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 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, say Clinton, and try to find the length of the longest prefix
in Clinton’s name that is a suffix in my name. That’s how close I am to being a politician like Clinton
Marge: Why on earth choose the longest prefix that is a suffix???
Homer: Well, our talents are deeply hidden within ourselves, Marge.
Marge: So how close are you?
Homer: 0!
Marge: I’m not surprised.
Homer: But you know, you must have some real math talent hidden deep in you.
Marge: How come?
Homer: Riemann and Marjorie gives 3!!!
Marge: Who the heck is Riemann?
Homer: Never mind.
Write a program that, when given strings s1 and s2, finds the longest prefix of s1 that is a suffix of s2.

Input

Input consists of two lines. The first line contains s1 and the second line contains s2. You may assume all letters are in lowercase.

Output

Output consists of a single line that contains the longest string that is a prefix of s1 and a suffix of s2, followed by the length of that prefix. If the longest such string is the empty string, then the output should be 0.
The lengths of s1 and s2 will be at most 50000.

Sample Input

clinton
homer
riemann
marjorie

Sample Output

0
rie 3

Source

HDU 2010-05 Programming Contest

Recommend

lcy

题解:

      ①题目要求求出a串的最长前缀满足是b串的后缀

      ②两种解法,

            ZJ的方法是直接将a作为模板串匹配,最后指针j的值就是匹配长度也等于最长长度。

            Me的方法是将两个串拼起来转化为求最长公共前后缀(意义就是next[])

            不过呢不能超界(比如b偷走了a的后面部分充当自己的),加一些判断就好了。

            其实就是拼起来后进行一次next的预处理然后不断j=f[j]就可以了。

#include<stdio.h>
#include<cstring>
#define go(i,a,b) for(int i=a;i<=b;i++)
const int N=3000003;int m,n1,n2,j,f[N];char a[N],b[N],P[N];
int main()
{
while(m=0,~scanf("%s%s",a,b))
{
n1=strlen(a);go(i,0,n1-1)P[m++]=a[i];
n2=strlen(b);go(i,0,n2-1)P[m++]=b[i];f[1]=f[0]=0;
go(i,1,m-1){j=f[i];while(j&&P[i]!=P[j])j=f[j];f[i+1]=P[i]==P[j]?j+1:0;} j=m;while(j>n1||j>n2)j=f[j];
if(j){go(i,0,j-1)printf("%c",P[i]);putchar(' ');};printf("%d\n",j);
}
return 0;
}//Paul_Guderian

可其它地方都没满,该满的地方都没满。————汪峰《满》

【HDU 2594 Simpsons' Hidden Talents】的更多相关文章

  1. HDU 2594 Simpsons’ Hidden Talents(辛普森一家的潜在天赋)

    HDU 2594 Simpsons’ Hidden Talents(辛普森一家的潜在天赋) Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 3 ...

  2. HDU 2594 Simpsons’ Hidden Talents(KMP的Next数组应用)

    Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java ...

  3. hdu 2594 Simpsons’ Hidden Talents KMP

    Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java ...

  4. hdu 2594 Simpsons’ Hidden Talents(KMP入门)

    Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java ...

  5. hdu 2594 Simpsons’ Hidden Talents KMP应用

    Simpsons’ Hidden Talents Problem Description Write a program that, when given strings s1 and s2, fin ...

  6. hdu 2594 Simpsons’ Hidden Talents 【KMP】

    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: ...

  7. hdu 2594 Simpsons’ Hidden Talents

    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 ...

  8. hdu 2594 Simpsons’ Hidden Talents(扩展kmp)

    Problem Description Homer: Marge, I just figured out a way to discover some of the talents we weren’ ...

  9. HDU 2594 Simpsons’ Hidden Talents (KMP)

    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. ...

随机推荐

  1. 提高篇(1):RMQ问题与ST表

    RMQ是英文Range Minimum/Maximum Query的缩写,是询问某个区间内的最值,这里讲一种解法:ST算法 ST算法通常用在要多次(10^6级别)询问区间最值的问题中,相比于线段树,它 ...

  2. 【ODT】cf896C - Willem, Chtholly and Seniorious

    仿佛没用过std::set Seniorious has n pieces of talisman. Willem puts them in a line, the i-th of which is ...

  3. leetcode笔记(二)94. Binary Tree Inorder Traversal

    题目描述 (原题目链接) Given a binary tree, return the inorder traversal of its nodes' values. For example:Giv ...

  4. percona-zabbix-templates插件安装监控MySQL

    前期准备:被监控机已经安装好php 1.在zabbix客户端安装mysql监控插件rpm包 rpm -ivh https://www.percona.com/downloads/percona-mon ...

  5. ofbiz研究

    近段时间,刚有有时间研究了下ofbiz ; 目前还是刚开始,后期会记录过程 有一起研究的没

  6. python3 练习题100例 (二十)

    #!/usr/bin/env python3# -*- coding: utf-8 -*-"""练习二十:判断一个年份是否是闰年公历闰年计算方法:1.普通年能被4整除且不 ...

  7. 笔记-爬虫-XPATH

    笔记-爬虫-XPATH 1.      xpath XPath是W3C的一个标准.它最主要的目的是为了在XML1.0或XML1.1文档节点树中定位节点所设计.目前有XPath1.0和XPath2.0两 ...

  8. 2,Python常用库之二:Pandas

    Pandas是用于数据操纵和分析,建立在Numpy之上的.Pandas为Python带来了两种新的数据结构:Pandas Series和Pandas DataFrame,借助这两种数据结构,我们能够轻 ...

  9. webstrom Certificate validation failed

    今天好烦.因为装了一个webstrome,我的svn不管用了. 为了防止webstrom,我的日期改了,日期改了,csdn登不上去了.告诉我时期不对. 草草哦哦. 这就是那个svn出问题之后的画面. ...

  10. 使用Windows SFC和DISM工具来解决服务器OS问题

    TechTarget中国原创] 随着使用时间的越来越多,Windows服务器安装的系统文件可能会被损坏或损毁.管理员一般可以通过系统自带的System File Checker (SFC) 或者更健壮 ...