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Description
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 Input
Hint
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].
读入数据时如果当前的和之前的相同则记录，处理时维护两个数字，题意的一个序列里最多两个不同数字，如果不符合就跳出，符合则判断下一个（j=b[j])
#include<stdio.h>
#include<algorithm>
using namespace std;
long long n,ans,len,cont,num1,num2=100005,a[100005],b[100005];
int main()
{
scanf("%lld",&n);
int i,j;
for(i=1; i<=n; i++)
{
scanf("%lld",&a[i]);
if(a[i]==a[i-1])
{
if(cont==0) //cont=0前一个不重复
cont=b[i-1];
b[i]=cont;
}
else
b[i]=i-1;
cont=0;
}
for(i=n; i>0; i--)
{
num1=a[i];
num2=100005;
len=1;
j=i-1;
while(j>0)
{
if(num1==a[j]||num2==a[j])
{
len+=j-b[j];
j=b[j];
continue;
}
else if(num2==100005)
{
num2=a[j];
len+=j-b[j];
j=b[j];
}
else break;
}
ans=max(ans,len);
if(i<ans)break;
}
printf("%lld\n",ans);
return 0;
}
【Virtual Judge】G - 一般水的题2-Approximating a Constant Range
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原文地址：http://www.cnblogs.com/flipped/p/5046941.html
