ZCMU2993 Fibonacci-ish(排序+二分)

最新推荐文章于 2020-01-02 15:35:28 发布

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本文介绍了一种算法，旨在通过重新排列给定整数序列，使其具有尽可能长的斐波那契前缀。首先对序列进行排序，然后选择两个数作为斐波那契序列的起始值，使用二分查找在原序列中寻找后续的斐波那契数，确保已选数不再重复使用。此方法能找出最长的斐波那契似序列。

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

1 2 -1

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.

思路

斐波那契数列的第一项、第二项影响整个数列，如何选择第一二两项决定数列的长度。

斐波那契数列是一个非递减数列，不妨现将所给的n个数排序。然后从中选择两个数作为第一、二项，然后根据其特性用二分在原序列中查找F(n)(F(n)=F(n-1)+F(n-2))...F(n+1)...F(n+2)...

注意，已选择过的数不能再次选择

代码

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
typedef long long LL;
#define MAX 1005
LL Array[MAX],Mark[MAX],x,y,n;
int Binary_search(int v,int l,int r)
{
    int left=l,right=r,mid,temp;
    while(left<=right)
    {
        mid=(left+right)/2;
        if(Array[mid]==v)
        {
            if(Mark[mid]==1)//若Array[mid]已被选择，则尝试在剩余的数中找与它大小相同的数
            {
                temp=Binary_search(v,left,mid-1);
                if(temp!=-1) return temp;
                temp=Binary_search(v,mid+1,right);
                if(temp!=-1) return temp;
                return -1;
            }
            else
                return mid;
        }
        else if(v>Array[mid])
            left=mid+1;
        else if(v<Array[mid])
            right=mid-1;
    }
    return -1;
}
int solve()
{
    int re=2,flag=1;
    while(flag)
    {
        int pos=Binary_search(x+y,0,n-1);
        if(pos!=-1&&Array[pos]==x+y)
        {
            x=y;
            y=Array[pos];
            Mark[pos]=1;
            re+=1;
        }
        else
            flag=0;
    }
    return re;
}
int main()
{
    int i,j;
    while(~scanf("%d",&n))
    {
        for(i=0;i<n;i++)
            scanf("%lld",&Array[i]);
        sort(Array,Array+n);
        int ans=2;
        memset(Mark,0,sizeof(Mark));
        for(i=0;i<n;i++)
        {
            for(j=0;j<n;j++)
            {
                if(i!=j)
                {
                    x=Array[i]; y=Array[j];
                    Mark[i]=Mark[j]=1;
                    int Return=solve();
                    memset(Mark,0,sizeof(Mark));
                    if(Return>ans) ans=Return;
                }
                if(ans==n) break;
            }
            if(ans==n) break;
        }
        printf("%d\n",ans);
    }
    return 0;
}