Training work:

The first week, basically 8: 00-11: 00, 14:00 - 18:00, 19:00 - 22:00

Training content:

Mainly, training 5 games, 4 cf, a cow

Recommended topic:

E2. Asterism (Hard Version)

Topic address:https://codeforces.com/contest/1371/problem/E2
This question can go to see the inscription, I also see the official question solvinghttps://codeforces.com/blog/entry/79624
First understand this formula

wherein BI indicates a number of times the entire sequence is less than the number equal to I.
When we choose the first number, it must be less than or equal to the number equal to X, that is, B_x
The second number, must be less than or equal to X + 1, ie B_x+1But the previous choosed a number, so only B_x+1 - 1 kind
Pushing in this class, can be available
Set max as the maximum of the entire sequence
$x+n-1<max$x+n−1<maxAt the time, the maximum of the largest, there is no sequence, definitely can't, that is, f (x) = 0, not satisfied
$x\geq max$x≥maxWhen you choose, you can choose, you can f (x) = n!
Other situations are $max-n+1 \leq x<max$max−n+1≤x<max
We only need to meet

The range does not exceed 2E5, so we can record i-b directly_iValue, then directly map

int n,m;
char str[N];
int f[N];
unordered_map<int,int> mm;
int main()
{
    rr(n,m);
    FOR(i,1,n){
        r(f[i]);
    }
    sort(f+1,f+n+1);
    FOR(i,f[n]-n+1,f[n]+n-2){
        mm[i]=(i%m-(upper_bound(f+1,f+n+1,i)-f-1)%m+m)%m;
    }
    unordered_map<int,int> mmm;
    int st=max(0,f[n]-n+1);
    FOR(i,1,n-1){
        int now=st+i-1;
        mmm[mm[now]]++;
    }
    vector<int> ans;
    FOR(i,st,f[n]-1){
        mmm[mm[i+n-1]]++;
        if(mmm.count(i%m)&&mmm[i%m]>0){
//            cout<<i<<endl;
        }
        else{
            ans.pb(i);
        }
        mmm[mm[i]]--;
    }
    cout<<ans.size()<<endl;
    for(int x:ans){
        cout<<x<<' ';
    }
    cout<<endl;
    return 0;
}