Нужна очень решить Решить задачи на темы "циклы с условием" и "циклы с переменной"

Varm: set of byte;u,s,r,a,p,e,c: integer;beginm := [];for u := 1 to 6 doif not (u in m) thenbeginInclude(m, u);for s := 0 to 6 doif not (s in m) thenbeginInclude(m, s);for r := 0 to 6 doif not (r in m) thenbeginInclude(m, r);for a := 0 to 6 doif not (a in m) thenbeginInclude(m, a);for p := 1 to 6 doif not (p in m) thenbeginInclude(m, p);for e := 0 to 6 doif not (e in m) thenbeginInclude(m, e);for c := 0 to 6 doif not (c in m) thenbeginInclude(m, c);if u * 1000 + s * 100 + s * 10 + r +u * 100 + s * 10 + a =p * 10000 + e * 1000 + a * 100 + c * 10 + e thenWriteLn(u, s, s, r, ' + ', u, s, a, ' = ', p, e, a, c, e);Exclude(m, c)end;Exclude(m, e);end;Exclude(m, p);end;Exclude(m, a)end;Exclude(m, r)end;Exclude(m, s)end;Exclude(m, u)end;WriteLn(u, s, s, r, ' + ', u, s, a, ' = ', p, e, a, c, e);end.

{попробуй это}

Var  i: integer;  begin
for i:=1 to 9 do  write(i);  writeln(10);
for i:=0 to 8 do  write(i);  writeln(9);
for i:=1 to 2 do  write('0');  for i:=1 to 7 do  write(i);  writeln('8');
for i:=1 to 3 do  write('0');  for i:=1 to 6 do  write(i);  writeln('7');
for i:=1 to 4 do  write('0');  for i:=1 to 5 do  write(i);  writeln('6');
for i:=1 to 5 do  write('0');  for i:=1 to 4 do  write(i);  writeln(5);
for i:=1 to 6 do  write('0');  for i:=1 to 3 do   write(i);  writeln('4');
for i:=1 to 7 do  write('0');  for i:=1 to 2 do  write(i);  writeln(3);
for i:=1 to 8 do  write('0');  write('1');writeln('2');
for i:=1 to 9 do  write('0');  writeln(1);  
end.

Как сделать более кратко не знаю. Но эта программа работает.