The first p

      m = 1000
      i = 0
      do j = 1, m
        do k = 1, m
          i = i + 1
          p(j + m * (k - 1)) = i
        end do
      end do

or in C

      i = 0;
      m = 1000;
      for (j = 0; j < m; j++)
        for (k = 0; k < m; k++)
          p[j + m * k] = i++;

The second p

      do j = 1, n
        p(j) = j
      end do

or in C

      for (j = 0; j < n; j++)
        p[j] = j;