1019: Arithmetic Sequence

Arithmetic Sequence

Time Limit: 1 Sec Memory Limit: 128 MB

Submit: 78 Solved: 29

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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.

Source

题意大概是从一组数中无序的取出m个数构成等差数列，求m的最大值，用dp求解，dp[i][d]为i以前包括i以d为公差的构成的最大个数的等差数列，j从0到i-1遍历所有节点，考察每一个节点是否与i构成公差为d的数列

废话少说，上代码==

#include <iostream>
#include <cstdio>
#include <algorithm>
#include<cstring>
using namespace std;

const int MAXN = 2016;

int n;
int dp[MAXN][MAXN];
int a[MAXN];

int main()
{
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; ++i) {
scanf("%d", a+i);
}
sort(a+1, a+1+n);
memset(dp, 0, sizeof dp);
int res = 0;
for (int i = 1; i <= n; ++i) {
for (int j = 1; j < i; ++j) {
dp[i][a[i]-a[j]] = max(dp[i][a[i]-a[j]], dp[j][a[i]-a[j]]+1);
res = max(res, dp[i][a[i]-a[j]]);
}
}
printf("%d\n", res+1);
}
}