CodeForces 633D Fibonacci-ish（容器暴力）

Fibonacci-ish time limit per test 3 seconds memory limit per test 512 megabytes input standard input output standard output

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. fn + 2 = fn + 1 + fn for all n ≥ 0.

You are given some sequence of integers a1, a2, ..., an. 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 ai.

The second line contains n integers a1, a2, ..., an (|ai| ≤ 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 ai would be Fibonacci-ish.

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

, 

, 

, 

, 28.

自己寫了，幾次都不對，最後看的題解，心酸！！

AC代碼： 

    強行暴力進行周遊！

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
using namespace std;
typedef long long LL;
const int N=1000000+5;
map<LL,int>mp;
LL a[1005];
int main()
{
    int n;
    LL res=0,mx=-1;
    scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
       scanf("%I64d",&a[i]);
       mx=max(a[i],mx);
      if(a[i]==0)++res;
      mp[a[i]]++;
    }
    LL ans=2;
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=n;++j)
        {
           if(j==i)continue;
           if(a[i]==0&&a[j]==0)continue;
           LL x=a[i],y=a[j],cnt=2;
           vector<LL>t;
           t.clear();
           t.push_back(x);
           t.push_back(y);
           mp[x]--;
           mp[y]--;
           while(x+y<=mx&&mp[x+y]>0)
           {
              LL tmp=x+y;
              mp[tmp]--;
              t.push_back(tmp);
              x=y;
              y=tmp;
              ++cnt;
           }
           for(int k=0;k<t.size();k++)
            mp[t[k]]++;
           ans=max(ans,cnt);
        }
    }
    printf("%I64d\n",max(ans,res));
    return 0;
}