TP

Interactive question: add after each output

fflush(stdout);

 

Question meaning: give a set: 44, 88, 1515, 1616, 2323, 4242. Now there is an array, which is an array of sets. Now you can ask judge J I four questions at most, "how much is a[i]*a[j]?" and then judge Ji will tell you the number. After that, you will output the array

Idea: since there are only four numbers and the product of two is unique, you can ask a[1]*a[2], a[1]*a[3], a[4]*a[5], a[4]*a[6]. Where a[1]*a[2] and a[1]*a[3] can determine a[1], a[2] and a[3]. a[4]*a[5] and a[4]*a[6] can determine a[4], a[5] and a[6]. In practice, write a pile of if

In fact, as long as I know to ask "a[1]*a[2], a[1]*a[3], a[4]*a[5], a[4]*a[6].", then we can enumerate all permutations, and the appropriate one is the answer

How to write a pile of if:

#include<bits/stdc++.h>
#define fuck(x) std::cout<<"["<<#x<<"->"<<x<<"]"<<endl;
using namespace std;
typedef long long ll;

const int M=2e5+5;
const int inf=1e9+5;
const int mod=1e9+7;
map<int,pair<int,int> >mp;

int x[]={4 , 8, 15, 16, 23, 42};
int main() {
    for(int i=0;i<6;i++){
        for(int j=i+1;j<6;j++){
                mp[x[i]*x[j]]=make_pair(i,j);
        }
    }
    int a,b,c,d,e,f;
    int ab;
    printf("? 1 2\n");
    fflush(stdout);
    scanf("%d",&ab);
    int _a,_b;
    _a=mp[ab].first,_b=mp[ab].second;
    int ac;
    printf("? 1 3\n");
    fflush(stdout);
    scanf("%d",&ac);
    int __a,_c;
    __a=mp[ac].first,_c=mp[ac].second;
    if(_a==__a){
        a=_a;
        b=_b;
        c=_c;
    }else if(_a==_c){
        a=_a;
        b=_b;
        c=__a;
    }else if(_b==__a){
        a=_b;
        b=_a;
        c=_c;
    }else{
        a=_b;
        b=_a;
        c=__a;
    }

    int de;
    printf("? 4 5\n");
    fflush(stdout);
    scanf("%d",&de);
    int _d,_e;
    _d=mp[de].first,_e=mp[de].second;
    int df;
    printf("? 4 6\n");
    fflush(stdout);
    scanf("%d",&df);
    int __d,_f;
    __d=mp[df].first,_f=mp[df].second;
    if(_d==__d){
        d=_d;
        e=_e;
        f=_f;
    }else if(_d==_f){
        d=_d;
        e=_e;
        f=__d;
    }else if(_e==__d){
        d=_e;
        e=_d;
        f=_f;
    }else{
        d=_e;
        e=_d;
        f=__d;
    }

    printf("! %d %d %d %d %d %d\n",x[a],x[b],x[c],x[d],x[e],x[f]);
    return 0;
}

Enumeration arrangement:

#include<bits/stdc++.h>
#define fuck(x) std::cout<<"["<<#x<<"->"<<x<<"]"<<endl;
using namespace std;
typedef long long ll;

int x[]= {4, 8, 15, 16, 23, 42};
int main() {
    int ab;
    printf("? 1 2\n");
    fflush(stdout);
    scanf("%d",&ab);
    int ac;
    printf("? 1 3\n");
    fflush(stdout);
    scanf("%d",&ac);
    int de;
    printf("? 4 5\n");
    fflush(stdout);
    scanf("%d",&de);
    int df;
    printf("? 4 6\n");
    fflush(stdout);
    scanf("%d",&df);
    do {
        if(x[0]*x[1]==ab&&x[0]*x[2]==ac&&x[3]*x[4]==de&&x[3]*x[5]==df) {
            printf("! %d %d %d %d %d %d\n",x[0],x[1],x[2],x[3],x[4],x[5]);
            break;
        }
    } while(next_permutation(x,x+6));
    return 0;
}