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

Fibonacci-ish

Time Limit: 3000MS Memory Limit: 524288KB 64bit IO Format: %I64d & %I64u

SubmitStatus

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 alln ≥ 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 sequencea_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

Sample Output

Hint

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