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。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<cmath>
5 #include<algorithm>
6 #define inf 0x3f3f3f3f
7 #define LL long long
8 #define maxn 110000
9 #define IN freopen("in.txt","r",stdin);
10 using namespace std;
11
12 int n;
13 int p[maxn];
14
15 int main(int argc, char const *argv[])
16 {
17 //IN;
18
19 while(scanf("%d",&n)!=EOF)
20 {
21 int ans = -1;
22 memset(p, 0, sizeof(p));
23
24 for(int i=1;i<=n;i++){
25 int x;scanf("%d",&x);
26
27 if(p[x-1]>p[x+1]) ans = max(ans, i-max(p[x+1],p[x-2]));
28 else ans = max(ans, i-max(p[x+2],p[x-1]));
29
30 p[x] = i;
31 }
32
33 printf("%d\n", ans);
34 }
35
36 return 0;
37 }