Description

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! 

Input

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. 

Output

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.

Sample Input

3025 27 30 34 39 45 52 60 69 79 69 60 52 45 39 34 30 26 22 1882 78 74 70 66 67 64 60 65 800

Sample Output

5

Hint

Use scanf instead of cin to reduce the read time.

Source

題意：有N(1 <= N <=20000)個音符的序列來表示一首樂曲，每個音符都是1..88範圍內的整數，現在要找一個重複的主題。“主題”是整個音符序列的一個子串，它需要滿足如下條件：

    1.長度至少為5個音符。

    2.在樂曲中重複出現。(可能經過轉調，“轉調”的意思是主題序列中每個音符都被加上或減去了同一個整數值)

    3.重複出現的同一主題不能有公共部分。

思路：這是論文裡一道題，那麼我們通過進行二分來求論文裡的k值，求出height[i]大於k的時候，然後求出這個區間裡面起始位置最大的差值，要求差值也要大於k，這樣才保證沒有重疊，這是因為height裡的都是已經按照字典序排好的了

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;#define LS 2*i#define RS 2*i+1#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 20005#define MOD 1000000007#define INF 0x3f3f3f3f#define EXP 1e-8int wa[N],wb[N],wsf[N],wv[N],sa[N];int rank[N],height[N],s[N],a[N],n;char str1[N],str2[N];//sa:字典序中排第i位的起始位置在str中第sa[i]//rank:就是str第i個位置的字尾是在字典序排第幾//height：字典序排i和i-1的字尾的最長公共字首int cmp(int *r,int a,int b,int k){    return r[a]==r[b]&&r[a+k]==r[b+k];}void getsa(int *r,int *sa,int n,int m)//n要包含末尾新增的0{    int i,j,p,*x=wa,*y=wb,*t;    for(i=0; i<m; i++)  wsf[i]=0;    for(i=0; i<n; i++)  wsf[x[i]=r[i]]++;    for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];    for(i=n-1; i>=0; i--)  sa[--wsf[x[i]]]=i;    p=1;    j=1;    for(; p<n; j*=2,m=p)    {        for(p=0,i=n-j; i<n; i++)  y[p++]=i;        for(i=0; i<n; i++)  if(sa[i]>=j)  y[p++]=sa[i]-j;        for(i=0; i<n; i++)  wv[i]=x[y[i]];        for(i=0; i<m; i++)  wsf[i]=0;        for(i=0; i<n; i++)  wsf[wv[i]]++;        for(i=1; i<m; i++)  wsf[i]+=wsf[i-1];        for(i=n-1; i>=0; i--)  sa[--wsf[wv[i]]]=y[i];        t=x;        x=y;        y=t;        x[sa[0]]=0;        for(p=1,i=1; i<n; i++)            x[sa[i]]=cmp(y,sa[i-1],sa[i],j)? p-1:p++;    }}void getheight(int *r,int n)//n不儲存最後的0{    int i,j,k=0;    for(i=1; i<=n; i++)  rank[sa[i]]=i;    for(i=0; i<n; i++)    {        if(k)            k--;        else            k=0;        j=sa[rank[i]-1];        while(r[i+k]==r[j+k])            k++;        height[rank[i]]=k;    }}int ans;int fun(int k){    int i,maxn,minn;    maxn = minn = sa[1];    UP(i,2,n)    {        if(height[i]>=k && i<n)        {            minn = min(minn,sa[i]);            maxn = max(maxn,sa[i]);            continue;        }        if(maxn-minn>=k) return 1;        maxn = minn = sa[i];    }    return 0;}int main(){    int i,j,k;    W((~scanf("%d",&n),n))    {        UP(i,0,n-1)        {            scanf("%d",&s[i]);        }        UP(i,0,n-2)        {            s[i] = s[i+1]-s[i]+100;        }        s[--n] = 0;        getsa(s,sa,n+1,200);        getheight(s,n);        int l = 4,r = n;        W(l<=r)        {            int mid = (l+r)/2;            if(fun(mid))            {                ans = mid;                l=mid+1;            }            else r = mid-1;        }        ans++;        printf("%d\n",ans<5?0:ans);    }    return 0;}