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

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…

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…

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…

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…

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…

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 (…

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…

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…

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…

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…

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] !…

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 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…