poj 1961 Period

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KMP POJ 1961 Period

题目传送门 /* 题意:求一个串重复出现(>1)的位置 KMP:这简直和POJ_2406没啥区别 */ /************************************************ * Author :Running_Time * Created Time :2015-8-10 9:13:24 * File Name :POJ_1961.cpp ************************************************/ #include <cstd…

Period http://poj.org/problem?id=1961 Time Limit: 3000MS Memory Limit: 30000K Description 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 pref…

POJ 1961 Period( KMP )*

Period Time Limit: 3000MSMemory Limit: 30000K Total Submissions: 12089Accepted: 5656 Description 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 prefi…

POJ 1961 Period（KMP）

http://poj.org/problem?id=1961 题意 :给你一个字符串,让你输出到第几个字符时,循环结的个数. 思路 :这个题和2409差不多,稍微修改一下,加一个循环就行了,用的也是KMP. #include <string.h> #include <stdio.h> #include <iostream> using namespace std ; ; char ch[maxn] ; int next[maxn] ; int main() { int…

POJ 1961 Period KMP算法next数组的应用

题目: http://poj.org/problem?id=1961 很好的题,但是不容易理解. 因为当kmp失配时,i = next[i],所以错位部分就是i - next[i],当s[0]...s[i]是一个周期串时,i-next[i]显然就是一个循环节,这时(i+1) % (i-next[i]) == 0,(因为下标是从0开始的,所以i+1表示前i个字符,也就是字符个数),判断上式是否为0就可以判断是否是周期串,如果是周期串时,(i+1) / (i-next[i]) 显然就是重复的次数.…

(简单) POJ 1961 Period，扩展KMP。

Description 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 larges…

poj 1961 Period 【KMP-next前缀数组的应用】

题目地址:http://poj.org/problem?id=1961 Sample Input 3 aaa 12 aabaabaabaab 0 Sample Output Test case #1 2 2 3 3 Test case #2 2 2 6 2 9 3 12 4 题目分析:给你一个字符串,最大长度1百万.输出是:以第1组样例解释,在aaa的字符串中,长度为2时,存在2个循环节a.当长度为3时,存在3个循环节a.以第二组样例解释,当长度为2时,存在2个循环节a.当长度为6时,存在2个循…