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". If we think of concatenation as multiplication, exponentiation by a non-negative inte…
Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 30069 Accepted: 12553 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 = "…
Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 37685 Accepted: 15590 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 = "…
题目链接:http://poj.org/problem?id=2406 题目大意:如果n%(n-next[n])==0,则存在重复连续子串,长度为n-next[n]. 例如: a b a b a b next:-1 0 0 1 2 3 4 next[n]==4,代表着,前缀abab与后缀abab相等的最长长度,这说明,ab这两个字母为一个循环节,长度=n-next[n]; #include <iostream> #inc…
poj2406 Power Strings(kmp) 给出一个字符串,问这个字符串是一个字符串重复几次.要求最大化重复次数. 若当前字符串为S,用kmp匹配'\0'+S和S即可. #include <cstdio> #include <cstring> using namespace std; const int maxn=2e6+5; char s1[maxn], s2[maxn]; int n1, n2, nxt[maxn], ans; int main(){ while (~…
Power Strings 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". If we think of concatenation as multiplication, exponentiation by a non…
Power Strings Time Limit: 3000MS Memory Limit: 65536K Total Submissions: 45008 Accepted: 18794 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 = "…