CF#213DIV2：B The Fibonacci Segment

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

10

Input

5

1 1 1 1 1

Output

2

 

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;

}