codeforces 213B div2 The Fibonacci Segment-优快云博客

本文介绍了一个算法问题，即在给定数组中找到最长的连续子序列，该子序列满足斐波那契数列特性。输入包括数组长度及元素值，输出为满足条件的最长子序列长度。

codeforces 213B DIV2 The Fibonacci Segment

Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Description

You have array a₁, a₂, ..., a_n. Segment [l, r] (1 ≤ l ≤ r ≤ n) is good if a_i = a_i - 1 + a_i - 2, for all i(l + 2 ≤ i ≤ r).

Let's define len([l, r]) = r - l + 1, len([l, r]) is the length of the segment [l, r]. Segment [l₁, r₁], is longer than segment [l₂, r₂], if len([l₁, r₁]) > len([l₂, r₂]).

Your task is to find a good segment of the maximum length in array a. Note that a segment of length 1 or 2 is always good.

Input

The first line contains a single integer n (1 ≤ n ≤ 10⁵) — the number of elements in the array. The second line contains integers: a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁹).

Output

Print the length of the longest good segment in array a.

Sample Input

Input

10
1 2 3 5 8 13 21 34 55 89

Output

Input

5
1 1 1 1 1

Output

找满足斐波那契数列的连续子序列最长长度

<span style="font-size:18px;">#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

struct node
{
    long long a,b;
};
node a[100005];
int main()
{
    int i,n,t,ans;
    while(~scanf("%d",&n))
    {

        for(i=0; i<n; i++)
        {
            scanf("%I64d",&a[i].a);
            a[i].b=0;
        }
        for(i=2; i<n; i++)
            if(a[i].a==a[i-1].a+a[i-2].a) a[i].b=1;
        t=0,ans=0;
        for(i=0; i<n; i++)
        {
            if(a[i].b==1) t++;
            else t=0;
            if(t>ans) ans=t;
        }
if(n<3) printf("%d\n",n);
   else     printf("%d\n",ans+2);
    }
    return 0;
}
</span>