We can create an array consisting of $d-1$ zeroes, the integers from $1$ to $n$, and then, if necessary, some more zeroes, so that the length of the array is divisible by $7$. The zeroes represent the missing days, i.e. the ones we will replace with ....

The length of this array divided by $7$ is the number of rows the calendar will have. We can now iterate through the array and create the rows: depending on the element of the arrays, we will determine which three characters should we add.

The above described solution is without decorations. We can add them easily: the first and last row are always the same, and all the other rows can be made with a little addition to the solution from above.