https://www.51nod.com/Challenge/Problem.html#!#problemId=1055
dp[i][j]代表以a[i] a[j]为最后两项的等差数列的长度 对于等差数列有2*b[i]=b[i-1]+b[i+1] 如果对于某个等差数列 我们已经找到了b[i-1]+d=b[i] 如果再找到b[i]+d=b[i+1]那就可以把b[i+1]加到该等差数列末尾 然后抓住b[i-1] b[i] b[i+1]的中间项b[i]向两边扩展转移即可
#include <bits/stdc++.h>
using namespace std;
const int maxn=1e4+10;
short dp[maxn][maxn];
int ary[maxn];
int n;
int main()
{
int i,j,k,ans;
scanf("%d",&n);
for(i=1;i<=n;i++) scanf("%d",&ary[i]);
sort(ary+1,ary+n+1);
for(i=1;i<=n;i++){
for(j=i+1;j<=n;j++){
dp[i][j]=2;
}
}
ans=2;
for(j=n-1;j>=2;j--){
i=j-1,k=j+1;
while(1<=i&&k<=n){
if(ary[i]+ary[k]<2*ary[j]) k++;
else if(ary[i]+ary[k]>2*ary[j]) i--;
else{
dp[i][j]=dp[j][k]+1;
ans=max(ans,(int)(dp[i][j]));
i--,k++;
}
}
}
printf("%d\n",ans);
return 0;
}