hzauoj Problem J: Arithmetic Sequence (dp)

Problem J: Arithmetic Sequence

Time Limit: 1 Sec  Memory Limit:
128 MB
Submit: 1806  Solved: 308

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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.
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define MAXN 2020
int dp[MAXN][MAXN];
int main()
{
int n,num[MAXN];
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
sort(num+1,num+n+1);
for(int i=1;i<=n;i++)
for(int j=1;j<MAXN;j++)
{
dp[i][j]=1;
}
int ans=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<i;j++)
if(num[i]>num[j])
{
int d=num[i]-num[j];
dp[i][d]=max(dp[i][d],dp[j][d]+1);
if(dp[i][d]>ans)
ans=dp[i][d];
}
}
int ans2=1,pre=num[1];
for(int i=2;i<=n;i++)
{
if(pre==num[i])
{
ans2++;
ans=max(ans,ans2);
}
else
{
pre=num[i];
ans2=1;
}
}
if(n==1)
printf("1\n");
else
printf("%d\n",ans);
}
return 0;
}