Hotaru's problem

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 907    Accepted Submission(s): 322

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
 
解题:manacher暴力搞,fread真是TLE的救星啊
 
 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
int s[maxn],p[maxn];
int manacher(int n) {
int id = ,maxlen = ;
s[] = -;
for(int i = n; i >= ; --i) {
s[i + i + ] = s[i];
s[i + i + ] = ;
}
n = (n<<|);
for(int i = ; i < n; ++i) {
if(p[id] + id > i) p[i] = min(p[*id-i],p[id]+id-i);
else p[i] = ;
while(s[i-p[i]] == s[i+p[i]]) ++p[i];
if(id + p[id] < i + p[i]) id = i;
if(maxlen < p[i]) maxlen = p[i];
}
return maxlen - ;
}
char *ch, *ch1, buf[*+], buf1[*+];
void read(int &x) {
for (++ch; *ch <= ; ++ch);
for (x = ; '' <= *ch; ch++) x = x * + *ch - '';
}
int main() {
int kase,n,cs = ;
ch = buf - ;
ch1 = buf1 - ;
fread(buf, , * * , stdin);
read(kase);
while(kase--) {
read(n);
for(int i = ; i < n; ++i)
read(s[i]);
manacher(n);
int ret = ;
n = (n<<|);
for(int i = ; i < n; i += ) {
if(p[i] - > ret) {
int r = p[i] + ;
while(r > ret && p[i + r] < r) --r;
if(ret < r) ret = r;
}
}
printf("Case #%d: %d\n",cs++,(ret>>)*);
}
return ;
}
/*
1
11
1 1 2 1 1 2 1 1 2 1 1
*/
 

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