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.

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].

我有点惊讶。。B题就考dp了，虽然是简单的dp

题意是说给一个序列，保证相邻两项差值不超过1，求一个最长子串长度，要求满足子串中最大值减最小值小于2

意思就是串中只能有相邻的两个数字咯

令f[i][1]表示以第i个数开头，只包含a[i]和a[i]+1两种数字的最长子串

令f[i][2]表示以第i个数开头，只包含a[i]和a[i]-1两种数字的最长子串

然后

a[i]==a[i+1] 则 f[i][1]=f[i+1][1] f[i][2]=f[i+1][2]

a[i]==a[i+1]+1 则 f[i][1]=1 f[i][2]=f[i+1][1]+1

a[i]==a[i+1]-1 则  f[i][2]=1 f[i][1]=f[i+1][2]+1

 #include<cstdio>

 #include<iostream>

 #include<cstring>

 #include<cstdlib>

 #include<algorithm>

 #include<cmath>

 #include<queue>

 #include<deque>

 #include<set>

 #include<map>

 #include<ctime>

 #define LL long long

 #define inf 0x7ffffff

 #define pa pair<int,int>

 #define pi 3.1415926535897932384626433832795028841971

 using namespace std;

 inline LL read()

 {

     LL x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 inline void write(LL a)

 {

     if (a<){printf("-");a=-a;}

     if (a>=)write(a/);

     putchar(a%+'');

 }

 inline void writeln(LL a){write(a);printf("\n");}

 int n,ans;

 int a[];

 int s1[];

 int s2[];

 int main()

 {

     n=read();

     for (int i=;i<=n;i++)a[i]=read();

     s1[n]=s2[n]=ans=;

     for (int i=n-;i>=;i--)

     {

         if (a[i]==a[i+])s1[i]=s1[i+]+,s2[i]=s2[i+]+;

         if (a[i]>a[i+])s1[i]=,s2[i]=s1[i+]+;

         if (a[i]<a[i+])s2[i]=,s1[i]=s2[i+]+;

         ans=max(ans,max(s1[i],s2[i]));

     }

     printf("%d\n",ans);

 }