CodeForces - 633D Fibonacci-ish 大数标记map+pair的使用

Fibonacci-ish

Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if

1. the sequence consists of at least two elements
2. f₀ and f₁ are arbitrary
3. f_n + 2 = f_n + 1 + f_n for all n ≥ 0.

You are given some sequence of integers a₁, a₂, ..., a_n. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 1000) — the length of the sequence a_i.

The second line contains n integers a₁, a₂, ..., a_n (|a_i| ≤ 10⁹).

Output

Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement.

Examples

Input

3
1 2 -1

Output

3

Input

5
28 35 7 14 21

Output

4

Note

In the first sample, if we rearrange elements of the sequence as  - 1, 2, 1, the whole sequence a_i would be Fibonacci-ish.

In the second sample, the optimal way to rearrange elements is , , , , 28.

从已知元素中寻找最长斐波那契长度。

个人非常喜欢的一道题，比较综合。用到了一些套路，比如用map标记大数，以及加上pair来标记一组数。

开始有往dp方向考虑，想先给数列排个序，然后依次往右找。但是两数相加不一定会变大，和可能会在左边，比如本题负数情况。

再一个每次寻找下一个值时，都要重新使用标记，为了防止后效性，使用递归来处理。

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<set>
#include<map>
#include<algorithm>
#define MAX 1005
#define INF 0x3f3f3f3f
using namespace std;

int a[MAX],c;
map<int,int> mp;
map<pair<int,int>,int> mmp;
void dfs(int x,int y){     //递归寻找

    if(mp[x+y]>) {
        c++;
        mp[x+y]--;
        dfs(y,x+y);
        mp[x+y]++;
    }
}
int main()
{
    int n,i,j;
    scanf("%d",&n);
    for(i=;i<=n;i++){
        scanf("%d",&a[i]);
        mp[a[i]]++;      //大数标记
    }
    int max=;
    for(i=;i<=n;i++){
        for(j=;j<=n;j++){    //考虑负数情况
            if(i!=j){
                if(mmp[make_pair(a[i],a[j])]) continue;
                mmp[make_pair(a[i],a[j])]=;     //标记一对数，优化防T
                c=;
                mp[a[i]]--;
                mp[a[j]]--;
                dfs(a[i],a[j]);
                mp[a[i]]++;
                mp[a[j]]++;
                if(c>max) max=c;
            }
        }
    }
    printf("%d\n",max);
    return ;
}