cf-1272d-Remove One Element

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本文解析了CD-1272D题目的算法解决方案，通过使用两个数组sum和sum1来记录数组中递增段的长度，最终通过模拟删除任意元素操作，找出最大严格递增子段的长度。

题目链接：cd-1272d

题面：
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.

题目大意:
给出数组a，长度为n。有去除其中任意一个数的操作，问得到的严格递增的子段长度为多少。

思路:很多人都是dp写的，我是先用两个数组sum和sum1，z在左右边界下标，存下数组中递增段的长度length。(严格递增段左边界:l，右边界r，sum[l]=sum1[r]=length)。最后模拟一遍删除任意一个数字的操作，得出最大的严格递增子段的长度。

#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+10;
int a[maxn],sum[maxn],sum1[maxn];
int main(){
	int n;
 	scanf("%d",&n);
 	for(int i=0;i<n;i++){
  		scanf("%d",&a[i]);
 	}
 	int ans=1; 
 	sum[0]=1;
 	for(int i=1;i<n;i++){
  		if(a[i]>a[i-1]){
   			sum[i]=sum[i-1]+1;
   			ans=max(ans,sum[i]);
  		}else{
   			sum[i]=1;
   			}
   	}
   	sum1[n-1]=1;
 	for(int i=n-2;i>=0;i--){
  		if(a[i+1]>a[i]){
   			sum1[i]=sum1[i+1]+1;
   			ans=max(ans,sum1[i]);
  		}else{
   			sum1[i]=1;
   			}
	}
	for(int i=1;i<n-1;i++){
  		if(a[i+1]>a[i-1]){
   			ans=max(ans,sum[i-1]+sum1[i+1]);
  		}		
 	}
 	printf("%d\n",ans);
}