Hotaru ' s problem

Hotaru ' s problem

Time limit:4000/2000 MS (java/others) Memory limit:65536/65536 K (java/others)
Total submission (s): 3314 Accepted Submission (s): 1101

Problem Descriptionhotaru Ichijou recently is addicated to math problems. Now she's playing with N-sequence.
Let's define N-sequence, which is composed with three parts and satisfied with the following condition:
1. The first part of the same as the thrid part,
2. The first part and the second part is symmetrical.
For example, the sequence 2,3,4,4,3,2,2,3,4 are a n-sequence, which the first part 2,3,4 are the same as the Thrid Part 2, 3, 4, the first part 2,3,4 and the second part 4,3,2 is symmetrical.

Give you n positive intergers, your task was to find the largest continuous sub-sequence, which is n-sequence.

Inputthere is multiple test cases. The first line of input contains an integer T (t<=20), indicating the number of test cases.

For each test case:

The first line of input contains a positive integer N (1<=n<=100000), the length of a given sequence

The second line includes N non-negative integers, each interger is no larger than9 , descripting a sequence.

Outputeach case contains only one line. Should start with ' case #i: ', with I implying the case number, followed by a integer, the largest length of n-se Quence.

We guarantee the sum of all answers are less than 800000.

Sample Input1102 3 4 4 3 2 2 3 4 4

Sample outputcase #1:9

Authoruestc

Source2015 multi-university Training Contest 7

//On the basis of Manachar, enumerate the center of Palindrome String, then find the third part. #include <cstdio>#include<cstring>#include<iostream>using namespacestd;intRead () {registerintx=0;BOOLf=1; RegisterCharCh=GetChar ();  while(ch<'0'|| Ch>'9'){if(ch=='-') f=0; ch=GetChar ();}  while(ch>='0'&&ch<='9') {x=x*Ten+ch-'0'; ch=GetChar ();} returnf?x:-x;}Const intn=3e5+Ten;intN,ans,cas,l,t,s[n],s[n],p[n];voidManacher () {intId=0, mx=-1;  for(intI=1; i<l;i++){        if(id+mx>i) P[i]=min (p[id*2-i],id+mx-i);  while(i-p[i]>=0&&i+p[i]<=l&&s[i-p[i]]==s[i+p[i]]) p[i]++; if(Id+mx<i+p[i]) id=i,mx=P[i]; }}voidinit () {L=0; memset (P,0,sizeofp);  for(intI=0; i<n;i++) s[++l]=-1, s[++l]=S[i]; s[++l]=-1; }intMain () { for(T=read (), cas=1; ans=0, cas<=t;cas++) {n=read ();  for(intI=0; i<n;i++) s[i]=read ();        Init (); Manacher ();  for(intI=1; i<=n*2+1; i+=2){             for(intj=i+p[i]-1; j-i>ans;j-=2){                if(j-i+1<=P[j]) {ans=max (ans,j-i);  Break; }}} printf ("Case #%d:%d\n", cas,ans/2*3); }    return 0;}

Hotaru ' s problem