Substrings Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8183 Accepted Submission(s): 3752 Problem Description You are given a number of case-sensitive strings of alphabetic characters, fin…
Codeforces Round #258 (Div. 2) Count Good Substrings D. Count Good Substrings time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output We call a string good, if after merging all the consecutive eq…
Substrings Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7205 Accepted Submission(s): 3255 Problem Description You are given a number of case-sensitive strings of alphabetic characters, find the…
试题 算法提高 Substrings 问题描述 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. 输入格式 The first line of the input…
题面 该场 Div. 3 最"难"的一道题:Funny Substrings O I D \tt OID OID 队长喜欢玩字符串,因为 " O n e I n D a r k " \tt "OneInDark" "OneInDark" 是一个望而生畏的字符串. O I D \tt OID OID 会进行 N \tt N N 次操作,每次操作形如以下两种之一: x := S : x 是变量名称,S 是一个具体的字符串,:= 符…
求不相同子串个数 该问题等价于求所有后缀间不相同前缀的个数..也就是对于每个后缀suffix(sa[i]),将贡献出n-sa[i]+1个,但同时,要减去那些重复的,即为height[i],故答案为n-sa[i]+1-height[i]的累计. ; var x,y,rank,sa,h,c:..maxn] of longint; s:ansistring; t,q,n:longint; function max(x,y:longint):longint; begin if x>y then e…
http://poj.org/problem?id=3415 Common Substrings Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 5805 Accepted: 1911 Description A substring of a string T is defined as: T(i, k)=TiTi+1...Ti+k-1, 1≤i≤i+k-1≤|T|. Given two strings A, B an…