cf goodbye2019 d

Only the first inquiry $k+1$ the number, ask $k+1$ st, $i$ once abandoned are $i$ elements, before assuming $k+1$ rear element array is sorted $b_{1}<b_{2}<...<b_{k+1}$For the former $m$ elements, $pos$ and $a_{pos}$They are $b_{m+1}$,and so $b_{m+1}$Appeared in a total $m$ times, $b_{m}$Appeared in a total $k+1-m$ times. thus, $m$ readily available.

#include<bits/stdc++.h>
using namespace std;
int main(){
    int times1=0,times2=0;
    int n,k;
    cin>>n>>k;
    int pos1,pos2;
    int num1,num2;
    cout<<"?";
    for(int i=2;i<=k+1;i++){
        cout<<' '<<i;
    }
    cout<<endl;
    cin>>pos1>>num1;
    times1++;
    fflush(stdout);
    for(int i=2;i<=k+1;i++){
        cout<<"?";
        for(int j=1;j<=k+1;j++){
            if(j!=i) cout<<" "<<j;
        }
        cout<<endl;
        int tmp_pos,tmp_num;
        cin>>tmp_pos>>tmp_num;
        if(tmp_pos==pos1) times1++;
        else {times2++;pos2=tmp_pos;num2=tmp_num;}
        fflush(stdout);
    }
    if(num1>num2){
        cout<<"! "<<times1<<'\n';
    }else {
        cout<<"! "<<times2<<'\n';
    }
}