| First we need to determine how many bracelets Ishiko is going to exhibit. When \(N = K\) (that is, when all of the beads are green) there is of course just \(1\) bracelet to display. Otherwise, there are \(\frac{(N-1)!}{K!}\) ways to arrange \(N\) beads in a ring when \(K\) are green beads and the rest are unique colors. Since there's always at least \(1\) unique bead, we can say that that unique bead is in a fixed position in each bracelet. | |
| Next, we observe that a display case of height \(H\) holds exactly \(H^2\) bracelets. The problem then is to determine the minimum number of perfect squares needed to sum to \(\frac{(N-1)!}{K!}\). According to [Lagrange's four-square theorem](https://en.wikipedia.org/wiki/Lagrange%27s_four-square_theorem), we will never need more than \(4\) squares, so Ishiko only needs at most \(4\) display cases. | |
| Let \(X\) be the number of bracelets to be displayed. As \(X\) can be as large as \(1{,}000{,}000{,}000!\), we still need a quick way to determine whether the answer is \(1\), \(2\), \(3\), or \(4\). The answer is \(1\) if \(X\) is a square number, which is something we can determine from the prime factorization of \(X\). If every prime in the factorization appears an even number of times, then \(X\) is a square number. | |
| [Legendre's formula](https://en.wikipedia.org/wiki/Legendre%27s_formula) gives us an easy way to compute the exponent on a given prime \(p\) in the factorization of \(n!\) by summing \(\lfloor n / p^i \rfloor\) for all \(i\), which is quick since \(i\) will never be larger than \(32\) for a \(32\)-bit integer. If \(a\) is the exponent on \(p\) in \(N-1!\), and \(b\) is the exponent on \(p\) in \(K\), then the exponent on \(p\) in \(\frac{(N-1!)}{K!}\) is just \(a - b\). | |
| How can we tell if the answer is \(2\)? Why, we can just apply the aptly-named [two-square theorem](https://en.wikipedia.org/wiki/Sum_of_two_squares_theorem), which says that \(X\) can be represented as a sum of two squares if and only if its prime decomposition contains no factor \(p^k\), where \(p \equiv 3\ (\text{mod}\ 4)\) and \(k\) is odd. Since we already have the prime factorization of \(X\), this is easy to determine. | |
| To determine if the answer is \(3\), we'll appeal to Legendre once more. [Legendre's three-square theorem](https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem) says that \(X\) can be represented as a sum of three squares as long as it is *not* of the form \(X = 4^a(8b + 7)\), for non-negative \(a\) and \(b\). We can check if \(X\) has a factor of \(4\) by checking if the exponent on \(2\) in the prime factorization is even. If so, we can then check if \(X / 4^a \equiv 7\ (\text{mod}\ 8)\) by multiplying all prime factors higher than \(2\), modulo \(8\). Once again, \(X\) can't have more than \(32\) distinct factors as \(N\) and \(K\) are both \(32\)-bit integers, so this computation is also quick. | |
| In the end, if the answer isn't \(1\), \(2\), or \(3\), then it must be \(4\). | |