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

Fibonacci-ish

Time Limit: 3000MS

Memory Limit: 524288KB

64bit IO Format: %I64d & %I64u

Submit

Status

Description

Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if

the sequence consists of at least two elements
f0 and
f1 are arbitrary
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.

Sample Input

Input

3
1 2 -1

Output

3

Input

5
28 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 sequence
ai 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;
}