Codeforces round # ff (Div. 2)

Another Chinese field is destined to be abused...

A and B: They are all very watery, just like a simulation;

C: CF's rare question is concise,

We can pre-process the number of ascending columns from left to right,

For example: 1 3 2 4 5 7

L [I] Starting from left 1 2 1 2 3 4

Number of descending numbers from right to left: R [I] 2 1 1 3 2 1

We found that for I: A [I-1] <A [I + 1]-1 it is possible to change a [I] to find more numbers,

Ans = max (ANS, max (L [I-1], R [I + 1]) + 1 );

There are a lot of details about hack, so be patient.

Code:

#include<iostream>#include<string>#include<string.h>#include<math.h>#include<algorithm>#include<vector>#include<set>#include<map>#define N 111111using namespace std;int a[N],l[N],r[N];int main(){    int n;    cin>>n;    for (int i=1;i<=n;i++){        cin>>a[i];        l[i]=r[i]=1;      }        int  ans=1;    for (int i=2;i<=n;i++)    {        if (a[i]>a[i-1]) l[i]+=l[i-1];        ans=max(ans,l[i]);    }    for (int i=n-1;i>=1;i--)        if (a[i]<a[i+1]) r[i]+=r[i+1];   l[0]=-1;   r[n+1]=-1;    a[0]=1111111111;    a[n+1]=-1;      for(int i=1;i<=n;i++)      {          if (a[i-1]<a[i+1]-1) ans=max(ans,l[i-1]+r[i+1]+1);          else ans=max(ans,max(r[i+1],l[i-1])+1);      }    cout<<ans<<endl;    return 0;}