codeforces#FF(div2) DZY Loves Sequences

n个数，能够随意改变当中一个数，求最长的上升子区间长度

思路：记录一个from[i]表示从位置i的数開始最长的上升区间长度

    记录一个to[i]表示到位置i的数所能达到的最长上升区间长度

枚举要改变的数的位置i，此时能达到的长度为to[i - 1] + from[i + 1] + 1，取最大值

//#pragma comment(linker, "/STACK:102400000,102400000")
//HEAD
#include <cstdio>
#include <cstring>
#include <vector>
#include <iostream>
#include <algorithm>

#include <queue>
#include <string>
#include <set>
#include <stack>
#include <map>
#include <cmath>
#include <cstdlib>

using namespace std;
//LOOP
#define FE(i, a, b) for(int i = (a); i <= (b); ++i)
#define FED(i, b, a) for(int i = (b); i>= (a); --i)
#define REP(i, N) for(int i = 0; i < (N); ++i)
#define CLR(A,value) memset(A,value,sizeof(A))
//STL
#define PB push_back
//INPUT
#define RI(n) scanf("%d", &n)
#define RII(n, m) scanf("%d%d", &n, &m)
#define RIII(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define RS(s) scanf("%s", s)
typedef long long LL;
const int INF = 0x3f3f3f3f;
const int MAXN = 1010;


#define FF(i, a, b) for(int i = (a); i < (b); ++i)
#define FD(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define CPY(a, b) memcpy(a, b, sizeof(a))
#define FC(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); it++)
#define EQ(a, b) (fabs((a) - (b)) <= 1e-10)
#define ALL(c) (c).begin(), (c).end()
#define SZ(V) (int)V.size()
#define RIV(n, m, k, p) scanf("%d%d%d%d", &n, &m, &k, &p)
#define RV(n, m, k, p, q) scanf("%d%d%d%d%d", &n, &m, &k, &p, &q)
#define WI(n) printf("%d\n", n)
#define WS(s) printf("%s\n", s)
typedef vector <int> VI;
typedef unsigned long long ULL;
const double eps = 1e-10;
const LL MOD = 1e9 + 7;

int a[100010];
int to[100010], from[100010];

int main()
{
    //freopen("0.txt", "r", stdin);
    int n;
    while (~RI(n))
    {
        FE(i, 1, n) RI(a[i]), to[i] = from[i] = 1;
        FE(i, 2, n)
            if (a[i] > a[i - 1])
                to[i] = to[i - 1] + 1;
        FED(i, n - 1, 1)
            if (a[i + 1] > a[i])
                from[i] = from[i + 1] + 1;
//        FE(i, 1, n)
//            cout << to[i] <<‘ ‘;;cout << endl;
//        FE(i, 1, n)
//            cout << from[i] <<‘ ‘ ; cout << endl;
        int ans = 1;
        if (n > 1)
            ans = max(from[2] + 1, to[n - 1] + 1);
        FE(i, 2, n - 1)
        {
            if (a[i + 1] - a[i - 1] >= 2)
            {
                if (to[i - 1] + from[i + 1] + 1 > ans)
                {
                    ans = to[i - 1] + from[i + 1] + 1;
//                    cout << i <<endl;
                }
            }
            else
            {
                ans = max(ans, to[i - 1] +1);
                ans = max(ans, from[i + 1] + 1);
            }

        }
        cout << ans << endl;
    }
    return 0;
}