Hotaru's problem
**Time Limit: 4000/2000 MS (Java/Others)
**Memory Limit:**
** 65536/65536 K (Java/Others)**
**Total Submission(s): 1806 Accepted Submission(s): 648**
**Problem Description**
Hotaru Ichijou recently is addicated to math problems. Now she is 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 is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
>Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
>Input
There are 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 than 109 , descripting a sequence.
>Output
Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.
>We guarantee that the sum of all answers is less than 800000.
>Sample Input
1
10
2 3 4 4 3 2 2 3 4 4
Sample Output
Case #1: 9
**Source**
**2015 Multi-University Training Contest 7** 不會的可以看下面的
淺談manacher算法
道理是一樣的，只不過因爲是偶數的迴文串所以直接跳過避過奇偶的插入
#include <cstdio>
#include <cstring>
#include <set>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
const int N=100005;
typedef pair<int,int> pa;
int a[N];
int p[N];
int main()
{
int T;
scanf("%d",&T);
for(int cas=1;cas<=T;cas++)
{
int n;
memset(p,0,sizeof p);
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int id=0,maxn=0;
for(int i=1;i<=n;i++)
{
if(p[id]+id>i+1){
p[i]=min(p[id*2-i],p[id]+id-i-1);
}
else
p[i]=1;
while(i+p[i]<=n && i+1-p[i]>=1 && a[i+p[i]]==a[i+1-p[i]]){
p[i]++;
}
if(p[i]+i>p[id]+id)
id=i;
}
for(int i=1;i<=n;i++){
p[i]=p[i]-1;
}
int mx=0;
int tt;
for(int i=1;i<=n;i++){
if(mx<p[i]){
tt=p[i];
while(tt>mx && p[i+tt]<tt)
tt--;
mx=max(mx,tt);
}
}
printf("Case #%d: %d\n",cas,3*mx);
}
}
/*
100
21
9 8 0 0 8 9 9 8 0 8 9 9 8 0 8 9 9 8 0 8 9
*/