Description
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
Lines 2.. N+1: N integers, one per line, the quality of the milk on day i appears on the ith line.
Output
Sample Input
8 2 1 2 3 2 3 2 3 1
Sample Output
4
思路：二分长度，判断符合条件不，后缀数组和HASH都行
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
#define LL unsigned long long
using namespace std;
const int mm=2e4+9;
const int mod=127;
LL h[mm],xp[mm],has[mm];
int rank[mm],f[mm];
int N,K;
bool cmp(const int&a,const int&b)
{
return has[a]<has[b]||(has[a]==has[b]&&a<b);
}
bool ok(int mid,int k)
{
for(int i=0;i<N-mid+1;++i)
{
rank[i]=i;
has[i]=h[i]-h[i+mid]*xp[mid];
}
sort(rank,rank+N-mid+1,cmp);
int num=0;
for(int i=0;i<N-mid+1;++i)
{
if(i==0||has[ rank[i] ]!=has[ rank[i-1] ])num=0;
if(++num>=k)return 1;
}
return 0;
}
int main()
{
while(~scanf("%d%d",&N,&K))
{
FOR(i,0,N-1)scanf("%d",&f[i]);
h[N]=0;
for(int i=N-1;i>=0;i--)
h[i]=h[i+1]*mod+f[i];
xp[0]=1;
FOR(i,1,N)xp[i]=xp[i-1]*mod;
int l=1,r=N,mid;
int ans=0;
//if(ok(4,2))puts("+++|");
while(l<=r)
{
mid=(l+r)/2;
if(ok(mid,K))
{ ans=max(ans,mid);
l=mid+1;
}
else r=mid-1;
}
printf("%d\n",ans);
}
}
