CodeForces - 255C Almost Arithmetical Progression dp 或 贪心

Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:

• a1 = p, where p is some integer;
• ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer.

Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.

Sequence s1,  s2,  ...,  sk is a subsequence of sequence b1,  b2,  ...,  bn, if there is such increasing sequence of indexes i1, i2, ..., ik (1  ≤  i1  <  i2  < ...   <  ik  ≤  n), that bij  =  sj. In other words, sequence s can be obtained from b by crossing out some elements.

Input

The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).

Output

Print a single integer — the length of the required longest subsequence.

Examples

Input

2
3 5

Output

2

Input

4
10 20 10 30

Output

3

Note

In the first test the sequence actually is the suitable subsequence.

In the second test the following subsequence fits: 10, 20, 10.

题解一：看到数据很多人会想到dp  我们用dp[i][j] 表示 第i这个位置 向前匹配到 j 这个数 所得到的最大值

那递推关系也就得到了  dp[i][ a[j] ] = dp[j][ a[i] ]  +  1 数据很大但数据量不大 离散化一下就可以了

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int N=1e6+10;
int n;
int a[N],id[N],dp[4100][4100];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		id[i]=a[i];
	}
	sort(id+1,id+1+n);
	int len=unique(id+1,id+1+n)-(id+1);
	for(int i=1;i<=n;i++)
		a[i]=lower_bound(id+1,id+1+n,a[i])-id;
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<i;j++)
		{
			dp[i][a[j]]=dp[j][a[i]]+1;
			ans=max(ans,dp[i][a[j]]);
		}
	}
	cout<<ans+1<<endl;
	return 0;
}

题解二：当然我们也可以贪心来做，先保存下有多少的不同的数，记录下每个数的位置，然后贪心一下就可以了

千万别用mp标记枚举的那两个数，会在68个超时的..因为我已经试过了

#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
const int N=1e6+10;
vector<int> v[N],vp;
int n;
int a[N],id[N];
map<int,int> mp;
int main()
{
	scanf("%d",&n);
	int ans=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		v[a[i]].push_back(i);
		if(v[a[i]].size()>ans) ans=v[a[i]].size();
		if(!mp[a[i]]) vp.push_back(a[i]),mp[a[i]]=1;
	}
	int len=vp.size();
	for(int i=0;i<len;i++)
	{
		for(int j=i+1;j<len;j++)
		{
			int x=vp[i],y=vp[j];
			int l1=v[x].size();
			int l2=v[y].size();
			int l=0,r=0;
			int cnt=1;
			while(l<l1&&r<l2)
			{
				while(r<l2&&v[y][r]<v[x][l])
					r++;
				if(r<l2) cnt++;
				else break;
				
				while(r<l2&&l<l1&&v[x][l]<v[y][r])
					l++;
				if(l<l1&&r<l2) cnt++;
				else break;
			}
			if(v[y][0]<v[x][0]) cnt++;
			ans=max(ans,cnt);
		}
	}
	printf("%d\n",ans);
	return 0;
}