Coderforce 602B 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].

题目大意：n个数的序列，问最长的区间，区间内最大的数和最小的数差小于等于1

ac代码

#include<stdio.h>#include<iostream>#include<algorithm>#include<stdlib.h>#include<string.h>#include<math.h>#define INF 0x3f3f3f3fusing namespace std;int a[100100];int  maxv[100010][20],minv[100010][20];int n;void init(){    int i,j,k;    for(i=1;i<=n;i++)    {        minv[i][0]=a[i];        maxv[i][0]=a[i];    }    for(j=1;(1<<j)<=n;j++)    {        for(k=0;k+(1<<j)-1<=n;k++)        {            minv[k][j]=min(minv[k][j-1],minv[k+(1<<(j-1))][j-1]);            maxv[k][j]=max(maxv[k][j-1],maxv[k+(1<<(j-1))][j-1]);        }    }}int q_max(int l,int r){    int k=(int)(log((double)(r-l+1))/log(2.0));    return max(maxv[l][k],maxv[r-(1<<k)+1][k]);}int q_min(int l,int r){    int k=(int)(log((double)(r-l+1))/log(2.0));    return min(minv[l][k],minv[r-(1<<k)+1][k]);}int main(){   // int n;    while(scanf("%d",&n)!=EOF)    {        int i;        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        init();        int ans=0;        int p=1;        for(i=1;i<=n;i++)        {            while(q_max(p,i)-q_min(p,i)>1&&p<=i)            {                p++;            }            ans=max(ans,i-p+1);        }        printf("%d\n",ans);    }}