Huazhong Agricultural University The fourth session of the Program Design Competition network synchronization game J

Problem j:arithmetic Sequence time limit:1 Sec Memory limit:128 MB
submit:1766 solved:299
[Submit] [Status] [Web Board] 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 is 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 is positive and not more than 2000. The input would end by EOF.

Output

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

Sample Input

51 3 5 7 1084 2 7 11 3 1 9 5

Sample Output

46

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 are 6.

Test instructions: gives you the length of the longest arithmetic progression in a sequence output sequence

Exercises

1. Violence sort and then violence enumerates each step of the pit point (note that the same number is equal to 0)

1#include <iostream>2#include <cstring>3#include <cstdio>4#include <map>5#include <queue>6#include <stack>7 using namespacestd;8 intN;9 intmp[2005];Ten intEXM; One intans; A intQ; - intMAXN; - intMain () the { -      while(SCANF ("%d", &n)! =EOF) -     { -ans=-1; +maxn=-1; -          for(intI=1; i<= -; i++) +mp[i]=0; A          for(intI=0; i<n;i++) at         { -scanf"%d",&EXM); -mp[exm]++; -           if(ans<MP[EXM]) -ans=MP[EXM]; -           if(maxn<EXM) inmaxn=EXM; -         } to           for(intI=1; i<=maxn;i++) +          { -             if(Mp[i]) the             { *                 for(intD=1;d <=maxn;d++) $               {Panax Notoginseng                 intgg=i+D; -q=1; the                   while(GG) +                { A                  if(gg>MAXN) the                     Break; +                  if(Mp[gg]) -q++; $                  Else $                     Break; -gg=gg+D;  -                } the                  if(q>ans) -ans=Q;Wuyi                  if(gg>MAXN) the                   Break; -               } Wu             } -          }   Aboutcout<<ans<<Endl;  $     } -     return 0; -  } 

2.DP processing (Yan code) dp[i][j] indicates the number of methods with the end of I as J

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cstring>5#include <cmath>6#include <vector>7#include <map>8 #pragmaComment (linker, "/stack:102400000,102400000")9 using namespacestd;Ten Const intN = -+ -, M = 1e6+Ten, mod = 1e9+7, INF =1e9; OnetypedefLong Longll; A   - intN,a[n]; - intDp[n][n]; the intMain () { -      while(SCANF ("%d", &n)! =EOF) { -          for(intI=1; i<=n;i++) scanf ("%d",&a[i]); -Sort (A +1, a+n+1); +          for(intj=1; j<=n;j++) -          for(intI=0; i<= -; i++) Dp[j][i] =1; +          for(intI=2; i<=n;i++) { A              for(intj=1; j<i;j++) { atDP[I][A[I]-A[J]] = max (dp[j][a[i]-a[j]]+1, dp[i][a[i]-A[j]]); -             } -         } -         intAns =0; -          for(intj=1; j<=n;j++) -          for(intI=0; i<= -; i++) { inAns =Max (ans,dp[j][i]); -          } toprintf"%d\n", ans); +     } -     return 0; the}

Huazhong Agricultural University fourth session of the Program Design Competition network synchronization game J