted: 10808 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the ra…

Musical Theme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 28435 Accepted: 9604 Description A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the pian…

Musical Theme 题意 给出n个1-88组成的音符,让找出一个最长的连续子序列,满足以下条件: 长度大于5 不重叠的出现两次(这里的出现可以经过变调,即这个序列的每个数字全都加上一个整数x) 思路 我们处理一下这个所谓的"变调":令\(a[i]=a[i+1]-a[i]\),这样就转化成了找最长的出现至少两次的不重叠子串.(这时长度变为n-1) 两种做法:1.二分+ hash 2. 二分+后缀数组 使用hash的时候,对于当前二分的长度x. 我们从x开始遍历到n,如果[i-x+…

题目链接: Poj  1743 Musical Theme 题目描述: 给出一串数字(数字区间在[1,88]),要在这串数字中找出一个主题,满足: 1:主题长度大于等于5. 2:主题在文本串中重复出现(或者经过调转出现,调转是主题同时加上或者减去同一个整数) 3:重复主题不能重叠 解题思路: 求调转重复出现的子串,那么主题之间的差值一定是不变的.可以求文本串s中相邻两个数的差值,重新组成一个新的文本串S,然后找S后缀串中最长公共不重叠前缀.rank相邻的后缀串,公共前缀一定最长,但是有可能重叠.…

[题目分析] 其实找最长的不重叠字串是很容易的,后缀数组+二分可以在nlogn的时间内解决. 但是转调是个棘手的事情. 其实只需要o(*￣▽￣*)ブ差分就可以了. 背板题. [代码] #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <map> #include <set> #include <queue> #…

利用后缀数组,先对读入整数处理str[i]=str[i+1]-str[i]+90这样可以避免负数,计算Height数组,二分答案,如果某处H<lim则将H数组分开,最终分成若干块,判断每块中是否存在SA[i]-SA[j]>lim(注意不是>=因为查分后如果相等那么两个公共串连接部分的元素是公用的,不符合题意),注意特判n=1的情况,因为查分后的结果是个空串,会导致RE. #include <iostream> #include <cstdio> #include…