Arithmetic Progressions
链接：https://ac.nowcoder.com/acm/contest/7831/B
来源：牛客网

题目描述

An arithmetic progression is a sequence of numbers a1, a2, ..., ak where the difference of consecutive members ai+1−ai is a constant (1 ≤ i ≤ k−1). For example, the sequence 5, 8, 11, 14, 17 is an arithmetic progression of length 5 with the common difference 3.
In this problem, you are requested to find the longest arithmetic progression which can be formed selecting some numbers from a given set of numbers. For example, if the given set of numbers is {0, 1, 3, 5, 6, 9}, you can form arithmetic progressions such as 0, 3, 6, 9 with the common difference 3, or 9, 5, 1 with the common difference −4. In this case, the progressions 0, 3, 6, 9 and 9, 6, 3, 0 are the longest.

输入描述:
示例1
输入
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6 
0 1 3 5 6 9

输出
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4

示例2
输入
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7 
1 4 7 3 2 6 5

输出
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7

示例3
输入
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5 
1 2 4 8 16

输出
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2

题意：

给定一些数，允许重新排列，求其中最长的等差数列的长度

题解：

队友做得。。。队友说是简单dp
p[i][j] 表示区间[i, j] 的最大等差序列长度，从当前位置往前往后找满足条件的区间，如果有就加一，边界dp[i][i + 1] = 2，不难理解，其实任意dp[i][j]（j！=i） 最小都为2.

代码：

#include<bits/stdc++.h>
#define maxn 6000
using namespace std;
int dp[maxn][maxn];
long long a[maxn];
int n;
int main(){scanf("%d",&n);for(int i=1;i<=n;i++){scanf("%lld",&a[i]);}sort(a+1,a+1+n);int ans=0;for(int i=1;i<=n;i++){int k=i-1;for(int j=i+1;j<=n;j++){dp[i][j]=2;int d=a[j]-a[i];while(k>=1&&a[i]-a[k]<d){k--;}if(k==0||a[i]-a[k]!=d)continue;dp[i][j]=max(dp[i][j],dp[k][i]+1);ans=max(ans,dp[i][j]);}}printf("%d",ans==0? 2 : ans);return 0;
}