题目链接：http://acm.hzau.edu.cn/problem.php?cid=1009&pid=9



Problem J: Arithmetic Sequence


Time Limit: 1 Sec   Memory Limit: 128 MB

Submit: 1823  

Solved: 317

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Description


Awith length n,mnumbers from A(ignore the order) which can form an arithmetic sequence and make mas large as possible.


Input



   There are multiple test cases. In each test case, the first line contains a positive integer n.nintegers separated by spaces, indicating the number sequence A. All the integers are positive and not more than 2000. The input will end by EOF.


Output



   For each test case, output the maximum  as the answer in one line.


Sample Input

5
1 3 5 7 10
8
4 2 7 11 3 1 9 5

Sample Output

4
6

HINT


   In the first test case, you should choose 1,3,5,7 to form the arithmetic sequence and its length is 4.



   In the second test case, you should choose 1,3,5,7,9,11 and the length is 6.




题意很简单，给我们一个序列，问我们从这个序列中选出一些数能构成的最长的等差序列的长度为多少。

这个问题我们用DP来解决，首先因为我们取数是不用关注它的顺序的，所以我们就可以先排个序，然后我们用dp[i][j]表示取到第i个数的时候间距为j的长度的等差数列最长的长度，所以我们有状态转移方程dp[i][dif] = dp[j][dif]+1


#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 2000+5;
int a[maxn];
int dp[maxn][maxn];
int main()
{
  int n;
  while(scanf("%d",&n)!=EOF)
  {
    for(int i=0; i<n; i++) scanf("%d",&a[i]);
    if(n <= 1)
    {
      printf("%d\n",n);
      continue;
    }
    sort(a,a+n);
    int ans = 1;
    for(int i=0; i<maxn; i++)
      for(int j=0; j<maxn; j++)
        dp[i][j] = 1;
    for(int i=1; i<n; i++)
    {
      for(int j=0; j<i; j++)
      {
        int dif = a[i]-a[j];
        dp[i][dif] = dp[j][dif]+1;
        ans = max(ans, dp[i][dif]);
      }
    }
    //for(int i=0; i<n; i++) printf("%d ",dp[i][0]);
    printf("%d\n",ans);
  }
}