Даны четыре числа a, с, k и b. Найти их сумму в том случае,

когда a>b и c<k. В противном случае найти их произведение.

b) Найдите периметр треугольника со сторонами a, f, g, если

a>g>f.

​

//Pascal
var
  A: array [1 .. 5, 1 .. 4] of integer;
  V: array [1 .. 20] of integer;
  i, j, z: byte;
  SUM1, SUM2, SUM3, MIN: integer;
begin
  randomize;
  writeln('***Сгенерированный массив***');
  SUM1 := 0; SUM2 := 0; SUM3 := 0;
  for i := 1 to 5 do
  begin
    for j := 1 to 4 do
    begin
      A[i, j] := random(30);
      if ((A[i, j] >= 0) and (A[i, j] <= 5)) then SUM1 := SUM1 + A[i, j];
      if ((A[i, j] >= 5) and (A[i, j] <= 15)) then SUM2 := SUM2 + A[i, j];
      write(A[i, j]:6)
    end;
    writeln;
  end;
  writeln('Сумма элементов [0 5] = ', SUM1);
  writeln('Сумма элементов [5 15] = ', SUM2);
  MIN := A[3, 1];
  for j := 1 to 4 do if A[3, j] < MIN then MIN := A[3, j];
  writeln('Минимальный элемент в третьей строке = ', MIN);
  writeln('***Вектор***');
  j := 1;
  z := 0;
  while (z < 20) do
  begin
    for i := 1 to 5 do
    begin
      z := z + 1;
      V[z] := A[i, j];
    end;
    j := j + 1;
  end;
  for i := 1 to 20 do writeln(V[i]:6);
  for i := 20 downto 10 do SUM3 := SUM3 + V[i];
  writeln('Сумма последних 10-ти элементов = ', SUM3);
  readln;
end.

ответ 31

код на Ruby 22

for i in 1..40
    s = i.to_s(2)
    k = 0
    s.chars().each {|c| k += 1 if c == "1" }
    s += (k % 2).to_s + "0"
    p [i, s, s.to_i(2)]
end

вывод
[1, "110", 6]
[2, "1010", 10]
[3, "1100", 12]
[4, "10010", 18]
[5, "10100", 20]
[6, "11000", 24]
[7, "11110", 30]
[8, "100010", 34]
[9, "100100", 36]
[10, "101000", 40]
[11, "101110", 46]
[12, "110000", 48]
[13, "110110", 54]
[14, "111010", 58]
[15, "111100", 60]
[16, "1000010", 66]
[17, "1000100", 68]
[18, "1001000", 72]
[19, "1001110", 78]
[20, "1010000", 80]
[21, "1010110", 86]
[22, "1011010", 90]
[23, "1011100", 92]
[24, "1100000", 96]
[25, "1100110", 102]
[26, "1101010", 106]
[27, "1101100", 108]
[28, "1110010", 114]
[29, "1110100", 116]
[30, "1111000", 120]
[31, "1111110", 126]
[32, "10000010", 130]
[33, "10000100", 132]
[34, "10001000", 136]
[35, "10001110", 142]
[36, "10010000", 144]
[37, "10010110", 150]
[38, "10011010", 154]
[39, "10011100", 156]
[40, "10100000", 160]