CF#213DIV2：B The Fibonacci Segment

本文介绍了一个寻找数组中最长斐波那契数列片段的算法问题，通过扫描数组并验证每个元素是否满足斐波那契数列条件，从而确定最长连续片段的长度。

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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 test(s)

Input

10
1 2 3 5 8 13 21 34 55 89

Output

Input

5
1 1 1 1 1

Output

找出符合a[i] = a[i-1]+a[i-2]的最长长度

没有考虑0的状况。。。。。。蛋疼，长久没做CF了，坑成球了。。。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

int a[100005],b[100005],len;

int main()
{
    int maxn,sum;
    int i,n;
    while(~scanf("%d",&n))
    {
        sum = 0;
        for(i = 0; i<n; i++)
        {
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        if(sum == 0)
        {
            printf("%d\n",n);
            continue;
        }
        if(n == 1)
        {
            printf("1\n");
            continue;
        }
        if(n == 2)
        {
            printf("2\n");
            continue;
        }
        len = 0;
        int l,r,flag = 1;
        for(i = 2; i<n; i++)
        {
            if(a[i-1]+a[i-2] == a[i])
                b[len++] = 1;
            else
                b[len++] = 0;
        }
        maxn = 0;
        l = 0;
        r = 0;
        for(i = 0; i<len; i++)
        {
            if(b[i] == 1)
            {
                if(flag)
                {
                    l = i;
                    flag = 0;
                }
                else
                    r = i;
            }
            else
            {
                if(r-l+1>maxn && r!=l)
                    maxn = r-l+1;
                flag = 1;
                l = 0;
                r = 0;
            }
        }
        if(r-l+1>maxn && r!=l)
            maxn = r-l+1;
        printf("%d\n",maxn+2);
    }

    return 0;
}