题目大意：给出n个数字。首先将这n个数前后做差，得到另一个长度是n-1的序列。求出这个序列的最长重复子串，且这些子串不能重叠。

PS：这题论文上有解析。

解题思路：先二分答案，把题目变成判断性问题：判断是否存在两个长度为k的字串是否相同的，且不重叠。解决这个问题的关键还是利用height数组。把排序后的后缀数组分成若干组，其中每组的后缀之间的height值都不小于k。

容易看出，有希望成为最长公共前缀不小于k的两个后缀一定在同一组。然后对于每组后缀，只须判断每个后缀的sa值的最大值和最小值之差是否不小于k。如果有一组满足，则说明存在，否则不存在。整个做法的时间复杂度为O(nlogn)

论文上写的很详细了啊，不过需要注意细节，比如二分的边界，以及字符串的长度等等。。。

Description

Input

Output

Sample Input

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

Sample Output

5

Hint

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <ctime>
#include <map>
#include <set>
#define eps 1e-9
///#define M 1000100
///#define LL __int64
#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)
#define mod 1000000007
#define Read() freopen("autocomplete.in","r",stdin)
#define Write() freopen("autocomplete.out","w",stdout)
#define Cin() ios::sync_with_stdio(false)

using namespace std;



inline int read()
{
    char ch;
    bool flag = false;
    int a = 0;
    while(!((((ch = getchar()) >= '0') && (ch <= '9')) || (ch == '-')));
    if(ch != '-')
    {
        a *= 10;
        a += ch - '0';
    }
    else
    {
        flag = true;
    }
    while(((ch = getchar()) >= '0') && (ch <= '9'))
    {
        a *= 10;
        a += ch - '0';
    }
    if(flag)
    {
        a = -a;
    }
    return a;
}
void write(int a)
{
    if(a < 0)
    {
        putchar('-');
        a = -a;
    }
    if(a >= 10)
    {
        write(a / 10);
    }
    putchar(a % 10 + '0');
}

const int maxn = 20010;
int wa[maxn], wb[maxn], wv[maxn], ws1[maxn];
int sa[maxn];
int cmp(int *r, int a, int b, int l)
{
    return r[a]==r[b]&&r[a+l]==r[b+l];
}
void da(int *r,int *sa,int n,int m)
{
    int i, j, p, *x = wa,*y = wb;

    for(i = 0; i<m; i++) ws1[i]=0;
    for(i = 0; i<n; i++) ws1[x[i]=r[i]]++;
    for(i = 1; i<m; i++) ws1[i]+=ws1[i-1];
    for(i = n-1; i>=0; i--) sa[--ws1[x[i]]]=i;

    for(j = 1, p = 1; 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++) ws1[i]=0;
        for(i = 0; i<n; i++) ws1[wv[i]]++;
        for(i = 1; i<m; i++) ws1[i]+=ws1[i-1];
        for(i = n-1; i>=0; i--) sa[--ws1[wv[i]]]=y[i];
        for(swap(x,y),p=1,x[sa[0]]=0,i=1; i<n; i++)
            x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
    }
    return;
}

int rank[maxn],height[maxn];
void calheight(int *r,int *sa,int n)
{
    int i,j,k=0;
    for(i = 1; i<=n; i++) rank[sa[i]]=i;
    for(i = 0; i<n; height[rank[i++]]=k)
        for(k?k--:0, j=sa[rank[i]-1]; r[i+k] == r[j+k]; k++);
    return;
}

int sqe[maxn];


bool judge(int mid, int n)
{
    int Max, Min;
    for(int i = 2; i <= n; i++)
    {
        Max = sa[i-1];
        Min = sa[i-1];
        while(height[i] >= mid)
        {
            Max = max(sa[i], Max);
            Min = min(sa[i], Min);
            i++;
        }
        if(Max - Min > mid) return true;
    }
    return false;
}

int main()
{
    int n;
    Cin();
    while(cin >>n)
    {
        if(!n) break;
        for(int i = 0; i < n; i++) cin >>sqe[i];
        if(n <= 9)
        {
            puts("0");
            continue;
        }
        for(int i = 0; i < n-1; i++) sqe[i] = sqe[i+1]-sqe[i]+88;
        sqe[n-1] = 0;

        da(sqe, sa, n, 200);
        calheight(sqe, sa, n-1);

        int l = 4;
        int r = n / 2;
        int ans = -1;
        while(l <= r)
        {
            int mid = (l+r)>>1;
            if(judge(mid, n))
            {
                l = mid+1;
                ans = mid;
            }
            else r = mid-1;
        }
        cout<<ans+1<<endl;
    }
    return 0;
}

POJ 1743 Musical Theme(后缀数组)