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 (~…

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 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 = "…

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: 39291 Accepted: 16315 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 = "abcd…

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 = "…

题目描述 PDF 输入输出格式 输入格式: 输出格式: 输入输出样例 输入样例#1: 复制 abcd aaaa ababab . 输出样例#1: 复制 1 4 3 题解 Luogu的题解 这里是对目前最高赞题解结论的证明. 结论:设字符串长度为$n$,最长相同前后缀的长度为$next[i]$, 如$n$%$(n-next[n])=0$,则答案为$n/(n-next[n])$,否则为$1$. 证明: 我们求$next$数组的时候,相当于每次把当前串这样对齐了一下↓ 而$next$求到$n$时,上面…

题意:重复子串次数 思路:kmp #include<iostream> #include<stdio.h> #include<string.h> using namespace std; #define MaxSize 1000005 int next[MaxSize]; void GetNext(char t[]){//求next数组 int j,k,len; j=; k=-; next[]=-; len=strlen(t); while(j<len){ ||t…

题意: n<=1000000,cas较大 思路:这是一道论文题 后缀数组已弃疗,强行需要DC3构造,懒得(不会)写 ..]of longint; n,m,i,j,len,ans,st:longint; ch:ansistring; flag:boolean; function min(x,y:longint):longint; begin if x<y then exit(x); exit(y); end; function cmp(a,b,l:longint):boolean; begin…