CodeForces - 633D Fibonacci-ish (map&暴力)

64bit IO Format: %I64d & %I64u

CodeForces - 633D

Fibonacci-ish

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Description

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
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.

Sample Input

Input

31 2 -1

Output

3

Input

528 35 7 14 21

Output

4

Hint

In the first sample, if we rearrange elements of the sequence as  - 1, 2, 1, the whole sequence a_i

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

, 

, 

, 

, 28.

Source

Manthan, Codefest 16

//上次就遇见一个题，考试完听大神说是用map做的，这次又是一个map,看来是时候学一下map了。

这个是看大神的代码写的，还在理解中。。。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<map>
#define INF 0x3f3f3f3f
#define ull unsigned long long
#define IN __int64
#define ll long long
using namespace std;
int a[1010];
map<int,int>fp;
int dfs(int a,int b)
{
	int ans=0;
	if(fp[a+b])
	{
		fp[a+b]--;
		ans=dfs(b,a+b)+1;
		fp[a+b]++;
	}
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		memset(a,0,sizeof(a));
		fp.clear();
		for(int i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
			fp[a[i]]++;
		}
		sort(a,a+n);
		int ans=0;
		int m=unique(a,a+n)-a;
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(i==j&&fp[a[i]]==1)
					continue;
				fp[a[i]]--;fp[a[j]]--;
				ans=max(ans,dfs(a[i],a[j])+2);
				fp[a[i]]++;fp[a[j]]++;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}