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 = ; i < n; i++){
         scanf("%d", a + i);
         mp[a[i]]++;
         //该数出现的次数
     }
     int ans = , now = , cnt, x, y, z;
     for(int i = ; i < n; i++){
         for(int j = ; j < n; j++){
             if(i == j) continue;
             x = a[i], y = a[j];
             if( x ==  && y ==  ){
                 //两个都是0的话，直接判断0的个数，因为0+0=0
                 if(mp[] > ans) ans = mp[];
                 continue;
             }
             cnt = ;
             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] > ){
                 z = x + y;
                 mp[z]--;
                 temp[cnt++] = z;
                 x = y;
                 y = z;
             }
             for(int i = ; i < cnt; i++)
                 mp[temp[i]]++;
             if(cnt > ans) ans = cnt;
         }
     }
     printf("%d\n",ans);
     return ;
 }