[CodeForces - 447C] C - DZY Loves Sequences

Posted 2020-09-29

C - DZY Loves Sequences

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.

Example

Input

6
7 2 3 1 5 6

Output

5

Note

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

 

题目意思是,给出一个序列,只能改变一个数的值,使最长上升子序列的长度最长.

这题其实挺水.我们设f[i]为第i个数左边的最长上升子序列的长度,g[i]为第i个数右边的最长上升子序列的长度,

然后在满足a[i-1]+1<=a[i+1]-1的情况下,求一下max(g[i]+f[i]+1)就行了.然后居然WA了......

后来发现,还有种情况没发现,就是以i结尾或开头的子序列,也有可能是最优的→_→

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 int a[100005],f1[100005],f2[100005],n,ans;
 6 int read(){
 7     int x=0,f=1; char ch=getchar();
 8     while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘) f=-f; ch=getchar();}
 9     while (ch>=‘0‘&&ch<=‘9‘) x=x*10+ch-‘0‘,ch=getchar();
10     return x*f;
11 }
12 int main(){
13     n=read(); for (int i=1; i<=n; i++) a[i]=read();
14     if (n==1){puts("1"); return 0;}
15     if (n==2){puts("2"); return 0;}
16     f1[1]=0,f1[2]=1;
17     for (int i=3; i<=n; i++) if (a[i-1]>a[i-2]) f1[i]=f1[i-1]+1; else f1[i]=1;
18     f2[n]=0,f2[n-1]=1;
19     for (int i=n-2; i>=1; i--) if (a[i+2]>a[i+1]) f2[i]=f2[i+1]+1; else f2[i]=1;
20     ans=max(ans,f2[1]+1); ans=max(ans,f1[n]+1);
21     for (int i=2; i<n; i++){
22         if (a[i-1]+1<=a[i+1]-1) ans=max(ans,f1[i]+f2[i]+1);
23         ans=max(ans,f1[i]+1);
24         ans=max(ans,f2[i]+1);
25     }
26     printf("%d",ans);
27     return 0;
28 }

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