codeforce 447C

http://codeforces.com/contest/447/problem/C

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

67 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.

题目大意：这道题是最长公共上升子序列的变形，给定的条件是在可以任意改变其中一个数的大小，当然也可以不改动。

大体思路：我的思路是先找出以每个a[i]为开头和结尾的最长公共子序列的长度，然后依次枚举每个a[i],具体见代码。

#include <stdio.h>#include <iostream>#include <algorithm>#include <string.h>using namespace std;int a[100006];int dp1[100006],dp2[100006];int n;int main(){    while(~scanf("%d",&n))    {        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        int maxn=1;        dp1[1]=1;        for(int i=2;i<=n;i++)        {            if(a[i]>a[i-1])                maxn++;            else                maxn=1;            dp1[i]=maxn;        }        maxn=1;        dp2[n]=1;        for(int i=n-1;i>=1;i--)        {            if(a[i]<a[i+1])                maxn++;            else                maxn=1;            dp2[i]=maxn;        }        int ans=0;        for(int i=1;i<=n;i++)        {            if(i==1)                ans=max(ans,dp2[i]);            else if(i==n)                ans=max(ans,dp1[i]);            else if(a[i+1]-a[i-1]>1)                ans=max(ans,dp1[i-1]+dp2[i+1]);            else                ans=max(ans,max(dp1[i],dp2[i]));        }        if(ans<n)            ans++;        printf("%d\n",ans);    }    return 0;}