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最新推荐文章于 2019-09-28 17:05:28 发布

爱吃老谈酸菜的DV

最新推荐文章于 2019-09-28 17:05:28 发布

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分类专栏：动态规划（DP）

本文链接：https://blog.youkuaiyun.com/weixin_43460224/article/details/99778215

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本文探讨了一个算法问题，即从给定的整数序列中找出最长的子序列，该子序列遵循特定的几乎等差数列规则。通过使用动态规划的方法，我们能够有效地解决这个问题，找到满足条件的最大子序列长度。

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Discription
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.

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.

题意
给定n个数，要求从这n个数中找一个最大的子串，满足：

a1 = p,；
ai = ai - 1 + ( - 1)i + 1·q (i > 1)；//其中p,q任取

思路
DP
状态转移方程为

dp[j][i]=dp[k][j]+1;

AC代码

#include<bits/stdc++.h>
using namespace std;
int n,ans,p,k;
int a[4010];
int dp[4010][4010];
int main()
{
    cin>>n;
    for(int i=1; i<=n; i++)
        cin>>a[i];
    int i,j;
    for(i=1;i<=n;i++)
    {
        for(j=0,k=0;j<i;j++)
        {
            dp[j][i]=dp[k][j]+1;
            if(a[i]==a[j])
               k=j;
            ans=max(dp[j][i],ans);
        }
    }
    cout<<ans<<endl;
    return 0;
}