Farmer John has noticed that the quality of milk given by his cows varies from day to day. On further investigation, he discovered that although he can't predict the quality of milk from one day to the next, there are some regular patterns in the daily milk quality.

To perform a rigorous study, he has invented a complex classification scheme by which each milk sample is recorded as an integer between 0 and 1,000,000 inclusive, and has recorded data from a single cow over N

 (1 ≤ N ≤ 20,000) days. He wishes to find the longest pattern of samples which repeats identically at least K (2 ≤ K ≤ N) times. This may include overlapping patterns -- 1 2 3 2 3 2 3 1 repeats 2 3 2 3 twice, for example.

Help Farmer John by finding the longest repeating subsequence in the sequence of samples. It is guaranteed that at least one subsequence is repeated at least K

times.

Input

Line 1: Two space-separated integers:  N and  K 
Lines 2..  N+1:  N integers, one per line, the quality of the milk on day  i appears on the  ith line.

Output

Line 1: One integer, the length of the longest pattern which occurs at least  K times

Sample Input

8 2
1
2
3
2
3
2
3
1

Sample Output

4
題解：
題目意思是讓求出現次數超過k次的子串的最長長度；
思路：字尾陣列，然後二分答案，求次數大於k的最長的子串;
參考程式碼：

  1 //#include<bits/stdc++.h>
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstring>
  5 #include<string>
  6 #include<cstdlib>
  7 #include<cmath>
  8 #include<algorithm>
  9 using namespace std;
 10 #define clr(a,val) memset(a,val,sizeof(a))
 11 #define lowbit(x) x&-x
 12 typedef long long ll;
 13 const int INF=0x3f3f3f3f;
 14 inline int read()
 15 {
 16     int x=0,f=1;char ch=getchar();
 17     while(ch<'0'||ch>'9') {if(ch=='-') f=-1; ch=getchar();}
 18     while(ch>='0'&&ch<='9') { x=x*10+ch-'0'; ch=getchar();}
 19     return x*f;
 20 }
 21 const int maxn=1e6+10;
 22 int N,K;
 23 struct SuffixArray{
 24     int s[maxn];
 25     int sa[maxn],height[maxn],rank[maxn],n;
 26     int t[maxn*2],t2[maxn*2];
 27     int cnt[maxn];
 28     void build_sa(int m)//字元都屬於0~m-1範圍 
 29     {
 30         int i,*x=t,*y=t2;
 31         for(i=0;i<m;i++) cnt[i]=0;
 32         for(i=0;i<n;i++) cnt[x[i]=s[i]]++;
 33         for(i=1;i<m;i++) cnt[i]+=cnt[i-1];
 34         for(i=n-1;i>=0;i--) sa[--cnt[x[i]]]=i; 
 35         for(int k=1,p;k<=n;k <<=1)//k<=n
 36         {
 37             p=0;
 38             for(i=n-k;i<n;i++) y[p++]=i;
 39             for(i=0;i<n;i++) if(sa[i]>=k) y[p++]=sa[i]-k;
 40             for(i=0;i<m;i++) cnt[i]=0;
 41             for(i=0;i<n;i++) cnt[x[y[i]]]++;
 42             for(i=1;i<m;i++) cnt[i]+=cnt[i-1];
 43             for(i=n-1;i>=0;i--) sa[--cnt[x[y[i]]]]=y[i];
 44             swap(x,y);
 45             p=1;x[sa[0]]=0;
 46             for(i=1;i<n;i++)
 47                    x[sa[i]]=y[sa[i-1]]==y[sa[i]]&&y[sa[i-1]+k]==y[sa[i]+k]? p-1:p++;
 48             if(p>=n) break;
 49             m=p;
 50         }
 51     }
 52     void build_height()
 53     {
 54         int k=0;
 55         for(int i=0;i<n;i++) rank[sa[i]]=i;
 56         for(int i=0;i<n-1;i++)
 57         {
 58             if(k) k--;
 59             int j=sa[rank[i]-1];
 60             while(s[i+k]==s[j+k]) k++;
 61             height[rank[i]]=k;
 62         }
 63     }
 64 } SA;
 65 
 66 inline bool check(int ans)
 67 {
 68     int cnt=1;
 69     for(int i=2;i<SA.n;++i)
 70     {
 71         if(SA.height[i]>=ans) cnt++;
 72         else
 73         {
 74             if(cnt>=K) return true;
 75             cnt=1;
 76         }
 77     }
 78     return false;
 79 }
 80 
 81 int main()
 82 {
 83     N=read();K=read();
 84     for(int i=0;i<N;++i) 
 85     {
 86         SA.s[i]=read();SA.s[i]++;
 87     }
 88     SA.s[N++]=0;SA.n=N;
 89     SA.build_sa(maxn);
 90     SA.build_height();
 91     int l=0,r=N;
 92     while(l<=r)
 93     {
 94         int mid=l+r>>1;
 95         if(check(mid)) l=mid+1;
 96         else r=mid-1;
 97     }
 98     cout<<r<<endl;
 99     return 0;
100 }

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