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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断了，但是发现，现在长大了却不知道怎么去判断亲和串了，于是他只好又再一次来请教聪明且乐于助人的你来解决这个问题。亲和串的定义是这样的：给定两个字符串s1和s2,如果能通过s1循环移位，使s2包含在s1中，那么

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 532    Accepted Submission(s): 220Problem DescriptionLadies and gentlemen, please

Problem DescriptionClairewd 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 e

DescriptionGiven 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, exponenti

Problem DescriptionIt 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.

DescriptionThe 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 w

DescriptionThe 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 sa

题目描述There is a string A. The length of A is less than 1,000,000. I rewrite it again and again. Then I got a new string: AAAAAA...... Now I cut it from two different position and get a new string B.

Problem DescriptionThe 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

Problem DescriptionGiven 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] =

题目链接：点击打开链接题意：给你一个串，让你在串后面添加尽可能少的字符使得这个串变成回文串。思路：这题可以kmp,manacher,后缀数组三种方法都可以做，kmp和manacher效率较高，时间复杂度是O(n)，后缀数组时间复杂度是O(nlogn).思路是求出元串的后缀和反串的前缀匹配的最大长度。用后缀数组的时候求出l=lcp(i,len+1)，判断l+i是不是等于len,如果等于那么就