华中农业大学第四届程序设计大赛网络同步赛 J

Problem J: Arithmetic Sequence

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1766 Solved: 299
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Description

Giving a number sequence A with length n, you should choosing m numbers from A(ignore the order) which can form an arithmetic sequence and make m as large as possible.

Input

There are multiple test cases. In each test case, the first line contains a positive integer n. The second line contains n integers 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：给你一个序列输出序列中最长等差数列的长度

题解：

1.暴力 sort排序一下然后暴力枚举每一个步长坑点（注意相同的数也就是等差可以为0）

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<map>

 #include<queue>

 #include<stack>

 using namespace std;

 int n;

 int mp[];

 int exm;

 int ans;

 int q;

 int maxn;

 int main()

 {

     while(scanf("%d",&n)!=EOF)

     {

         ans=-;

         maxn=-;

         for(int i=;i<=;i++)

          mp[i]=;

         for(int i=;i<n;i++)

         {

           scanf("%d",&exm);

           mp[exm]++;

           if(ans<mp[exm])

           ans=mp[exm];

           if(maxn<exm)

           maxn=exm;

         }

          for(int i=;i<=maxn;i++)

          {

             if(mp[i])

             {

                for(int d=;d<=maxn;d++)

               {

                 int gg=i+d;

                 q=;

                  while(gg)

                {

                  if(gg>maxn)

                    break;

                  if(mp[gg])

                    q++;

                  else

                    break;

                   gg=gg+d;

                }

                  if(q>ans)

                  ans=q;

                  if(gg>maxn)

                  break;

               }

             }

          }

          cout<<ans<<endl;

     }

     return ;

  }

2.dp处理（yan代码） dp[i][j] 表示以i为结尾 j为等差的方法数目

 #include <iostream>

 #include <algorithm>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #include<vector>

 #include<map>

 #pragma comment(linker, "/STACK:102400000,102400000")

 using namespace std;

 const int N = +, M = 1e6+, mod = 1e9+,inf = 1e9;

 typedef long long ll;

 int n,a[N];

 int dp[N][N];

 int main() {

     while(scanf("%d",&n)!=EOF) {

         for(int i=;i<=n;i++) scanf("%d",&a[i]);

         sort(a+,a+n+);

         for(int j=;j<=n;j++)

         for(int i=;i<=;i++) dp[j][i] = ;

         for(int i=;i<=n;i++) {

             for(int j=;j<i;j++) {

                 dp[i][a[i]-a[j]] =  max(dp[j][a[i]-a[j]]+,dp[i][a[i]-a[j]]);

             }

         }

         int ans = ;

         for(int j=;j<=n;j++)

         for(int i=;i<=;i++) {

             ans = max(ans,dp[j][i]);

          }

         printf("%d\n",ans);

     }

     return ;

 }