Description：
You are given an array a consisting of
You can remove at most one element from this array. Thus, the final length of the array is 
Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.
Recall that the contiguous subarray a with indices from
Input
The first line of the input contains one integer
The second line of the input contains n integers
Output
Print one integer — the maximum possible length of the strictly increasing contiguous subarray of the array a after removing at most one element.
Examples
input
5 1 2 5 3 4
output
4
input
2 1 2
output
2
input
7 6 5 4 3 2 4 3
output
2 Note In the first example, you can delete
题意： 给出一串序列，可以删除序列中的一个元素，也可以不删，求操作后的最大上升连续子序列的长度。 我们先找到每段上升连续子序列的长度是多少，既然可以删除一个数字，那么要使得最后得上升子序列的长度最大那就删除两段子序列的交界的那个数字，分为两种情况：
- 前一个序列的右端点比后一个序列的左端点大比左端点下一个数字小，就把后一个序列的左端点删除。
- 前一个序列的右端点的前一个数比后一个序列的左端点小，就把前一个序列的右端点删除。 然后扫一遍统计最大的长度。
AC代码：
#include <cstdio>#include <vector>#include <queue>#include <cstring>#include <cmath>#include <map>#include <set>#include <string>#include <iostream>#include <algorithm>#include <iomanip>#include <stack>#include <queue>using namespace std;#define sd(n) scanf("%d", &n)#define sdd(n, m) scanf("%d%d", &n, &m)#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)#define pd(n) printf("%d\n", n)#define pc(n) printf("%c", n)#define pdd(n, m) printf("%d %d", n, m)#define pld(n) printf("%lld\n", n)#define pldd(n, m) printf("%lld %lld\n", n, m)#define sld(n) scanf("%lld", &n)#define sldd(n, m) scanf("%lld%lld", &n, &m)#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)#define sf(n) scanf("%lf", &n)#define sc(n) scanf("%c", &n)#define sff(n, m) scanf("%lf%lf", &n, &m)#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)#define ss(str) scanf("%s", str)#define rep(i, a, n) for (int i = a; i <= n; i++)#define per(i, a, n) for (int i = n; i >= a; i--)#define mem(a, n) memset(a, n, sizeof(a))#define debug(x) cout << #x << ": " << x << endl#define pb push_back#define all(x) (x).begin(), (x).end()#define fi first#define se second#define mod(x) ((x) % MOD)#define gcd(a, b) __gcd(a, b)#define lowbit(x) (x & -x)typedef pair<int, int> PII;typedef long long ll;typedef unsigned long long ull;typedef long double ld;const int MOD = 1e9 + 7;const double eps = 1e-9;const ll INF = 0x3f3f3f3f3f3f3f3fll;const int inf = 0x3f3f3f3f;inline int read(){ int ret = 0, sgn = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') sgn = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { ret = ret * 10 + ch - '0'; ch = getchar(); } return ret * sgn;}inline void Out(int a){ if (a > 9) Out(a / 10); putchar(a % 10 + '0');}ll gcd(ll a, ll b){ return b == 0 ? a : gcd(b, a % b);}ll lcm(ll a, ll b){ return a * b / gcd(a, b);}///快速幂m^k%modll qpow(int m, int k, int mod){ ll res = 1, t = m; while (k) { if (k & 1) res = res * t % mod; t = t * t % mod; k >>= 1; } return res;}// 快速幂求逆元int Fermat(int a, int p) //费马求a关于b的逆元{ return qpow(a, p - 2, p);}///扩展欧几里得int exgcd(int a, int b, int &x, int &y){ if (b == 0) { x = 1; y = 0; return a; } int g = exgcd(b, a % b, x, y); int t = x; x = y; y = t - a / b * y; return g;}///使用ecgcd求a的逆元xint mod_reverse(int a, int p){ int d, x, y; d = exgcd(a, p, x, y); if (d == 1) return (x % p + p) % p; else return -1;}///中国剩余定理模板ll china(int a[], int b[], int n) //a[]为除数，b[]为余数{ int M = 1, y, x = 0; for (int i = 0; i < n; ++i) //算出它们累乘的结果 M *= a[i]; for (int i = 0; i < n; ++i) { int w = M / a[i]; int tx = 0; int t = exgcd(w, a[i], tx, y); //计算逆元 x = (x + w * (b[i] / t) * x) % M; } return (x + M) % M;}const int N = 2e5 + 10;int a[N] , l[N] , r[N];int main(){ int n ; sd(n); rep(i,1,n) sd(a[i]); int now=1,cnt=0; int maxn=0 ; while(now<=n) { int i=now; now++; while(now<=n&&a[now]>a[now-1]) now++; l[cnt]=i; r[cnt]=now-1; cnt++; maxn=max(maxn,now-i); } rep(i,0,cnt-2) { int res=l[i+1]+1; if(a[r[i]]<a[res]) maxn=max(maxn,r[i]-l[i]+1+r[i+1]-l[i+1]); else if(a[r[i]-1]<a[l[i+1]]) maxn=max(maxn ,r[i]-l[i]+r[i+1]-l[i+1]+1); } pd(maxn); return 0;}