int main()

{

long long int n,m,t,x;

cin>>t;

while(t–)

{

cin>>n>>m;

if(n>=m)

{

```
x=(n/2);
cout<<x<<"\n";
}
else
{
x=(m/2);
cout<<x<<"\n";
}
}
return 0;
```

}

int main()

{

long long int n,m,t,x;

cin>>t;

while(t–)

{

cin>>n>>m;

if(n>=m)

{

```
x=(n/2);
cout<<x<<"\n";
}
else
{
x=(m/2);
cout<<x<<"\n";
}
}
return 0;
```

}

1 Like

@chaseme your logic is incorrect.

In Question mentioned " the minimum number of boxes he would have to open in the worst case scenario to infer the position of the box that contains Modak."

so given N*M matrix we will take maximum of (N,M)

Then answer will be simple (int)math.log(max(N,M),2)