2015 Multi-University Training Contest 7 hdu 5371 Hotaru's problem

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

2015 Multi-University Training Contest 7 1003

 

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
 */