HZAU--21--Arithmetic Sequence（二维dp）



Arithmetic Sequence

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 1815  Solved: 315

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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.
找出最长的等差序列，输出长度
二维dp，第二维记录公差
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int num[2020],dp[2020][2020];
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(num,0,sizeof(num));
for(int i=0;i<2020;i++)
{
for(int j=0;j<2020;j++)
dp[i][j]=1;
}
for(int i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
sort(num+1,num+n+1);
int ans=1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<i;j++)
{
int d=num[j]-num[i];
dp[i][d]=max(dp[i][d],dp[j][d]+1);
ans=max(dp[i][d],ans);
}
}
printf("%d\n",ans);
}
return 0;
}