C. DZY Loves Sequences

C. DZY Loves Sequences

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

DZY has a sequence a, consisting of
n integers.

We'll call a sequence ai, ai + 1, ..., aj
(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, ..., an (1 ≤ ai ≤ 109).

Output
In a single line print the answer to the problem — the maximum length of the required subsegment.

Sample test(s)

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.

//题意：给你一个数列，你可以改变这个数列其中任意一个数的值（只能改变一个），找出一个最大子段，其中这个子段要严格递增（只要递增就行且不能出现前后相等）

//AC代码

#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<string>
#include<map>
#include<cmath>
const int MAX=100001;
long long xx[MAX];
long long yy[MAX];
using namespace std;
int main()
{
long long n,i,j,m,t,k,As,Ae,Bs,Be,B1,B2,Max,num,a,b,p;
//while(1)
//{
cin>>n;
memset(xx,0,sizeof(xx));
memset(yy,0,sizeof(yy));
As=0;
Ae=0;
Bs=0;
Be=0;
B1=0;
B2=0;
Max=0;
a=1;
b=1;
k=0;
p=0;
cin>>t;
As=t;
xx[a]=t;
a+=1;
for(i=2;i<=n;i++)
{
cin>>m;
//cout<<"OK1"<<i<<endl;
if(t<m)
{
t=m;
xx[a]=m;
a+=1;
}
else
{
Ae=xx[a-1];
As=xx[a-2];
t=m;
yy[b]=m;
b+=1;
break;
}
}
//cout<<"OK"<<endl;
//cout<<As<<" "<<Ae<<" "<<a<<endl;
for(j=i+1;j<=n;j++)
{
cin>>m;
//cout<<m<<" zheshi"<<endl;
//cout<<"OK2 "<<j<<endl;
if(t<m)
{
t=m;
yy[b]=m;
b+=1;
}
else
{
if(j==n)
p=1;
Bs=yy[b-2];
Be=yy[b-1];
B1=yy[1];
B2=yy[2];
//cout<<Bs<<" "<<Be<<" "<<B1<<" "<<B2<<" "<<b<<" acm"<<endl;
if((a-1)==1)
{
if((b-1)==1)
{
num=2;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
else
{
num=b;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
As=Bs;
Ae=Be;
a=b;
b=1;
yy[1]=m;
yy[2]=0;
t=m;
b+=1;
//cout<<As<<" "<<Ae<<" "<<a<<endl;
}
else
{
//cout<<"000"<<endl;
if((b-1)==1)
{
num=a;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
else
{
//cout<<"111"<<endl;
if((B2-Ae>=2)||(B1-As>=2))
{
//cout<<"222"<<endl;
num=a+b-2;
if(Max<=num)
Max=num;
//cout<<As<<" "<<Ae<<" "<<a<<endl;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
else
{
if(a>b)
{
num=a;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
else
{
num=b;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
}
}
As=Bs;
Ae=Be;
a=b;
b=1;
yy[1]=m;
yy[2]=0;
t=m;
b+=1;
}
}
}
//--------------------------------------
if(p==0&&b==1)
{
num=a-1;
if(Max<=num)
Max=num;
}
if(p==0&&b!=1)
{
Bs=yy[b-2];
Be=yy[b-1];
B1=yy[1];
B2=yy[2];
//cout<<As<<" "<<Ae<<" "<<a<<" "<<B1<<" "<<B2<<" "<<b<<" yuweiwei"<<endl;
if((a-1)==1)
{
if((b-1)==1)
{
num=2;
if(Max<=num)
Max=num;
//cout<<"Max="<<Max<<endl;
}
else
{
num=b;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
//cout<<As<<" "<<Ae<<" "<<a<<endl;
}
else
{
//cout<<"000"<<endl;
if((b-1)==1)
{
num=a;
if(Max<=num)
Max=num;
//cout<<"Max="<<Max<<endl;
}
else
{
//cout<<"111"<<endl;
if((B2-Ae>=2)||(B1-As>=2))
{
// cout<<"222"<<endl;
num=a+b-2;
if(Max<=num)
Max=num;
//cout<<"num="<<num<<"Max="<<Max<<endl;
}
else
{
if(a>b)
{
num=a;
if(Max<=num)
Max=num;
//cout<<"Max="<<Max<<endl;
}
else
{
num=b;
if(Max<=num)
Max=num;
// cout<<"Max="<<Max<<endl;
}
}
}
}
}
//cout<<"OKOK"<<endl;
cout<<Max<<endl;
//}
return 0;
}
/*
6
7 7 7 7 7 7
6
7 8 7 7 7 8
6
7 7 8 7 7 8
6
7 3 4 1 6 7

*/