B. Approximating a Constant Range
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

When Xellos was doing a practice course in university, he once had to measure the intensity of an effect that slowly approached equilibrium. A good way to determine the equilibrium intensity would be choosing a sufficiently large number of consecutive data points that seems as constant as possible and taking their average. Of course, with the usual sizes of data, it's nothing challenging — but why not make a similar programming contest problem while we're at it?

You're given a sequence of n data points a1, ..., an. There aren't any big jumps between consecutive data points — for each 1 ≤ i < n, it's guaranteed that |ai + 1 - ai| ≤ 1.

A range [l, r] of data points is said to be almost constant if the difference between the largest and the smallest value in that range is at most 1. Formally, let M be the maximum and m the minimum value of ai for l ≤ i ≤ r; the range [l, r] is almost constant if M - m ≤ 1.

Find the length of the longest almost constant range.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 100 000) — the number of data points.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 100 000).

Output

Print a single number — the maximum length of an almost constant range of the given sequence.

Sample test(s)
Input
5
1 2 3 3 2
Output
4
Input
11
5 4 5 5 6 7 8 8 8 7 6
Output
5
Note

In the first sample, the longest almost constant range is [2, 5]; its length (the number of data points in it) is 4.

In the second sample, there are three almost constant ranges of length 4: [1, 4], [6, 9] and [7, 10]; the only almost constant range of the maximum length 5 is [6, 10].

题意：求最大区间长度  区间要求满足：区间最大值与最小值的差小于等于1

题解：

例如

5

1 2 3 3 2

差值分别为 2-1=1；

               3-2=1；

               3-3=0；

               2-3=-1；  另外  it's guaranteed that |ai + 1 - ai| ≤ 1.

可以判断 当连续的差值或相隔差值为0 的两个差值 相等时 该段区间结束 更新最大值

 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,a;
int main()
{
    scanf("%d",&n);
    scanf("%d",&a);
    int cha=0;
    int judge=0;
    int l=0,r=0;
    int ans=0,exm;
    for(int i=1; i<n; i++)
    {
        scanf("%d",&exm);
        cha=exm-a;//计算差值
        a=exm;
        if(cha==0)//差值为零 相等时 
            continue;
        if(cha!=judge)//当前差值与 之前一个差值比较
        {
            judge=cha;//更新到当前区间
            r=i;
        }
        else
        {
            if(i-l>ans)//更新区间大小
                ans=i-l;
            l=r;
            r=i;
        }
    }
    if(n-l>ans)//特列 后端 都相等
        ans=n-l;
    cout<<ans<<endl;
    return 0;
}