Almost Arithmetical Progression

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我是靠谱客的博主怕黑丝袜，最近开发中收集的这篇文章主要介绍Almost Arithmetical Progression，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

C - Almost Arithmetical Progression

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 255C

Description

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;
a_i = a_i - 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, ..., s_k is a subsequence of sequence b₁, b₂, ..., b_n, if there is such increasing sequence of indexesi₁, i₂, ..., i_k(1 ≤ i₁ < i₂ < ... < i_k ≤ n), that b_{i_j} = s_j. 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, ..., b_n(1 ≤ b_i ≤ 10⁶).

Output

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

Sample Input

Input

2
3 5

Output

Input

4
10 20 10 30

Output

Hint

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

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

dp的式子很简单，不过比较难想。

用一个二维数组记录。

——>>状态：dp[i][j] 表示满足条件的最后两个数是 ai 和 aj 的子序列长度

状态转移方程：dp[i][j] = dp[last][i] + 1;（last 是小于 i 的但离 i 最近的 a[last] == a[j] 成立的位置）

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const int MAXN=4000+10;
int a[MAXN],dp[MAXN][MAXN];
int main()
{
int n;
scanf("%d",&n);
int i,j,last;
for(i=1;i<=n;++i)
{
scanf("%d",&a[i]);
}
int ans=0;
dp[0][0]=0;
for(j=1;j<=n;++j)
for(i=0,last=0;i<j;++i)
{
dp[i][j]=dp[last][i]+1;//dp[0][j]都会默认是1，dp[1][j]默认就会是2
if(a[i]==a[j])last=i;
ans=max(ans,dp[i][j]);
}
printf("%dn",ans);
return 0;
}