ac1160.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 23 行

program tju1160;
var
  s:string;
  c:char;
  x:extended;
  k,a:longint;
begin
  repeat
    s:='';
    repeat
      read(c);
      if c=' ' then break;
      s:=s+c;
    until false;
    while s[length(s)]='0' do delete(s,length(s),1);

    val(s,x,k);readln(k);
    if x<1 then a:=1 else a:=trunc(ln(x)/ln(10)*k)+1;
    if pos('.',s)>0 then inc(a,(length(s)-pos('.',s))*k);
    writeln(a);
  until seekeof;
end.