Codeforces--633D--Fibonacci-ish（暴力搜索+去重）（map）

Fibonacci-ish

Time Limit: 3000MS

Memory Limit: 524288KB

64bit IO Format: %I64d & %I64u

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Status

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 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.

Sample Input

Input

3
1 2 -1

Output

3

Input

5
28 35 7 14 21

Output

4

Sample Output

Hint

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.

Source

Manthan, Codefest 16

刚开始想直接暴力搜索的，可是数据的范围有点大，数组记录无法实现，需要用map，用map的时候要去重

#include<cstdio>
#include<cstring>
#include<map>
#include<algorithm>
using namespace std;
map<int,int>fp;
int a[1010];
int f(int a,int b)
{
	int ans=0;
	if(fp[a+b])
	{
		fp[a+b]--;
		ans=f(b,a+b)+1;
		fp[a+b]++;
	}
	return ans;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		fp.clear();
		for(int i=0;i<n;i++)
		scanf("%d",&a[i]),fp[a[i]]++;
		sort(a,a+n);
		int N=unique(a,a+n)-a;
		int ans=0;
		for(int i=0;i<N;i++)
		{
			for(int j=0;j<N;j++)
			{
				if(i==j&&fp[a[i]]==1) continue;
				fp[a[i]]--,fp[a[j]]--;
				ans=max(ans,f(a[i],a[j])+2);
				fp[a[i]]++,fp[a[j]]++;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}