cf446A DZY Loves Sequences

A. DZY Loves Sequences

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

DZY has a sequence a, consisting of n integers.

We'll call a sequence ai, ai + 1, ..., aj (1 ≤ i ≤ j ≤ n) a subsegment of the sequence a. The value (j - i + 1) denotes the length of the subsegment.

Your task is to find the longest subsegment of a, such that it is possible to change at most one number (change one number to any integer you want) from the subsegment to make the subsegment strictly increasing.

You only need to output the length of the subsegment you find.

Input

The first line contains integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In a single line print the answer to the problem — the maximum length of the required subsegment.

Sample test(s)

input

6
7 2 3 1 5 6

output

5

Note

You can choose subsegment a2, a3, a4, a5, a6 and change its 3rd element (that is a4) to 4.

题意是有一串序列，可以把其中的一个数修改成任意值，求这样改完以后最长上升子序列长度

三维dp:前i个、是否用掉第i个、是否和以前的相连的最长长度

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#define inf 0x7fffffff
#define ll long long
using namespace std;
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,ans;
int a[100005],f[100005][2][2];//used? last?
int main()
{
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    for(int i=1;i<=n;i++)
    {
    	f[i][0][0]=f[i][1][1]=1;
		if(i!=1)f[i][1][0]=2;
        if(a[i]>a[i-1])
		{
		    f[i][0][0]=max(f[i][0][0],f[i-1][0][0]+1);
		    f[i][1][0]=max(f[i][1][0],f[i-1][1][0]+1);
		    f[i][1][1]=max(f[i][1][1],f[i-1][0][0]+1);
		}
		else 
		{
			f[i][1][1]=max(f[i][1][1],f[i-1][0][0]+1);
		}
		if(i!=1&&a[i]>a[i-2]+1)f[i][1][0]=max(f[i][1][0],f[i-1][1][1]+1);
		ans=max(ans,f[i][0][0]);
		ans=max(ans,f[i][1][0]);
		ans=max(ans,f[i][1][1]);
	}
	printf("%d",ans);
}