#include
using namespace std;
int possible_tiling(int n)
{
if(n<=4)
{
return n;
}
return possible_tiling(n-1) + possible_tiling(n-4);
}
int main()
{
int n;
cin>>n;
cout<<possible_tiling(n);
return 0;
}
#include
using namespace std;
int possible_tiling(int n)
{
if(n<=4)
{
return n;
}
return possible_tiling(n-1) + possible_tiling(n-4);
}
int main()
{
int n;
cin>>n;
cout<<possible_tiling(n);
return 0;
}
hello @sroy3166
if u have taken ur tile size as 1X4.then ur recurrence relation is correct.
but some changes are needed in base case.
it should be
if(n<4)
return 1;
if(n==4)
return 2;
if(n<4) we can place n tiles size(4X1)
yeah and that is the only one way.