ac1103.pas

来自「同济大学 Online在线题库 AC源代码合集 程序设计竞赛必看资料」· PAS 代码 · 共 29 行

program tju1103;
const
  primes=10;
  prime:array[1..primes]of byte=(2,3,5,7,11,13,17,19,23,29);
    //The last prime is never used, but necessary
var
  n,best,bcount:longint;
procedure search(now:int64;count,p,max:word);
  var
    k:byte;
  begin
    if (count>bcount) or (count=bcount) and (now<best) then begin
      best:=now;bcount:=count;
    end;
    for k:=1 to max do begin
      now:=now*prime[p];
      if now>n then exit;
      search(now,count*(k+1),p+1,k);
    end;
  end;
begin
  repeat
    bcount:=0;
    readln(n);
    search(1,1,1,99);
    writeln(best);
  until seekeof;
end.