题解 | #牛牛的数列#

#include <iostream>
using namespace std;
#define N 100001
int s[N], s1[N], s2[N], s3[N], s4[N], pre[N], nex[N];
int main()
{
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> s[i];
        s1[i] = 1;
        s2[i] = 1;
        s3[i] = 1;
        s4[i] = 1;
        pre[i] = -1;
        nex[i] = -1;
    }
    int maxv = 0;
    
    for (int i = 1; i < n; i++)
    {
        if (s[i] > s[i - 1])
            s1[i] = s1[i - 1] + 1;
        else
        {
            nex[i] = s[i - 1] + 1;
            pre[i - 1] = s[i] - 1;
            s2[i] = s1[i - 1];
            if (i >= 2 && s[i] - 1 > s[i - 2])
                s3[i - 1] = s1[i - 2] + 1;
        }
        if (nex[i - s1[i] + 1] < s[i - s1[i] + 2])
            s4[i] = max(s1[i] + s2[i - s1[i] + 1], s4[i]);
        if (pre[i - s1[i]] < s[i - s1[i] + 1])
            s4[i] = max(s1[i] + s3[i - s1[i]], s4[i]);
        maxv = max(maxv, s4[i]);
        maxv = max(maxv, s2[i]);
    }
    cout<<maxv;
}