Arithmetic Progressions

Posted 2023-02-14 地球太危险了

一、问题

An arithmetic progression is a sequence of numbers a_1, a_2, . . . , a_ka1​,a2​,...,ak​ where the difference of consec- utive members ai+1 − ai is a constant (1 ≤ i ≤ k − 1). For example, the sequence 5, 8, 11, 14, 17 is an arithmetic progression of length 5 with the common difference 3.In this problem, you are requested to find the longest arithmetic progression which can be formed selecting some numbers from a given set of numbers. For example, if the given set of numbersis 0, 1, 3, 5, 6, 9, you can form arithmetic progressions such as 0, 3, 6, 9 with the common difference 3, or 9, 5, 1 with the common difference −4. In this case, the progressions 0, 3, 6, 9 and 9, 6, 3, 0 are the longest.

题目大意：给定 N(1<N<=5000) 个不同元素组成的集合，求从中选出若干数字组成的最长等差数列的长度是多少。

二、解题思路

（1）最优子结构

有一串数组为<x1，x2，x3，x4，x5，x6，...>，设xi，xj分别为某个最长等差数列的倒数第2个元素和倒数第1个元素，则长度记为L<xi，xj>，则这个等差数列的长度L<xi, xj> = L<xt, xi>，其中0<=t<i。

（2）构造递归表达式

          d[0][j] = 2, 1<j<n;

          d[i][j] = max2, d[t][i] + 1, 0<t<i;

三、代码

#include <iostream>
#include <algorithm>

using namespace std;
const int N =  5010;
int d[N][N] = 0;
int a[N] = 0;

int main ()
    int n;
    cin>>n;
    for(int i = 0; i < n; i++)
        cin>>a[i];
    
    sort(a, a + n);

    for(int j = 1; j < n; j++)
        d[0][j] = 2;
    
    int max = 2;
    for(int i = 1; i < n - 1; i++)
        for(int j = i + 1; j < n; j++)
            d[i][j] = 2;
            for(int t = 0; t < i; t++)
                if(a[j] - a[i] == a[i] - a[t])
                    d[i][j] = d[t][i] + 1;
                    if(d[i][j] > max)
                        max = d[i][j];
                    
                
            
        
    

    cout<<max<<endl;

    return 0;