【32.22%】【codeforces 602B】Approximating a Constant Range

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard 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.

Examples

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].

【题目链接】:http://codeforces.com/contest/602/problem/B

【题解】



用ST算法先搞出各个区间的最大、最小值

然后枚举区间的左端点、二分右端点.

判断依据就是最大值和最小值的差;

如果小于等于1则右端点可以再往右点。否则往左

一般用二分都能过吧。



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n;
int a[MAXN];
int ma[MAXN][18],mi[MAXN][20];
int pre2[20];
int need[MAXN];

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]),mi[i][0] = ma[i][0] = a[i];
    pre2[0] = 1;
    for (int i = 1; i <= 18; i++)
        pre2[i] = pre2[i - 1] << 1;
    need[1] = 0; need[2] = 1;
    int temp = 2;
    for (int i = 3; i <= n; i++)
        if (pre2[temp] == i)
            need[i] = need[i - 1] + 1, temp++;
        else
            need[i] = need[i - 1];
    for (int l = 1; pre2[l] <= n; l++)
        for (int i = 1;i <= n;i++)
            if (i + pre2[l] - 1 <= n)
            {
                ma[i][l] = max(ma[i][l - 1], ma[i + pre2[l - 1]][l - 1]);
                mi[i][l] = min(mi[i][l - 1], mi[i + pre2[l - 1]][l - 1]);
            }
    int ans = 0;
    rep1(i,1,n)
    {
        int l = i,r = n,tempr=l;
        while (l <= r)
        {
            int m = (l+r)>>1;
            int le = need[m-i+1];
            int mi1 = min(mi[i][le],mi[m-pre2[le]+1][le]);
            int ma1 = max(ma[i][le],ma[m-pre2[le]+1][le]);
            if (abs(mi1-ma1)<=1)
            {
                tempr = m;
                l = m+1;
            }
            else
                r = m-1;
        }
        if(tempr-i+1> ans)
            ans = tempr-i+1;
    }
    cout << ans << endl;
    return 0;
}
close