B. The Fibonacci Segment

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You have array a1, a2, ..., an. Segment [l, r] (1 ≤ l ≤ r ≤ n) is good if ai = ai - 1 + ai - 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, ..., an (0 ≤ ai ≤ 109).

Output

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

Sample test(s)

input

101 2 3 5 8 13 21 34 55 89

output

10

input

51 1 1 1 1

output

2

解题说明：此题就是判断一个数列中满足连续两个数之和等于后面一个数的最长串，用暴力遍历即可。

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int main(){int i,n,ans,k;int a[100000];scanf("%d",&n);for( i=0;i<n;i++) {scanf("%d",&a[i]);}ans=min(2,n);k=2;for( i=2;i<n;i++){if(a[i]==a[i-1]+a[i-2]) {k++;}else{k=2;}ans=max(ans,k);}printf("%d\n",ans);return 0;}