Almost Arithmetical Progression(dp)

阿新 • • 發佈：2018-11-11

Almost Arithmetical Progression

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:

a

₁ = 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 indexes i₁, 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.

Examples

Input

2
3 5

Output

Input

4
10 20 10 30

Output

Note

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

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

題意：找一個最長的鋸齒形子串例如10 30 10 30 10

思路：%大佬%大佬%大佬大佬tql

　　因為資料在1e6而n只有4000，所以大佬說可以先搞一波離散化，然後開二維dp。dp[x][y]中x表示當前數的位置，y表示x這個位置上的數字前面的數，例如1 4 1 4 dp[3][4]代表當數字為1且其前面的數字為4時的最大子串長度。dp[3][4]=dp[2][1]+1,具體看程式碼。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 const int maxn=4010;
 7 int a[maxn],id[maxn],b[maxn];
 8 int dp[maxn][maxn];
 9 int n;
10 int main()
11 {
12     scanf("%d",&n);
13     for(int i=1;i<=n;i++)
14     {
15         scanf("%d",&a[i]);
16         id[i]=a[i];    
17     }
18     sort(id+1,id+1+n);
19     int len=unique(id+1,id+1+n)-(id+1);
20     for(int i=1;i<=n;i++)
21     {
22         a[i]=lower_bound(id+1,id+1+n,a[i])-id;
23     }
24     int ans=0;
25     for(int i=1;i<=n;i++)
26     {
27         for(int j=1;j<i;j++)
28         {
29             dp[i][a[j]]=dp[j][a[i]]+1;
30             ans=max(ans,dp[i][a[j]]);
31         }
32     }
33     printf("%d\n",ans+1);
34     return 0;
35 }

Almost Arithmetical Progression(dp)