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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].
1 #include<cstdio> 2 #include<vector> 3 #include<cmath> 4 #include<queue> 5 #include<map> 6 #include<cstring> 7 #include<algorithm> 8 using namespace std; 9 typedef long long ll; 10 typedef unsigned long long ull; 11 const int maxn=100005; 12 int main() 13 { 14 int n; 15 int a[maxn],s[maxn]; 16 scanf("%d",&n); 17 for(int i=1;i<=n;i++) 18 scanf("%d",&a[i]); 19 int ans=1; 20 for(int i=1,l=1,r=0;i<=n;i++) 21 { 22 while(l<=r&&a[i]<a[s[r]])r--; 23 s[++r]=i; 24 while(a[s[r]]>a[s[l]]+1)l++; 25 ans=max(ans,s[r]-s[l]+1); 26 } 27 for(int i=1,l=1,r=0;i<=n;i++) 28 { 29 while(l<=r&&a[i]>a[s[r]])r--; 30 s[++r]=i; 31 while(a[s[l]]>a[s[r]]+1)l++; 32 ans=max(ans,s[r]-s[l]+1); 33 } 34 printf("%d\n",ans); 35 return 0; 36 }
codeforces 602B Approximating a Constant Range
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原文地址：http://www.cnblogs.com/homura/p/4995639.html
