Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 11602   Accepted: 5680 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…

The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such boring job, t…

链接:http://poj.org/problem? id=2752 题意:对于一个字符串S,可能存在前n个字符等于后n个字符,从小到大输出这些n值. 思路:这道题加深了对next数组的理解.next[i+1]相当于以第i位结尾的长度为next[i+1]的子串与前next[i+1]个字符组成的子串同样.理解之后就比較好做了,首先字符串的长度len肯定是一个答案.然后next[len]也是一个答案,原因如红字所写,如此迭代直到next下标值等于0停止.这是从大到小得到了答案.再反序输出就可以. 由…

一.题目 POJ2752 二.分析 比较明显的KMP运用. 但是这题不是只找一个,仔细看题后可以发现相当于是在找到最大的满足条件的后缀后,再在这个后缀里面找满足条件的后缀. 可以不断的运用KMP得出答案,但是会超时. 寻找优化,发现答案在处理过的next数组中,因为题目中的条件就是前缀和后缀交集,那么前缀的串肯定与后缀的串相同,那么我们只需要改变长度继续分析就可以了. 三.AC代码 1 #include <cstdio> 2 #include <iostream> 3 #inclu…

Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14106   Accepted: 7018 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…

Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names to their newly-born babies. They seek the name, and at the same time seek the fame. In order to escape from such b…

题目地址:http://poj.org/problem?id=3006 刷了好多水题,来找回状态...... Dirichlet's Theorem on Arithmetic Progressions Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 16803   Accepted: 8474 Description If a and d are relatively prime positive integers, t…

题目链接:http://poj.org/problem?id=2976 题意: 共有n场考试,每场考试你得的分数为a[i],总分为b[i]. 你可以任意去掉k场考试. 问你最大的 100.0 * ( ∑ a[i] / ∑ b[i] )的值.(四舍五入) 题解: 相当于从n场考试中选n-k场. 二分: 二分最大答案 ∑ a[i] / ∑ b[i] >= L 即:∑ a[i] - ∑(b[i]*L) >= 0 check函数: 求数组val[i] = a[i] - b[i]*L 将val排序. 取…

题目链接:http://poj.org/problem?id=2185 题目大意:求一个二维的字符串矩阵的最小覆盖子矩阵,即这个最小覆盖子矩阵在二维空间上不断翻倍后能覆盖原始矩阵. 题目分析:next函数的应用.需要枚举每一行每一列的字符串所对应的的 nxt[] 值,然后通过分析计算出最小的宽和最小的高. 具体分析 参考链接:https://blog.csdn.net/u013686535/article/details/52197467 一看这题,容易想出一种很直观的做法:求出每一行的最小重复…

<题目链接> 题目大意: 给定一段序列,进行q次询问,输出每次询问区间的最大值与最小值之差. 解题分析: RMQ模板题,用ST表求解,ST表用了倍增的原理. #include <cstdio> #include <cstring> #include <cmath> #include <algorithm> using namespace std; typedef long long ll; ; int n,q; ll maxsum[M][],mi…