题设有两个要求：

Do not use any built-in date functions in your computer language.

Don't just precompute the answers, either, please. 

没理解，不管怎么样算是练基本功吧打了好长个程序。

procedure leapyear用于计算当前年是闰年。

procedure unleapyear用于计算当前年非闰年。

start表示当前年Jan. 1st是星期几。

注意一个小细节就是输出顺序：Saturday, Sunday, Monday, Tuesday, ..., Friday.

const day1:array[1..12] of integer=(31,29,31,30,31,30,31,31,30,31,30,31);      day2:array[1..12] of integer=(31,28,31,30,31,30,31,31,30,31,30,31);var a:array[1..7] of longint;    n,k,start,i:integer;procedure leapyear(s:integer);var i:integer;begin  s:=(s+12-1) mod 7+1;  inc(a[s]);  for i:=1 to 11 do    begin      s:=(s+day1[i]-1) mod 7+1;      inc(a[s]);    end;end;procedure unleapyear(s:integer);var i:integer;begin  s:=(s+12-1) mod 7+1;  inc(a[s]);  for i:=1 to 11 do    begin      s:=(s+day2[i]-1) mod 7+1;      inc(a[s]);    end;end;begin  assign(input,'friday.in');reset(input);  assign(output,'friday.out');rewrite(output);  readln(n);  k:=0;  start:=1;  while k
      
       0)) or ((k+1900) mod 400=0) then        begin          leapyear(start);          start:=(start+366-1) mod 7+1;        end      else        begin          unleapyear(start);          start:=(start+365-1) mod 7+1;        end;      inc(k);    end;  writeln(a[6],' ',a[7],' ',a[1],' ',a[2],' ',a[3],' ',a[4],' ',a[5]);  close(input);close(output);end.