51nod1055 最长等差数列

$n leq 10000$的不同数字，问能组成的最长等差数列。

$f(i,j)$--$i,j$结尾的最长等差数列。枚举中间$i$，往左右扩展，找$j,k$，用$f(j,i)$来更新$f(i,k)$。

$n leq 100000$的暂时不会。

//#include<iostream>
#include<cstring>
#include<cstdio>
//#include<time.h>
//#include<complex>
//#include<set>
//#include<queue>
//#include<vector>
#include<algorithm>
#include<stdlib.h>
using namespace std;

#define LL long long
int qread()
{
    char c; int s=0,f=1; while ((c=getchar())<'0' || c>'9') (c=='-') && (f=-1);
    do s=s*10+c-'0'; while ((c=getchar())>='0' && c<='9'); return s*f;
}

//Pay attention to '-' , LL and double of qread!!!!

int n;
short f[10002][10002];
int a[10011];
int main()
{
    n=qread();
    for (int i=1;i<=n;i++) a[i]=qread();
    sort(a+1,a+1+n);
    for (int i=1;i<=n;i++)    
        for (int j=i+1;j<=n;j++)
            f[i][j]=2;
    for (int i=1;i<=n;i++)
    {
        int j=i-1,k=i+1,v=a[i]*2;
        while (j && k<=n)
        {
            int u=a[j]+a[k];
            if (u<v) k++;
            else if (u>v) j--;
            else if (u==v) {f[i][k]=max(f[i][k],short(f[j][i]+1)); j--; k++;}
        }
    }
    int ans=0;
    for (int i=1;i<=n;i++)
        for (int j=i+1;j<=n;j++)
            ans=max(ans,(int)f[i][j]);
    printf("%d
",ans);
    return 0;
}