First of all. If you get WA#1, check how you scan input stream, cause they've put some trash in file.

I used the following idea. The whole number with base k divisible by ( k - 1 ), if sum( div( digit[ i ] ) ) is divisible by ( k - 1 ). Where digit[ i ] is k-based digit in input number ( char from input stream ); div( digit[ i ] ) is a remainder of division digit[ i ] * k by ( k - 1 ). Let see digit on i-th position, for example, it's A. So in decimal implementation it's equal to A * k ^ i.
A * k ^ i = A * k^i - A * k^(i-1) + A * k^(i-1) = A * k^(i-1) * ( k - 1 ) + A * k^(i-1)
A * k^(i-1) * ( k - 1 ) is divisible by ( k - 1 ), so we have to check only A * k^(i-1). If we repeat the same logic to this expression (i - 2) times, we see, that we have to check division A * k / ( k - 1 ).

I tried my best to explain the main idea, sorry if it can't be understood.

You don't need your "div()".

Isn't it okay to just find the digital sum of the given number and add 1?

Thank you. And we even can go a step further. Now see, A * k = A * k - A + A = A(k - 1) + A, So what we need to examine is just whether sum( digit[i]) is divisible by ( k - 1).