Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 27889		Accepted: 9413

A musical melody is represented as a sequence of N (1<=N<=20000)notes that are integers in the range 1..88, each representing a key on the piano. It is unfortunate but true that this representation of melodies ignores the notion of musical timing; but, this programming task is about notes and not timings.
Many composers structure their music around a repeating &qout;theme&qout;, which, being a subsequence of an entire melody, is a sequence of integers in our representation. A subsequence of a melody is a theme if it:

• is at least five notes long
• appears (potentially transposed -- see below) again somewhere else in the piece of music
• is disjoint from (i.e., non-overlapping with) at least one of its other appearance(s)

Transposed means that a constant positive or negative value is added to every note value in the theme subsequence.
Given a melody, compute the length (number of notes) of the longest theme.
One second time limit for this problem‘s solutions!

The input contains several test cases. The first line of each test case contains the integer N. The following n integers represent the sequence of notes.
The last test case is followed by one zero.

For each test case, the output file should contain a single line with a single integer that represents the length of the longest theme. If there are no themes, output 0.

30
25 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 18
82 78 74 70 66 67 64 60 65 80
0

5

Use scanf instead of cin to reduce the read time.

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=20005;
int wa[N<<1],wb[N<<1],to[N],sa[N],h[N],sr[N];
bool sf(int n,int a)
{
    int mx=0,mi=n;
    for(int i=1;i<n;++i)
    {
        if(h[i]>=a)
        {
            mx=max(mx,max(sa[i],sa[i-1]));
            mi=min(mi,min(sa[i],sa[i-1]));
        }
        else mx=0,mi=n;
        if(mx-mi>=a) return 1;    
    }
    return 0;
}
int main()
{
    int n;
    while(scanf("%d",&n)&&n)
    {
        int *x=wa,*y=wb,m=200;
        for(int i=0;i<n;++i) scanf("%d",&y[i]);
        --n;
        for(int i=0;i<m;++i) to[i]=0;
        for(int i=0;i<n;++i) ++to[x[i]=y[i+1]-y[i]+88];
        for(int i=1;i<m;++i) to[i]+=to[i-1];
        for(int i=n-1;i>=0;--i) sa[--to[x[i]]]=i;
        for(int i=0;i<n;++i) sr[i]=x[i];
        for(int j=1,p=0;p<n;m=p,j<<=1)
        {
            p=0;
            for(int i=n-j;i<n;++i) y[p++]=i,x[i+j]=-1;
            for(int i=0;p<n;++i) if(sa[i]>=j) y[p++]=sa[i]-j;
            for(int i=0;i<m;++i) to[i]=0;
            for(int i=0;i<n;++i) ++to[x[y[i]]];
            for(int i=1;i<m;++i) to[i]+=to[i-1];
            for(int i=n-1;i>=0;--i) sa[--to[x[y[i]]]]=y[i];
            int *t=x;x=y,y=t;
            x[sa[0]]=0;
            for(int i=p=1;i<n;++i)
                y[sa[i]]!=y[sa[i-1]]||y[sa[i]+j]!=y[sa[i-1]+j]?x[sa[i]]=p++:x[sa[i]]=p-1;
        }
        for(int i=0,j,k=0;i<n;h[x[i++]]=k)
            if(!x[i]) k=0;
            else for(k?--k:0,j=sa[x[i]-1];i+k<n&&j+k<n&&sr[i+k]==sr[j+k];++k);
        int l=0,r=n/2+1;
        while(l<r-1)
        {
            int mid=(l+r)>>1;
            if(sf(n,mid)) l=mid; 
            else r=mid;
        }
        printf("%d\n",l<4?0:l+1);
    }
    return 0;
}