function [tnk,dnk]=nik(k,nn,t1);
% [tnk,dnk]=nik(k,nn,t1)
% B樣條曲線的基函數(shù)
% n=nn+2為控制點(diǎn)數(shù)目
% k-1為曲線次數(shù)
n=nn+2;
eps=1.0e-3;
t(1:k)=0.0;

if n>k
   for i=k+1:n;
      t(i)=(i-k)/(n-(k-1));
   end
end
t(n+1:n+k)=1.0;


for i=1:n+k
   tn(i,1)=0.0;
end

for i=k:n
   if and((t(i+1)-t1)>eps,(t1-t(i))>eps)    
      tn(i,1)=1.0;
   elseif(abs(t1-t(i))<eps)
      tn(i,1)=1.0;
   else
      tn(i,1)=0.0;
   end
end

   if abs(t1)<eps
      tn(k,1)=1.0;
   elseif abs(t1-1.0)<eps
      tn(n,1)=1.0;
   end

for kk=2:k
   for i=1:n+k-kk
      if (t(i+kk-1)-t(i))<eps
         delt1=0.0;
      else
         delt1=(t1-t(i))/(t(i+kk-1)-t(i));
      end
      if (t(i+kk)-t(i+1))<eps
         delt2=0.0;
      else
         delt2=(t(i+kk)-t1)/(t(i+kk)-t(i+1));
      end
      tn(i,kk)=tn(i,kk-1)*delt1+tn(i+1,kk-1)*delt2;
      
      if kk==k
         if(t(i+k-1)-t(i))<eps
            part1=0.0;
         else
            part1=tn(i,kk-1)/(t(i+k-1)-t(i));
         end
         if(t(i+k)-t(i+1))<eps
            part2=0.0;
         else
            part2=tn(i+1,kk-1)/(t(i+k)-t(i+1));
         end
            dn(i,kk)=(k-1)*(part1-part2);
         end
      end
     
end
tnk=tn(1:n,k);
dnk=dn(1:n,k);