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.
Examples
inputCopy
3
1 2 -1
outputCopy
3
inputCopy
5
28 35 7 14 21
outputCopy
4
Note
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
题意：
给你n个数，问你可以构造的最长类斐波那契数列是多少，什么是类斐波那契数列？f[1]和f[2]随机，f[n]=f[n-1]+f[n-2]。
题解：
这道题算是2000分中比较水的题了吧，毕竟数据范围只有1e3，那么我们暴力枚举f[1]和f[2]就好了，然后用一个map存经过的状态，那么下一次就不会以这个状态为开头，这样就不会t。
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pa pair<ll,ll>
const int N=1e3+5;
map<ll,int>mp;
map<pa,int>used;
int vis[N];
ll a[N];
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%lld",&a[i]),mp[a[i]]++;
int ans=2,cnt=0;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(i==j||used[{a[i],a[j]}]==1)
continue;
used[{a[i],a[j]}]=1;
int len=2;
ll x=a[i],y=a[j];
vis[++cnt]=x,vis[++cnt]=y;
mp[x]--,mp[y]--;
while(mp[x+y]>0)
{
swap(x,y);
y+=x;
mp[y]--;
len++;
vis[++cnt]=y;
used[{x,y}]=1;
}
ans=max(ans,len);
if(ans==n)
break;
for(int k=1;k<=cnt;k++)
mp[vis[k]]++;
cnt=0;
}
if(ans==n)
break;
}
printf("%d\n",ans);
return 0;
}