cf602B Approximating a Constant Range

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B. Approximating a Constant Rangetime limit per test 2 secondsmemory limit per test 256 megabytesinput standard inputoutput standard output

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.

Examplesinput

5
1 2 3 3 2

output

4

input

11
5 4 5 5 6 7 8 8 8 7 6

output

5

Note

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].

 

我有點驚訝。。B題就考dp了，雖然是簡單的dp

題意是說給一個序列，保證相鄰兩項差值不超過1，求一個最長子串長度，要求滿足子串中最大值減最小值小於2

意思就是串中只能有相鄰的兩個數字咯

令f[i][1]表示以第i個數開頭，只包含a[i]和a[i]+1兩種數位最長子串

令f[i][2]表示以第i個數開頭，只包含a[i]和a[i]-1兩種數位最長子串

然後

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

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

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

 

 1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set>10 #include<map>11 #include<ctime>12 #define LL long long13 #define inf 0x7ffffff14 #define pa pair<int,int>15 #define pi 3.141592653589793238462643383279502884197116 using namespace std;17 inline LL read()18 {19     LL x=0,f=1;char ch=getchar();20     while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}21     while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}22     return x*f;23 }24 inline void write(LL a)25 {26     if (a<0){printf("-");a=-a;}27     if (a>=10)write(a/10);28     putchar(a%10+‘0‘);29 }30 inline void writeln(LL a){write(a);printf("\n");}31 int n,ans;32 int a[100010];33 int s1[100010];34 int s2[100010];35 int main()36 {37     n=read();38     for (int i=1;i<=n;i++)a[i]=read();39     s1[n]=s2[n]=ans=1;40     for (int i=n-1;i>=1;i--)41     {42         if (a[i]==a[i+1])s1[i]=s1[i+1]+1,s2[i]=s2[i+1]+1;43         if (a[i]>a[i+1])s1[i]=1,s2[i]=s1[i+1]+1;44         if (a[i]<a[i+1])s2[i]=1,s1[i]=s2[i+1]+1;45         ans=max(ans,max(s1[i],s2[i]));46     }47     printf("%d\n",ans);48 }

cf602B

 

cf602B Approximating a Constant Range