D. Fibonacci-ish
题目连接：
http://www.codeforces.com/contest/633/problem/D
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
3
1 2 -1
Sample Output
3
Hint
题意
给你n个数，然后让你随便取一些数出来，使得能够组成的广义fib序列最长
题解：
暴力枚举第一个数和第二个数就好了，然后再暴力往下走，开一个map，记录一下每个数还有多少个
注意回溯。
然后还得注意第一个数和第二个数是0的情况，这样的话，你在暴力枚举的时候，可能复杂度就变成n^3了
所以再开一个map<pair<int,int>,int> 记录一下你枚举了哪些数就好了，这样复杂度就不可能变成n^3了。
当然，你预先处理出第一个数是0，第二个数是0的也可以
其他的fib长度感觉不会超过100吧？
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e3+5;
int n;
map<int,int> H;
map<pair<int,int>,int>H2;
int a[maxn];
int deal(int x,int y)
{
if(H[x+y]==0)return 0;
H[x+y]--;
int ans = deal(y,x+y);
H[x+y]++;
return ans+1;
}
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",&a[i]),H[a[i]]++;
sort(a,a+n);
int ans = 0;
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
if(i==j)continue;
if(H2[make_pair(a[i],a[j])])continue;
H2[make_pair(a[i],a[j])]=1;
H[a[i]]--;
H[a[j]]--;
ans=max(ans,deal(a[i],a[j]));
H[a[i]]++;
H[a[j]]++;
}
}
cout<<ans+2<<endl;
}