Approximating a Constant Range（RMQ）

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

51 2 3 3 2

output

4

input

115 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].

AC代码：

#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<cstdio>#include<cmath>using namespace std;#define CRL(a) memset(a,0,sizeof(a))typedef __int64 ll;#define T 100005#define mod 1000000007int v[T],n;int dp_max[T][32];int dp_min[T][32];void RMQ_init(){for(int i=0;i<n;++i)dp_max[i][0]=v[i],dp_min[i][0]=v[i];for(int j=1;(1<<j)<=n;++j)for(int i=0;i+(1<<j)-1<n;++i){dp_max[i][j] = max(dp_max[i][j-1],dp_max[i+(1<<(j-1))][j-1]);dp_min[i][j] = min(dp_min[i][j-1],dp_min[i+(1<<(j-1))][j-1]);}}int RMQ_max(int L,int R){int k=0;while((1<<(k+1))<=R-L+1)k++;return max(dp_max[L][k],dp_max[R-(1<<k)+1][k]);}int RMQ_min(int L,int R){int k=0;while((1<<(k+1))<=R-L+1)k++;return min(dp_min[L][k],dp_min[R-(1<<k)+1][k]);}int main(){#ifdef zsc    freopen("input.txt","r",stdin);#endifint i;while(~scanf("%d",&n)){for(i=0;i<n;++i){scanf("%d",&v[i]);}RMQ_init();int p=0,ma=0;for(i=1;i<n;++i){while(RMQ_max(p,i)-RMQ_min(p,i)>1 && p<=i ){p++;}ma = max(ma,i-p+1);}printf("%d\n",ma);}return 0;}