龙崎的专栏

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

A. Police Recruits#include#include#include#includeusing namespace std;int n;int main(){ //freopen("in.txt","r",stdin); cin>>n; int ans=0,x,p=0; for(int i=0;i<n;i++) {

题意：求最长的公共子串，使她后者是他的翻转串在每个

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

调了一上午终于调出来了，这都是什么错误，为什么

L-Gap SubstringsOutput: Standard Output If a string is in the form UVU, where Uis not empty, andV has exactly L characters, we say UVUis anL-Gap string. For example, abcbabc is a 1-Gapstri

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

Language:DefaultFacer’s stringTime Limit: 3000MS Memory Limit: 65536KTotal Submissions: 1766 Accepted: 529DescriptionMinifacer was very happy these days

E. Misha and LCP on Treetime limit per test8 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard outputMisha has a tree with characters written on

E. Mike and Friendstime limit per test3 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputWhat-The-Fatherland is a strange country! All pho

Best Cow Line, GoldTime Limit: 5000MS Memory Limit: 65536KTotal Submissions: 4488 Accepted: 1664DescriptionFJ is about to take his N (1 ≤ N ≤ 30,000) cows to th

思路：把两个串拼接起来然后求

题意：给定一个串，只能在后面添加字符，使其成为一个回文串。思路：反转后求rev的next函数，然后与原串进行匹配，找到已经是回文的位置，输出#include#include#include#include#includeusing namespace std;const int maxn=100100;int next[maxn];string a,rev;void ge

跟Life Forms很像

把两个串拼接起来，中间用神奇地‘$'字符隔开。然后用倍增算法求sa数组、rank数组、height数组，然后利用height数组统计第一个串称A串的所有后缀与第二个串B串的所有后缀所共有的长度大于k的子串。   上面的逻辑过程容易理解，用aa和aa就能推出为什么答案是5，子串a有2个，aa有1个，2*2 + 1*1 = 5.问题的关键是要找出各个子串总匹配数（子串相等称为匹配）。height数

详见训练指南，需要注意的细节太多了#include#include#include#include#includeusing namespace std;const int MAXN=100010;int n;struct Suffix{ int s[MAXN],sa[MAXN]; int height[MAXN],rank[MAXN]; int c[

调了一晚上死活没调出来，

也可以用KMP做具体参见

题意：找出一个最长的并且重复次数至少为k次的子串。

思路：二分答案，然后遍历height数组，看能不能找到

Online JudgeOnline ExerciseOnline TeachingOnline ContestsExercise AuthorF.A.QHand In HandOnline AcmersForum | DiscussStatistical ChartsProblem ArchiveRealtime Judge Statu

在后缀数组神文中也这题的题解。比较容易理解的部分就是枚举长度为L，然后看长度为L的字符串最多连续出现几次。既然长度为L的串重复出现，那么str[0],str[l],str[2*l]……中肯定有两个连续的出现在字符串中。那么就枚举连续的两个，然后从这两个字符前后匹配，看最多能匹配多远。即以str[i*l],str[i*l+l]前后匹配，这里是通过查询suffix(i*l),suffi

突然发现前面写的后缀数组不太对，拍出来的sa

A - 1001DescriptionAn Arc of Dream is a curve defined by following function:AoD(n)=∑n−1i=0ai∗biAoD(n)=\sum_{i=0}^{n-1}a_i*b_iwhere a0=A0a_0 = A_0 ai=ai−1∗AX+AYa_i = a_{i-1}*AX+AY b0=B0b_0 = B_0 bi=