题目链接:http://acm.acmcoder.com/showproblem.php?pid=2594 题意:求最长的串 同一时候是s1的前缀又是s2的后缀.输出子串和长度. 思路:kmp 代码: #include <vector> #include <string> #include <algorithm> #include <iostream> #include <stdio.h> #include <string.h> us…

Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4597    Accepted Submission(s): 1671 Problem Description Homer: Marge, I just figured out a way to discover some of the…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 6888    Accepted Submission(s): 2461 Problem Description Homer: Marge, I just figured out a way to discover some of the…

Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2798    Accepted Submission(s): 1055 Problem Description Homer: Marge, I just figured out a way to discover some of the…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 这题直接用KMP算法就能够做出来,只是我还尝试了用扩展的kmp,这题用扩展的KMP效率没那么高. KMP算法: #include<stdio.h> #include<iostream> #include<string.h> using namespace std; int next[50001]; char p[50000],s[50000]; void getnex…

Problem Description 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 so…

HDU 2594 Simpsons’ Hidden Talents(辛普森一家的潜在天赋) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) [Description] [题目描述] Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marge:…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4683    Accepted Submission(s): 1702 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

Simpsons’ Hidden Talents Problem Description Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2. Sample Input clinton homer riemann marjorie   Sample Output 0 rie 3   思路:要求的是s1的最长前缀是s2的后缀:那么kmp中的…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4543    Accepted Submission(s): 1648 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10513    Accepted Submission(s): 3671 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 题目大意:给两串字符串s1,s2,,找到最长子串满足既是s1的前缀又是s2的后缀,输出子串,及相应长度. 解题思路:这题是不是跟POJ 2752很像,没错,我们只要将s1.s2合并,不断递归直到找到长度小于等于s1.s2的公共前后缀即可. 代码 #include<iostream> #include<cstdio> #include<string> #include&…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 思路:将两个串连起来求一遍Next数组就行长度为两者之和,遍历时注意长度应该小于两个串中的最小值 #include<cstdio> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<vector> #include&…

题意:两个字符串s.t,求s和t的最长的相同的前缀和后缀 思路:先求s的next数组,再求t的next数组(即代码中ex数组,此时不是自己与自己匹配,而是与s匹配),最后看ex[len2]即可(len2为串t的长度). #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 50005 int _next[MaxSize],ex[MaxSiz…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 15015    Accepted Submission(s): 5151 Problem Description Homer: Marge, I just figured out a way to discover some of the…

求next数组,(一般有两种,求循环节用的见代码)求出循环节的长度. #include <cstdio> #define N 100005 int n,next[N]; char s[N]; int main(){ scanf("%d",&n); while(n--){ scanf("%s",s); int i=0,k=-1; next[0]=k; while(s[i]){ if(k==-1||s[i]==s[k]) { i++; k++; ne…

http://acm.hdu.edu.cn/showproblem.php?pid=2594 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 9919    Accepted Submission(s): 3418 Problem Description Homer: Marge, I just figured out a way to…

http://acm.hdu.edu.cn/showproblem.php?pid=2594 Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4756    Accepted Submission(s): 1732 Problem Description Homer: Marge, I j…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1746 Accepted Submission(s): 637 Problem Description Homer: Marge, I just figured out a way to discover some of the talents…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5763 题目大意: T组数据,给两个字符串s1,s2(len<=100000),s2可以被解读成2种意思,问s1可以解读成几种意思(mod 1000000007). 题目思路: [动态规划][KMP] 题目有点绕,看看样例就懂了.其实不用KMP直接用substr就能做. 首先不解读成另一个意思的话,f[i]=f[i-1],接着如果当前位置能够与s2匹配,那么f[i]+=f[i-strlen(s2)]…

题目链接:https://vjudge.net/problem/HDU-2594 Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 10647    Accepted Submission(s): 3722 Problem Description Homer: Marge, I just f…

地址:http://acm.hdu.edu.cn/showproblem.php?pid=2594 题目: Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8709    Accepted Submission(s): 3051 Problem Description Homer: Mar…

Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2875    Accepted Submission(s): 1095 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

Simpsons’ Hidden Talents Time Limit: 1000ms Memory Limit: 32768KB This problem will be judged on HDU. Original ID: 259464-bit integer IO format: %I64d      Java class name: Main   Homer: Marge, I just figured out a way to discover some of the talents…

Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Problem Description 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? Hom…

题目链接: http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1794 题目大意: 两个无刻度的钟面,每个上面有N根针(N<=200000),每个针都是相同的,分别指向Ai,Bi(360°被分成360000小份),问能否将其中一个旋转和另一个重合. 题目思路: [KMP][最小表示法] 循环同构问题.可以写KMP,我懒得写KMP了就写了循环同构的最小表示法. 首先将Ai排序,然后求差(记得取模360000,WA了一次),接下来复制一遍开始匹配. A…

HDOJ 2203 亲和串 [KMP] Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 16573 Accepted Submission(s): 7273 Problem Description 人随着岁数的增长是越大越聪明还是越大越笨,这是一个值得全世界科学家思考的问题,同样的问题Eddy也一直在思考,因为他在很小的时候就知道亲和串如何判…

[KMP]Censoring 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 rat…

[KMP]OKR-Periods of Words 题目描述 串是有限个小写字符的序列,特别的,一个空序列也可以是一个串.一个串P是串A的前缀,当且仅当存在串B,使得A=PB.如果P≠A并且P不是一个空串,那么我们说P是A的一个proper前缀.定义Q是A的周期,当且仅当Q是A的一个proper前缀并且A是QQ的前缀(不一定要是proper前缀).比如串abab和ababab都是串abababa的周期.串A的最大周期就是它最长的一个周期或者是一个空串(当A没有周期的时候),比如说,ababab的…

问题 L: [KMP]Radio Transmission 题目描述 给你一个字符串,它是由某个字符串不断自我连接形成的.但是这个字符串是不确定的,现在只想知道它的最短长度是多少. 输入 第一行给出字符串的长度L,第二行给出一个字符串,全由小写字母组成. 输出 输出最短的长度. 样例输入 8 cabcabca 样例输出 3 提示 我们可以利用abc不断自我连接得到abcabcabc,读入的cabcabca是它的子串. 对于全部数据,1≤L≤1e6 [题意]: 题意花里胡哨,其实就是问,最小循环串…