Almost Arithmetical Progression(dp)

Posted 2021-01-19 1013star

Almost Arithmetical Progression

 CodeForces - 255C 

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 indexes i₁,?i₂,?...,?i_k (1??≤??i₁??<??i₂??<?... ??<??i_k??≤??n), that b_ij??=??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

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：找一个最长的锯齿形子串 例如10 30 10 30 10

思路：%大佬%大佬%大佬 大佬tql 

　　  因为数据在1e6而n只有4000，所以大佬说可以先搞一波离散化，然后开二维dp。dp[x][y]中x表示当前数的位置，y表示x这个位置上的数字前面的数，例如1 4 1 4 dp[3][4]代表当数字为1且其前面的数字为4时的最大子串长度。dp[3][4]=dp[2][1]+1,具体看代码。

 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
",ans+1);
34     return 0;
35 }