Codeforces 256A Almost Arithmetical Progression【dp】

最新推荐文章于 2019-08-20 09:30:31 发布

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最新推荐文章于 2019-08-20 09:30:31 发布

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分类专栏： dp 文章标签： Codeforces 256A

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本文介绍了一种算法，用于解决寻找给定序列中最长的几乎等差子序列的问题。这种子序列由一个整数p开始，后续项遵循加减q的规律。通过动态规划方法，文章详细阐述了如何在O(n^2)的时间复杂度内找到该子序列。

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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 analmost arithmetical progression, if its elements can be represented as:

a₁ = p, wherep 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 ofn 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 sequenceb₁, b₂, ..., b_n, if there is such increasing sequence of indexesi₁, i₂, ..., i_k(1 ≤ i₁ < i₂ < ... < i_k ≤ n), thatb_{i_j} = s_j. In other words, sequences 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 integersb₁, 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.

题目大意：

给你一个长度为N的序列，然你从中找出一个最长子序列a1，a2，.......an，使得a1=p，a2=a1-q，a3=a2+q...............

思路：

1、其实就是找一个最长子序列，使得其为：p，q，p，q......................

2、设定dp【i】【j】表示第i个数前边的数为j，那么不难想到其状态转移方程：

dp【i】【j】=dp【j】【pre】，这里pre表示上一个a【i】的位子。

时间复杂度O（n^2）

Ac代码：

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
int a[5000];
int dp[5005][5005];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        int ans=0;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            int pre=0;
            for(int j=0;j<i;j++)
            {
                dp[i][j]=dp[j][pre]+1;
                if(a[j]==a[i])pre=j;
                ans=max(ans,dp[i][j]);
            }
        }
        printf("%d\n",ans);
    }
}