kuangbin字符串专题传送门--http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70325#overview 算法模板: KMP: ; ; int a[MAXN],b[MAXM],Next[MAXM]; int n,m; void getNext(int b[],int Next[]) { ,k=-; Next[]=-; ) { ||b[j]==b[k]) //匹配 { j++,k++; Next[j]=k; } else k=Ne…

最近做完了kuangbin的一套关于kmp的题目(除了一道字典树的不会,因为还没学字典树所以先放放),做个总结.(kuangbin题目的链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70325#problem/A) 说实话,kmp这个东西真的理解了很久,因为网上模板的下标都不一样,而且讲解的部分和代码实现又有点差别.话说当初看到一个比较好的博客忘记保存了..= = 首先给个最简单的题目:http://oj.acm.zstu.ed…

首先是几份模版 KMP void kmp_pre(char x[],int m,int fail[]) { int i,j; j = fail[] = -; i = ; while (i < m) { && x[i] != x[j]) j = fail[j]; fail[++i] = ++j; } } int kmp_count(char x[],int m,char y[],int n) { ,j = ; ; while (i < n) { && y[i] !…

[KMP] 学习KMP,我们先要知道KMP是干什么的. KMP?KMPLAYER?看**? 正如AC自动机,KMP为什么要叫KMP是因为它是由三个人共同研究得到的- .- 啊跑题了. KMP就是给出一个母串S和串T,然后看T是不是S的子串. 易想到朴素算法,且时间复杂度是明显的O(NM). 那么为什么KMP的复杂度会这么高呢? 因为每次失配的时候,指针只是简单的把在S串的指针向后移动一位,T串回到开头,其中对于子串T已匹配过的信息没有充分利用. KMP是干嘛的? 利用一个next数组使得失配时T…

kmp: KMP的主要目的是求B是不是A的子串,以及若是,B在A中所有出现的位置 写的很详细的大佬的博客:http://www.matrix67.com/blog/archives/115 模板: /* pku3461(Oulipo), hdu1711(Number Sequence) 这个模板 字符串是从0开始的 Next数组是从1开始的 */ #include <iostream> #include <cstring> using namespace std; ; int ne…

Give you n ( n < 10000) necklaces ,the length of necklace will not large than 100,tell me How many kinds of necklaces total have.(if two necklaces can equal by rotating ,we say the two necklaces are some). For example 0110 express a necklace, you can…

题意:真难懂.. 给出26个英文字母的加密表,明文中的'a'会转为加密表中的第一个字母,'b'转为第二个,...依次类推. 然后第二行是一个字符串(str1),形式是密文+明文,其中密文一定完整,而明文可能不完整(也可能没有). 求出最短的完整的字符串(密文+明文). 思路: 1.用kmp来做: 首先肯定的是,给定的串中明文长度一定小于等于密文.也就是说明文长度小于等于总长的一半. 于是,取总长的后一半作为主串,然后把串反翻译一遍得到str2,然后用str2与str1的后一半进行匹配.首次把st…

来刷kuangbin字符串了,字符串处理在ACM中是很重要的,一般比赛都会都1——2道有关字符串处理的题目,而且不会很难的那种,大多数时候都是用到一些KMP的性质或者找规律. 点击标题可跳转至VJ比赛题目链接. A - Number Sequence 题意就是让你去找在串A找串B首次出现的位置,现在串不是字符串,而是数字串,所以用int数组存储即可,然后就是裸KMP. 代码: #include <string> #include <algorithm> #include <i…

After an uphill battle, General Li won a great victory. Now the head of state decide to reward him with honor and treasures for his great exploit. One of these treasures is a necklace made up of 26 different kinds of gemstones, and the length of the…

Beside other services, ACM helps companies to clearly state their “corporate identity”, which includes company logo but also other signs, like trademarks. One of such companies is Internet Building Masters (IBM), which has recently asked ACM for a he…

You are given a number of case-sensitive strings of alphabetic characters, find the largest string X, such that either X, or its inverse can be found as a substring of any of the given strings. InputThe first line of the input file contains a single…

It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example: s: "abab" The prefixes are: "a", "ab&qu…

The Genographic Project is a research partnership between IBM and The National Geographic Society that is analyzing DNA from hundreds of thousands of contributors to map how the Earth was populated. As an IBM researcher, you have been tasked with wri…

CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the en…

一块花布条,里面有些图案,另有一块直接可用的小饰条,里面也有一些图案.对于给定的花布条和小饰条,计算一下能从花布条中尽可能剪出几块小饰条来呢? Input输入中含有一些数据,分别是成对出现的花布条和小饰条,其布条都是用可见ASCII字符表示的,可见的ASCII字符有多少个,布条的花纹也有多少种花样.花纹条和小饰条不会超过1000个字符长.如果遇见#字符,则不再进行工作. Output输出能从花纹布中剪出的最多小饰条个数,如果一块都没有,那就老老实实输出0,每个结果之间应换行. Sample In…

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book: Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, pu…

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are…

链接: https://vjudge.net/problem/HDU-2594#author=0 题意: 求S1的前缀和S2的后缀的<最大>匹配 思路: kmp方法: 将s1, s2首尾连接, 根据Next数组求, 注意长度要比s1, 和s2的长度小. ExtenKmp: 考虑以s1为模板串, 匹配s2, 对于第一个满足exten[i]+i == len的点, 就为最长的长度. 代码: #include <iostream> #include <cstdio> #inc…

One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps: First step: girls will write a long string (only contains lower case) on the paper. For example, "abcde", but 'a' in…

给出一个只由小写英文字符a,b,c...y,z组成的字符串S,求S中最长回文串的长度. 回文就是正反读都是一样的字符串,如aba, abba等 Input输入有多组case,不超过120组,每组输入为一行小写英文字符a,b,c...y,z组成的字符串S 两组case之间由空行隔开(该空行不用处理) 字符串长度len <= 110000Output每一行一个整数x,对应一组case,表示该组case的字符串中所包含的最长回文长度. Sample Input aaaa abab Sample Out…

Give you a string with length N, you can generate N strings by left shifts. For example let consider the string “SKYLONG”, we can generate seven strings: String Rank SKYLONG 1 KYLONGS 2 YLONGSK 3 LONGSKY 4 ONGSKYL 5 NGSKYLO 6 GSKYLON 7 and lexicograp…

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according…

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marge: Yeah, what is it? Homer: Take me for example. I want to find out if I have a talent in politics, OK? Marge: OK. Homer: So I take some politician’s…

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…

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defin…

For each prefix of a given string S with N characters (each character has an ASCII code between 97 and 126, inclusive), we want to know whether the prefix is a periodic string. That is, for each i (2 <= i <= N) we want to know the largest K > 1 (…

题意: 将一段字符串 分割成两个串 如果分割后的串为回文串,则该串的价值为所有字符的权值之和(字符的权值可能为负数),否则为0. 问如何分割,使得两个串权值之和最大 思路: 首先了解扩展kmp 扩展KMP:给出模板串A和子串B,长度分别为lenA和lenB,要求在线性时间内,对于每个A[i](0<=i<=lenA-1),求出A[i..lenA-1]与B的最长公共前缀长度,记为ex[i](或者说,ex[i]为满足A[i..i+z-1]==B[0..z-1]的最大的z值). 根据上一篇文章我们可知…

最长回文串问题 manacher算法是用来求解最长回文串的问题.最长回文串的解法一般有暴力法.动态规划.中心扩展法和manacher算法. 暴力法的时间复杂度为\(O(n^3)\),一般都会超时: 动态规划的时间复杂度和空间复杂度均为\(O(n^2)\),通过矩阵压缩存储,空间复杂度常数可以降低为0.5,但时间复杂度较高,基本不能再优化: 中心扩展法在性能上优于动态规划,空间复杂度为\(O(1)\),但时间复杂度仍然是\(O(n^2)\): manacher性能最好,时间复杂度和空间复杂度均为\…

首先,在谈到Manacher算法之前,我们先来看一个小问题:给定一个字符串S,求该字符串的最长回文子串的长度.对于该问题的求解.网上解法颇多.时间复杂度也不尽同样,这里列述几种常见的解法. 解法一       通过枚举S的子串.然后推断该子串是否为回文.因为S的子串个数大约为latex=\dpi{100}&space;\fn_jvn&space;N^2"> \dpi{100}&space;\fn_jvn&space;N^2" title="…

一. KMP 1 找字符串x是否存在于y串中,或者存在了几次 HDU1711 Number Sequence HDU1686 Oulipo HDU2087 剪花布条 2.求多个字符串的最长公共子串 POJ3080 Blue Jeans HDU1238 Substrings 3.next数组的应用 1) 求循环节 HDU3746 Cyclic Nacklace HDU1358 Period POJ2406 Power Strings HDU3374 String Problem 2) 利用next…