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…

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…

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…

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): 15015    Accepted Submission(s): 5151 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): 4543    Accepted Submission(s): 1648 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

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 name…

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…

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 tak…

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…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2594 思路: 其实就是求相同的最长前缀与最长后缀 KMP算法的简单应用: 假设输入的两个字符串分别是s1,s2.s1后面再上一个任意的大写字母,然后再将串s2连在s1的后面,再在后面加上一个任意的大写字母(注意应该与前面所加的字母不同),那么对当前的s1串求其失败函数f,那么f[n-1]即为最大的匹配数,是不是很简单啊!!!至于中间和后面为什么加大些字母,留给读者自己思考咯 代码: #include…

最近在学习字符串的知识,在字符串上我跟大一的时候是没什么区别的,所以恶补了很多基础的算法,今天补了一下字符串哈希,看的是大一新生的课件学的,以前觉得字符串哈希无非就是跟普通的哈希没什么区别,倒也没觉得有什么特别大的用处,敲一敲才发现其实讲究还是比较多的.哈希冲突是常有的事,换一下mod,换一下进制数才有可能过,另外一种说法是用两个互质的量做hash,如果两个都相等的话那冲突就会少很多,这个倒没有做过多大的尝试,侥幸地过了一下这道题 #pragma warning(disable:4996) #i…

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&…

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): 4597    Accepted Submission(s): 1671 Problem Description Homer: Marge, I just figured out a way to discover some of the…

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…

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): 2875    Accepted Submission(s): 1095 Problem Description Homer: Marge, I just figured out a way to discover some of the t…

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: 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…

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…

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…

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…

Sample Input clinton homer riemann marjorie Sample Output 0 rie 3 看输出才题意...拓展kmp特征很明显嘛....注意开始就匹配到尾的情况 #include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include &…

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…

题意: 有两个字符串,找一个最长子串是的该串既是第一个字的前缀,又是第二个串的后缀. 分析: 把两个串并起来然后在中间加一个无关字符,求next数组即可. #include <cstdio> #include <cstring> + ; ], s2[maxn]; ], l; void get_next() { , j = ; next[] = -; while(j < l) { || s1[k] == s1[j]) { k++; j++; next[j] = k; } els…

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…

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 8010 Accepted Submission(s): 2837 Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren't aware we had…

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…