Power Strings(kmp妙解)】的更多相关文章

Power Strings Time Limit : 6000/3000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 29   Accepted Submission(s) : 14 Problem Description Given two strings a and b we define a*b to be their concatenation. For example,…
Power Strings Time Limit: 3000MSMemory Limit: 65536K Total Submissions: 29663Accepted: 12387 Description 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…
点击打开链接 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27368   Accepted: 11454 Description 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 =…
Description Problem D: Power Strings Given two strings a and b we define a*b to be their concatenation. For example, ifa = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiatio…
题目传送门 Power Strings 格式难调,题面就不放了. 一句话题意,求给定的若干字符串的最短循环节循环次数. 输入样例#1: abcd aaaa ababab . 输出样例#1: 1 4 3 就这样. 分析: 一道思路神奇的题目,需要深入理解$KMP$的$next$数组. 如果自己写几个字符串推一下就可以发现,一个由循环节构成的字符串,从第二个循环节开始$next$值是依次递增的,因为$next$数组的本质是表示$0\~i-1$的最长公共前缀后缀长度.也就不难想到,只要判断一下$nex…
题目链接:https://vjudge.net/problem/POJ-2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 52631   Accepted: 21921 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc"…
Power Strings   Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 47748   Accepted: 19902 Description 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 = &quo…
http://poj.org/problem?id=2406 Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 27003   Accepted: 11311 Description Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = &q…
Power Strings Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 28102   Accepted: 11755 Description 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 = "…
给你一个串s,如果能找到一个子串a,连接n次变成它,就把这个串称为power string,即a^n=s,求最大的n. 用KMP来想,如果存在的话,那么我每次f[i]的时候退的步数应该是一样多的  譬如ababab  我每次退的一定是2步,检验一下这个串的失配指针是不是这个性质,如果是的话,那么n=strlen(s)/退的步数,否则就是直接1好了. #include<iostream> #include<cstring> #include<cstdio> #includ…