Almost Arithmetical Progression

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_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.

Sample Input

Input 2 3 5 Output 2 Input 4 10 20 10 30 Output 3

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("%d\n",ans); return 0; }