Coderforces 633D：Fibonacci-ish（map+暴力枚举）

http://codeforces.com/problemset/problem/633/D

D. Fibonacci-ish

 

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.

Examples

input

3 1 2 -1

output

3

input

5 28 35 7 14 21

output

4

Note

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.

 

题意：给出 n个数，求出他们任意排列之后最长的斐波那契数列是多长。

思路：这题也是听了师兄的讲解之后才会做的。因为斐波那契数的递增是巨大的，1e9里面的斐波那契数是比较少的，

因此暴力枚举，比较巧妙的是用map来存一个数有没有出现过，并且如果是很多个零的话要加一个特判，不然的话可能会超时。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
#define N 1010

map <int, int> mp;
int a[N];
int temp[N];

int main()
{
    int n ;
    cin >> n;
    mp.clear();
    for(int i = 0; i < n; i++){
        scanf("%d", a + i);
        mp[a[i]]++;
        //该数出现的次数
    }
    int ans = 2, now = 0, cnt, x, y, z;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i == j) continue;
            x = a[i], y = a[j];
            if( x == 0 && y == 0 ){
                //两个都是0的话，直接判断0的个数，因为0+0=0
                if(mp[0] > ans) ans = mp[0];
                continue;
            }
            cnt = 0;
            mp[x]--; mp[y]--;
            temp[cnt++] = x;
            temp[cnt++] = y;
            //先把要加的数从map里面删除，避免重复，用temp存放，之后再放回去
            while(mp[x+y] > 0){
                z = x + y;
                mp[z]--;
                temp[cnt++] = z;
                x = y;
                y = z;
            }
            for(int i = 0; i < cnt; i++)
                mp[temp[i]]++;
            if(cnt > ans) ans = cnt;
        }
    }
    printf("%d
",ans);
    return 0;
}