Codeforces Round #FF (Div. 2)：C. DZY Loves Sequences

最新推荐文章于 2022-12-16 16:19:21 发布

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最新推荐文章于 2022-12-16 16:19:21 发布

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本文探讨了一个问题：在给定的一系列整数中，找到一个最长的子序列，通过调整其中的一个元素，使得该子序列成为严格递增序列。通过输入一系列整数作为序列，并输出调整后的最长子序列长度。

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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

Note

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

题意就是，你能够改变字符串中的一个字符。就出其最长的连续字串

如案列， 7 2 3 1 5 6 —————7 2 3 4 5 6

输出即为5.



#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<sstream>
#include<cmath>

using namespace std;

#define f1(i, n) for(int i=0; i<n; i++)
#define f2(i, m) for(int i=1; i<=m; i++)
#define f3(i, n) for(int i=n; i>=1; i--)
#define f4(i, n) for(int i=1; i<=n; i++)
#define f5(i, n) for(int i=2; i<=n; i++)
#define M 1005

const int INF = 0x3f3f3f3f;
int n, a[100005], b[100005];

int main()
{
    cin>>n;
    f4(i, n)
    cin>>a[i];
    b[1]=1;
    f5(i, n)
    {
        b[i]=1;
        if (a[i]>a[i-1])
            b[i]=b[i-1]+1;
    }
    int ans=-INF;
    f3(i, n)
    {
        if (b[i]==n)
            ans=max(b[i], ans);
        else
            ans=max(ans, b[i]+1);
        if (a[i-b[i]+1]-1>a[i-b[i]-1])
            ans=max(ans,b[i]+b[i-b[i]]);
        if (a[i-b[i]+2]-1>a[i-b[i]])
            ans=max(ans,b[i]+b[i-b[i]]);
    }
    cout<<ans<<endl;
    return 0;
}