Codeforces Round #FF (Div. 2)446A - DZY Loves Sequences（最长上升子序列变形）

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本文介绍了一种算法，用于解决寻找最长子序列的问题，该子序列可以通过改变一个数变为严格递增。文章提供了一个AC代码示例，展示了如何通过预处理前后缀的最大递增长度来高效地解决这个问题。

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DZY has a sequence a, consisting of n integers.

We'll call a sequence a_i, a_i + 1, ..., a_j (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, ..., a_n (1 ≤ a_i ≤ 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.

AC代码：

#include <stdio.h>
#include <string.h>
#define Max 100100
#define ll long long

#define max(a,b) a>b?a:b;
ll a[Max],l[Max],r[Max];

int main(){
	ll n,p,i,j,ans;
	while(~scanf("%lld",&n)){
		memset(a,0,sizeof(a));
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
		ans=1;
		for(i=0;i<n;i++)
			scanf("%lld",&a[i]);
		if(n==1)
			printf("1\n");
		else if(n==2)
			printf("2\n");
		else{
			l[0]=1;
			for(i=1;i<n;i++){
				if(a[i]>a[i-1])
					l[i]=l[i-1]+1;
				else
					l[i]=1;
			}
			r[n-1]=1;
			for(i=n-2;i>=1;i--){
				if(a[i+1]>a[i])
					r[i]=r[i+1]+1;
				else
					r[i]=1;
			}
			for(i=n-2;i>=1;i--){
				if(a[i-1]+1<a[i+1])
					ans=max(ans,l[i-1]+1+r[i+1]);
			}
			for(i=1;i<n;i++){
				if(a[i]<=a[i-1])
					ans=max(ans,l[i-1]+1);
			}
			for(i=0;i<n-1;i++){
				if(a[i]>=a[i+1])
					ans=max(ans,r[i+1]+1);
			}
			printf("%lld\n",ans);
		}
	}
	return 0;
}