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 = "…
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…
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…
kmp,根据next数组的性质如果有答案的话就是n/(n-(ne[n]+1)),否则是1 搬来打算用SA后来发现必须用DC3就没写 #include<iostream> #include<cstdio> #include<cstring> using namespace std; const int N=1000005; int n,ne[N]; char s[N]; int main() { while(scanf("%s",s)&&…
