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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.
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).
Print a single number — the maximum length of an almost constant range of the given sequence.
5
1 2 3 3 2
4
11
5 4 5 5 6 7 8 8 8 7 6
5
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 }