(KMP)Simpsons’ Hidden Talents -- hdu -- 2594
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
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.
The lengths of s1 and s2 will be at most 50000.
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<stack>
using namespace std; #define N 100050 int Next[N]; void FindNext(char S[])
{
int i=, j=-;
int Slen = strlen(S); Next[] = -; while(i<Slen)
{
if(j==- || S[i]==S[j])
Next[++i] = ++j;
else
j = Next[j];
}
}
int main()
{
char s1[N], s2[N]; while(scanf("%s%s", s1, s2)!=EOF)
{
int len1=strlen(s1), len2=strlen(s2);
int Min = min(len1, len2);
char S[N], s[N]; strcpy(s, s1);
strcat(s1, s2); FindNext(s1); int len = strlen(s1);
if(Next[len]== && len>)
printf("0\n");
else if(Next[len]>Min)
{
if(len1>len2)
printf("%s %d\n", s2, len2);
else
printf("%s %d\n", s, len1);
}
else
{
memset(S, , sizeof(S));
strncpy(S, s1, Next[len]);
printf("%s %d\n", S, Next[len]);
} }
return ;
}
(KMP)Simpsons’ Hidden Talents -- hdu -- 2594的更多相关文章
- Simpsons’ Hidden Talents HDU - 2594(拓展kmp)
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 ...
- Simpsons’ Hidden Talents - HDU 2594(求相同的前缀后缀)
题目大意:给你两个字符串,找出一个最大的子串,这个子串要是前面串的前缀并且是后面串的后缀........... 分析:next的简单运用吧,可以把两个串进行合并,中间加一个不能被匹配的字符,然后求 ...
- kuangbin专题十六 KMP&&扩展KMP HDU2594 Simpsons’ Hidden Talents
Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had. Marg ...
- 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 ...
- 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 ...
- hdu 2594 Simpsons’ Hidden Talents KMP
Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java ...
- hdu 2594 Simpsons’ Hidden Talents KMP应用
Simpsons’ Hidden Talents Problem Description Write a program that, when given strings s1 and s2, fin ...
- hdu 2594 Simpsons’ Hidden Talents(KMP入门)
Simpsons’ Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java ...
- hdoj 2594 Simpsons’ Hidden Talents 【KMP】【求串的最长公共前缀后缀】
Simpsons' Hidden Talents Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java ...
随机推荐
- Jquery detect page refresh
first thing there are 3 functions we will use: function setCookie(c_name, value, exdays) { ...
- Conductor Task Workers
由远程worker执行的conductor任务通过HTTP端点进行通信以轮询任务并更新执行状态. conductor提供了轮询任务的框架,管理执行线程并将执行状态更新回服务器.该框架提供了Java和P ...
- 实例学习SSIS(一)
网址: http://www.cnblogs.com/tenghoo/archive/2009/10/archive/2009/10/archive/2009/10/archive/2009/10/a ...
- alter 和 update的用法和区别
alter的增加和删除alter table xs_kc add xuefen number;alter table xs_kc drop column xuefen; 删除的时候必须使用column ...
- 每月IT摘录201805
摘录自互联网的前辈心得: 一.技术:0.精通一个淘汰的技术对你没有任何价值.学习的精力有限,更应该花在值得学的技术上.比如网络.操作系统.数据结构.算法1.工作要有定力,更多的应该是关心问题如何更有效 ...
- 第五章 二叉树(e1)先序遍历
- 转)ubuntu安装clang
主要参考:http://blog.csdn.net/firebird321/article/details/48528569 1.下载源码 去http://llvm.org/releases/down ...
- python的paramiko模块简单应用
用法1,SSHClient 分别可以使用密码和秘钥登陆,然后执行命令,并且获取执行结果 import paramiko #创建一个SSH对象 ssh = paramiko.SSHClient() #允 ...
- 平衡二叉树Balanced Binary Tree
[抄题]: Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced b ...
- 4-js 函数
总是有些奇奇怪怪的问题: <div> <p class="productStatus"> <span>成交量 <em>${goods ...