ac1057.pas

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

program tju1057;
const
  mintime=480;
  maxtime=1260;
var
  count:array[mintime-1..maxtime]of byte;
  n,t,i,j,p,x,y:word;
  s:string;
  no:boolean;
procedure time2num(s:string;var x:word);
  var
    p,h,m:byte;
  begin
    p:=1;
    h:=0;repeat h:=h*10+ord(s[p])-48;inc(p);until s[p]=':';inc(p);
    m:=0;repeat m:=m*10+ord(s[p])-48;inc(p);until s[p]>'9';
    x:=(h mod 12+ord(s[p]='p')*12)*60+m;
  end;
procedure out(x:word);
  var
    h,m:byte;
    s:string[2];
  begin
    h:=x div 60;m:=x mod 60;
    if h<12 then s:='am' else s:='pm';
    h:=h mod 12;if h=0 then h:=12;
    if h<10 then write('0');write(h,':');if m<10 then write('0');write(m,s);
    no:=false;
  end;
begin
  repeat
    fillchar(count,sizeof(count),0);
    readln(n);
    for i:=1 to n do begin
      readln(t);
      for j:=1 to t do begin
        readln(s);
        p:=pos('-',s);
        time2num(copy(s,1,p-1),x);
        time2num(copy(s,p+1,length(s)-p),y);
        for p:=x to y-1 do
          inc(count[p]);
      end;
    end;

    no:=true;
    for i:=mintime to maxtime do
      if (count[i]=n) and (count[i-1]<n) then begin
        out(i);write('-');
      end
      else if (count[i]<n) and (count[i-1]=n) then begin
        out(i);writeln;
      end;
    if no then writeln('No');
  until seekeof;
end.