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;
}
@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)