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
5
1 2 3 3 2
4
11
5 4 5 5 6 7 8 8 8 7 6
5
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].
思路：
一开始各种想LIS，但是CF再次证明了前两道题目全是想法题（虽然想法并不比算法简单= =~）
首先是要在一整个数段中找到其中的一个子数段，在它的M-m<=1的情况下使其len最大
那么这里就涉及到两个思维的关键点：
（1）不同状态之间的转换：为每个状态设置一个M和m，对于每一个后进入的点判断是否可以继续维持前一状态，如果可以就count++
（2）要意识到新状态的开始点并不一定是旧状态的结束点：可能有一点使得原来的状态不能够持续下去而结束了，但是并不意味着这一点就是新状态的开始点，在维持旧状态的过程中可能就出现了新状态的开始点，这点主要到了以后只要根据不同的情况去找那个开始点就OK了
#include <iostream>
using namespace std; int m,M,n;
int num[]; bool ok(int t)
{
if(m==M) {
if(t==m)
return true;
else if(t == m-||t == m+)
return true;
else
return false;
}
else {
if(t==M||t==m)
return true;
else
return false;
}
} int main()
{
while(cin>>n)
{
cin>>num[];
m = num[];
M = num[];
int count = ;
int ans = ;
for(int i = ;i <= n;i++)
{
cin>>num[i];
if(ok(num[i]))
{
if(m==M && num[i] == m-)
m = num[i];
else if(m==M && num[i] == m+)
M = num[i];
count++;
}
else {
if(m == M) {
M = m = num[i];
count = ;
}
else {
int pos;
if(num[i] == num[i-]+) {
M = num[i];
m = num[i-];
for(int j = i-;j >= ;j--)
{
if(num[j] != num[i-])
break;
pos = j;
}
count = i-pos+;//此时的count应该=当前的坐标-s段开始的坐标
}
else if(num[i] == num[i-]-) {
M = num[i-];
m = num[i];
for(int j = i-;j >= ;j--)
{
if(num[j] != num[i-])
break;
pos = j;
}
count = i-pos+;//同上
}
else {
M = m = num[i];
count = ;
}
}
}
ans = max(ans,count);
}
cout<<ans<<endl;
}
return ;
}