cf 602 B(模拟)

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].

#include <bits/stdc++.h>
using namespace std;
multiset <int> s;
int a[100005];
int main()
{
int n,ans=0;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d",a+i);
int j=0,i=0;
while(j<n)
{
s.insert(a[j]);
if(abs(*s.begin()-*s.rbegin())>1)
{
s.erase(s.find(a[i]));
i++;
}
ans=max(ans,int(s.size()));
j++;
}
printf("%d\n",ans);
return 0;
}