C. DZY Loves Sequences

C. 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.

本题坑还算比較多的,我用的递推,dp1[i]表示以i结尾的递增序列的长度,dp2[i]表示已i開始的递增序列的长度,要找出最长的仅仅改变一个数就能构成递增序列的序列长度仅仅须要枚举i即可了,代码例如以下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
typedef long long LL;
using namespace std;
int A[100005];
int dp1[100005],dp2[100005];
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        memset(dp1,0,sizeof(dp1));
        memset(dp2,0,sizeof(dp2));
        scanf("%d",A+1);
        dp1[1]=1;
        for(int i=2;i<=n;i++)
        {
            scanf("%d",A+i);
            if(A[i]>A[i-1])dp1[i]=dp1[i-1]+1;
            else dp1[i]=1;
        }
        dp2[n]=1;
        for(int i=n-1;i>=1;i--)
        {
            if(A[i]<A[i+1])dp2[i]=dp2[i+1]+1;
            else dp2[i]=1;
        }
        int m=1;
        A[0]=0;
        A[n+1]=1000000005;
        for(int i=1;i<=n;i++){
                int t=dp1[i-1]+dp2[i+1]+1;
                if(A[i+1]-A[i-1]<2)t=max(dp1[i],dp2[i])+1;
                m=max(m,t);
        }
        cout<<m<<endl;
    }
    return 0;
}