1656D - K-good
1656D - K-good
n is k-good if and only if
- $ n \geq 1 + 2 + \ldots + k = \frac{k(k+1)}{2}.$
- \(n \equiv 1 + 2 + \ldots + k \equiv \frac{k(k+1)}{2} \pmod{k}.\)
It is clear that both conditions are necessary, and it turns out they're sufficient too since $\frac{k(k+1)}{2} + m \cdot k $ is attainable for any integer $ m \geq 0$ by repeatedly adding \(k\)
Note that, if \(k\) is even, the second condition is \(n \equiv \frac{k}{2} \pmod{k}\), which is true if and only if \(2n\) is a multiple of k but n is not a multiple of k. So all k which divide \(2n\) but do not divide n satisfy the second condition, and we want the smallest of them in order to have the best chance of satisfying the first condition. The smallest of such \(k\)
If it doesn't, consider \(k_2 = \frac{2n}{k_1}\). Note that \(k_2\)
So \(k_2\) satisfies the first condition.
Note that \(k_2\) is only a valid answer if \(k_2 \neq 1\). If \(k_2 = 1\), then we have that n is a power of 2, and in this case there is no answer since all odd candidates of k must be odd divisors of n, of which there is only 1, and the smallest even candidate for k was \(k_1 = 2n\), which does not work. So we have to answer -1.