Codeforces--633D--Fibonacci-ish（map+去重）（twice）



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 andf₁ 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 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

我去，以前做过的，就记得是dfs，但是忘记要用map跟去重了，真是人老了，记性不行咯

#include<iostream>#include<map>#include<cstring>#include<algorithm>using namespace std;int num[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 (cin >> n){memset(num, 0, sizeof(num));fp.clear();for (int i = 0; i < n; i++)cin >> num[i], fp[num[i]]++;sort(num, num + n);int ans = 0;int N = unique(num, num + n)-num;for (int i = 0; i < N; i++){for (int j = 0; j < N; j++){if (i == j&&fp[num[i]] == 1) continue;fp[num[i]]--, fp[num[j]]--;ans = max(ans, DFS(num[i], num[j]) + 2);fp[num[i]]++, fp[num[j]]++;}}cout << ans << endl;}return 0;}