My program run in 0.741 seconds and i saw most people only needed
0.09s. How can i solve this any faster?

var d:array[1..50000] of integer;
    n,m,i,j,k:longint;
begin
  fillchar(d,sizeof(d),0);
  readln(n,m);
  for i:=m+1 to n do begin
      k:=i;
      for j:=2 to trunc(sqrt(k)) do
          while k mod j=0 do begin
                inc(d[j]); k:=k div j;
                end;
      if k<>1 then inc(d[k]);
      end;
  for i:=1 to n-m do begin
      k:=i;
      for j:=2 to trunc(sqrt(k)) do
          while k mod j=0 do begin
                dec(d[j]); k:=k div j;
                end;
      if k<>1 then dec(d[k]);
      end;
  k:=0;
  for i:=2 to n do
      if d[i]>0 then inc(k);
  writeln(k);
end.

> My program run in 0.741 seconds and i saw most people only needed
> 0.09s. How can i solve this any faster?
>
> var d:array[1..50000] of integer;
>     n,m,i,j,k:longint;
> begin
>   fillchar(d,sizeof(d),0);
>   readln(n,m);
>   for i:=m+1 to n do begin
>       k:=i;
>       for j:=2 to trunc(sqrt(k)) do
>           while k mod j=0 do begin
>                 inc(d[j]); k:=k div j;
>                 end;
>       if k<>1 then inc(d[k]);
>       end;
>   for i:=1 to n-m do begin
>       k:=i;
>       for j:=2 to trunc(sqrt(k)) do
>           while k mod j=0 do begin
>                 dec(d[j]); k:=k div j;
>                 end;
>       if k<>1 then dec(d[k]);
>       end;
>   k:=0;
>   for i:=2 to n do
>       if d[i]>0 then inc(k);
>   writeln(k);
> end.
>

0.001s, 336kb.
Use following fact from the number theory:
Power of each prime in binomial coefficient's factorization equals to the number of borrows in the substraction n-k in base p. Iterate all primes less then N and count it if prime will greater then 0;