Codeforces 602B Approximating a Constant Range（想法题）

B. Approximating a Constant Range

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

来自 <http://codeforces.com/contest/602/problem/B>

Codeforces Round #333 (Div. 2)

【题意】：

n个数，相邻数的差不超过1.

求最长的区间，使得极差不超过1.

【解题思路】：

对于X，包含X的合法区间需要考虑X-1 X+1 X+2 X-2的位置：

用数组P[i]记录下至此 i 出现的最大位置；

若X-1的最大位置大于X+1的，则考虑X+1和X-2的位置即可；

相反，只需要考虑X-1和X+2的位置。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #define inf 0x3f3f3f3f
 #define LL long long
 #define maxn 110000
 #define IN freopen("in.txt","r",stdin);
 using namespace std;

 int n;
 int p[maxn];

 int main(int argc, char const *argv[])
 {
     //IN;

     while(scanf("%d",&n)!=EOF)
     {
         int ans = -;
         memset(p, , sizeof(p));

         for(int i=;i<=n;i++){
             int x;scanf("%d",&x);

             if(p[x-]>p[x+]) ans = max(ans, i-max(p[x+],p[x-]));
             else ans = max(ans, i-max(p[x+],p[x-]));

             p[x] = i;
         }

         printf("%d\n", ans);
     }

     return ;
 }