codeforce -602B Approximating a Constant Range(暴力)

B. Approximating a Constant Range

time limit per test

2 seconds

memory limit per test

256 megabytes

 

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].

题意是，找连续的并且任意两个数相差不超过1的最长串。

思路：题中说相邻的两个数相差不超过1；

那么cnt最小为2，cnt赋初值2；由于要相差不超过一，所以每个串的最大值最小值相差不能超过一，

那么从第三个元素开始，如果abs(a[i]-max)<=1&&abs(a[i]-min)<=1，就cnt++,表示该元素能加入上一个串，因为有新数字加入，所以更新max,和minn.

如果不符合的话cnt置为2就以a[i],开始向前找串。

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 #include<string.h>
 6 using namespace std;
 7 int a[100005];
 8 int main(void)
 9 {
10     int n,i,k,p,q,j;
11     while(scanf("%d",&n)!=EOF)
12     {
13         for(i=0; i<n; i++)
14         {
15             scanf("%d",&a[i]);
16         }
17         int maxx,minn;
18         maxx=max(a[0],a[1]);
19         minn=min(a[0],a[1]);
20         int cnt=2;
21         int sum=2;
22         int x=maxx,y=minn;
23         for(i=2; i<n; i++)
24         {
25             if(abs(a[i]-maxx)<=1&&abs(a[i]-minn)<=1)
26             {
27                 cnt++;
28                 maxx=max(maxx,a[i]);
29                 minn=min(a[i],minn);
30             }
31             else
32             {
33                 cnt=2;
34                 maxx=max(a[i],a[i-1]);
35                 minn=min(a[i],a[i-1]);
36                 for(j=i-2; j>=0; j--)//从后往前找
37                 {
38                     if(abs(a[j]-maxx)<=1&&abs(a[j]-minn)<=1)
39                     {
40                         cnt++;
41                         maxx=max(maxx,a[j]);
42                         minn=min(minn,a[j]);
43                     }
44                     else break;
45                 }
46             }
47             if(cnt>sum)
48             {
49                 sum=cnt;
50             }
51         }
52         printf("%d\n",sum);
53     }
54     return 0;
55 }