Manthan, Codefest 16-D. Fibonacci-ish

原题链接

D. 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

31 2 -1

output

3

input

528 35 7 14 21

output

4

某大牛思路:

:由于Fibonacci数列的长度是log级的,所以直接枚举f0,f1然后构造出这个数列,不断更新答案就可以了 
需要注意的是,特判f0=f1=0的情况,不然会退化到n3的 
需要注意的是,不能重新添加map,要记录用过的数回溯一下,不然也会退化到n3的 
由于用map记录数是否用过,时间复杂度为O(n2log2n)

#include <bits/stdc++.h>#define maxn 1005#define MOD 1000000007using namespace std;typedef long long ll;map<ll, int> m;int num[maxn];vector<int> v;int main(){//freopen("in.txt", "r", stdin);int n;scanf("%d", &n);for(int i = 0; i < n; i++){ scanf("%d", num+i); m[num[i]]++;   }   int ans = m[0];   for(int i = 0; i < n; i++)    for(int j = 0; j < n; j++){    if(i == j || (!num[i] && !num[j]))     continue;    v.clear();    v.push_back(num[i]);    v.push_back(num[j]);    m[num[i]]--;    m[num[j]]--;    int cnt = 2;    ll pre = num[j], k = (ll)num[i] + num[j];    while(m[k]){    cnt++;    m[k]--;    v.push_back(k);        ll d = k;        k += pre;        pre = d;    }    ans = max(ans, cnt);    for(int i = 0; i < v.size(); i++)     m[v[i]]++;    }    printf("%d\n", ans);return 0;} 

: