You are given an array aa consisting of nn integers.

You can remove at most one element from this array. Thus, the final length of the array is n?1n?1 or nn.

Your task is to calculate the maximum possible length of the strictly increasing contiguous subarray of the remaining array.

Recall that the contiguous subarray aa with indices from ll to rr is a[l…r]=al,al+1,…,ara[l…r]=al,al+1,…,ar. The subarray a[l…r]a[l…r] is called strictly increasing if al<al+1<?<aral<al+1<?<ar.

Input
The first line of the input contains one integer nn (2≤n≤2?1052≤n≤2?105) — the number of elements in aa.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the ii-th element of aa.

Output
Print one integer — the maximum possible length of the strictly increasing contiguous subarray of the array aa 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 a3=5a3=5. Then the resulting array will be equal to [1,2,3,4][1,2,3,4] and the length of its largest increasing subarray will be equal to 44.
思路：最多刪除一個，那么我們就記錄這一個數(shù)字前一個數(shù)字和后一個數(shù)字的最長連續(xù)序列。然后貪心的取最大值。事先用dfs預處理好每一個數(shù)字最遠可以到達哪里。具體看代碼。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=2e5+100; int a[maxx]; int dp[maxx]; int np[maxx]; int n;inline void dfs(int &i,int &cnt) {if(i==n) {dp[n]=1;i+=1;return ;}int ant=cnt,j=i;if(a[i+1]>a[i]) dfs(i+=1,cnt+=1);else i+=1;dp[j]=cnt-ant+1; } inline void dfs1(int &i,int &cnt) {if(i==1){np[i]=1;i-=1;return ;}int ant=cnt,j=i;if(a[i-1]<a[i]) dfs1(i-=1,cnt+=1);else i-=1;np[j]=cnt-ant+1; } int main() {scanf("%d",&n);int _max=0;a[n+1]=a[0]=0;for(int i=1;i<=n;i++) {scanf("%d",&a[i]);}int cnt;for(int i=1;i<=n;) dfs(i,cnt=0);//dp[i]代表的是第i位開始，最長的上升序列,i++for(int i=n;i>=1;) dfs1(i,cnt=0);//np[i]代表的是第i位開始，最長的下降序列,i--for(int i=1;i<=n;i++) _max=max(_max,dp[i]);for(int i=2;i<n;i++) if(a[i-1]<a[i+1]) _max=max(_max,np[i-1]+dp[i+1]);//貪心的取最大值cout<<_max<<endl;return 0; }//1 2 3 4 5 6

努力加油a啊，(^o)/~

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