ac1019.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 25 行

PAS
25
字号
program tju1019;
var
  n,x,y,i:shortint;
procedure print(x,y:shortint);
  begin
    write('(',x,',',y,')');
  end;
function min(a,b:shortint):shortint;
  begin
    if a<b then min:=a else min:=b;
  end;
function max(a,b:shortint):shortint;
  begin
    if a>b then max:=a else max:=b;
  end;
begin
  repeat
    read(n,x,y);
    for i:=1 to n do print(x,i);writeln;
    for i:=1 to n do print(i,y);writeln;
    for i:=max(1,x-y+1) to min(n,x-y+n) do print(i,y-x+i);writeln;
    for i:=min(n,x+y-1) downto max(1,x+y-n) do print(i,x+y-i);writeln;
  until seekeof;
end.

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