CF 156 C Almost Arithmetical Progression

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/47836217

本文探讨了如何在给定的整数序列中找到最长的几乎等差数列子序列，提供了详细的算法实现步骤和实例解析。

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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.

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.

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;

#define N 4004

int n,a[N];
int dp[N][N];

void solve()
{
    int i,j;
    memset(dp,0,sizeof(dp));
    dp[0][0]=0;
    int ans=1;
    for(i=1;i<=n;i++)
    {
        int last;
        for(j=last=0;j<i;j++)
          {
              dp[i][j]=dp[j][last]+1;
              //printf("%d %d %d %d\n",last,i,j,dp[i][j]);
              ans=max(ans,dp[i][j]);
              if(a[j]==a[i]) last=j;//last 代表坐标小于j的a[i]的坐标最大的位置
          }
    }
    printf("%d\n",ans);
}

int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {

        for(i=1;i<=n;i++)
            scanf("%d",&a[i]);
        solve();
    }
    return 0;
}