CodeForces 256A （dp）

A. Almost Arithmetical Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 indexesi1, 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.

Sample test(s)

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.

题意：从给定的有序的序列中找最多的p，q，p，q.

dp[i][j]:代表在第 i 个数前一个数是 j 的情况下的最多值，

则：

dp[i][j]=dp[j][pre] + 1,

#include<cstdio>
#include<stdlib.h>
#include<string.h>
#include<string>
#include<map>
#include<cmath>
#include<iostream>
#include <queue>
#include <stack>
#include<algorithm>
#include<set>
using namespace std;
#define INF 1e8
#define eps 1e-8
#define LL long long 
#define maxn 100105
#define mod  1000000009  

int n,a[4004],dp[4004][4004];
int main()
{
	while(scanf("%d",&n)==1)
	{
		int ans=-1;
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			int pre=0;
			for(int j=0;j<i;j++)
			{
				dp[i][j]=dp[j][pre]+1;
				if(a[i]==a[j]) pre=j;
				ans=max(ans,dp[i][j]);
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}
/*
8
10 10 20 30 20 10 20 40
*/