#include
using namespace std;
int tilling(int n,int m)
{
if(n==1 || n==0)
{
return n;
}
if(m==0 || m==1)
{
return m;
}
int ans=tilling(n-1,m);
int ans1=tilling(n-m,m-1);
int ans2=ans+ans1;
return ans2;
}
int main() {
int t;
cin>>t;
int q=1;
int n,m;
while(q<=t)
{
cin>>n>>m;
int ans=tilling(n,m);
cout<<ans<<endl;
q++;
}
return 0;
}
this solution will give you tle as there are two recursion calls at every level, so in worst case there will be 2^n recursion calls. change you solution and use bottom-up dynamic programming
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