CF446A DZY Loves Sequences 简单dp

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本文详细解析了CF446ADZYLovesSequences问题，通过动态规划方法求解最长严格递增子序列，允许修改一个数。介绍了算法思路，包括两个动态规划数组dp1和dp2的定义与更新过程，以及如何通过比较元素来确定子序列长度。

CF446A DZY Loves Sequences 简单dp

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.

Examples

Input

Copy

6
7 2 3 1 5 6

Output

Copy

Note

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

问最多修改一个数字，序列可获得地最大严格递增字段长度为多大；

考虑dp;

dp1 表示以 i 位置结尾的最长子段长度;

dp2 表示以 i 位置开头的最长子段长度；

特判一下当 n=1时，长度为1；

考虑拼接：当 x[ i+1 ]>=2+ x[ i-1 ]时，那么改变 x[ i ]即可拼接子段

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
    ll x = 0;
    char c = getchar();
    bool f = false;
    while (!isdigit(c)) {
        if (c == '-') f = true;
        c = getchar();
    }
    while (isdigit(c)) {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f ? -x : x;
}

ll gcd(ll a, ll b) {
    return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
    if (!b) {
        x = 1; y = 0; return a;
    }
    ans = exgcd(b, a%b, x, y);
    ll t = x; x = y; y = t - a / b * y;
    return ans;
}
*/


int n;
int x[maxn];


int main() {
    //ios::sync_with_stdio(0);
    cin >> n;
    vector<int>dp1(maxn, 1);
    vector<int>dp2(maxn, 1);
    for (int i = 0; i <= n; i++)dp1[i] = dp2[i] = 1;
    for (int i = 0; i < n; i++)rdint(x[i]);
    if (n < 2) {
        cout << 1 << endl; return 0;
    }
    for (int i = 1; i < n; i++)
        dp1[i] = (x[i] > x[i - 1]) ? dp1[i - 1] + 1 : 1;
    for (int i = n - 2; i >= 0; i--)
        dp2[i] = (x[i + 1] > x[i]) ? dp2[i + 1] + 1 : 1;
    int ans = 0;
    for (int i = 1; i < n; i++)ans = max(ans, dp1[i - 1] + 1);
    for (int i = 0; i < n; i++)ans = max(dp2[i + 1] + 1, ans);
    for (int i = 1; i <= n - 1; i++) {
        
        if (x[i + 1] - x[i - 1] >= 2) {
            ans = max(ans, dp1[i - 1] + 1 + dp2[i + 1]);
        }
    }
    cout << ans << endl;
    return 0;
}