Finding Power of Factorial Divisor¶
You are given two numbers $n$ and $k$. Find the largest power of $k$ $x$ such that $n!$ is divisible by $k^x$.
Prime $k$¶
Let's first consider the case of prime $k$. The explicit expression for factorial
Note that every $k$-th element of the product is divisible by $k$, i.e. adds $+1$ to the answer; the number of such elements is $\Bigl\lfloor\dfrac{n}{k}\Bigr\rfloor$.
Next, every $k^2$-th element is divisible by $k^2$, i.e. adds another $+1$ to the answer (the first power of $k$ has already been counted in the previous paragraph). The number of such elements is $\Bigl\lfloor\dfrac{n}{k^2}\Bigr\rfloor$.
And so on, for every $i$ each $k^i$-th element adds another $+1$ to the answer, and there are $\Bigl\lfloor\dfrac{n}{k^i}\Bigr\rfloor$ such elements.
The final answer is
This result is also known as Legendre's formula. The sum is of course finite, since only approximately the first $\log_k n$ elements are not zeros. Thus, the runtime of this algorithm is $O(\log_k n)$.
Implementation¶
int fact_pow (int n, int k) {
int res = 0;
while (n) {
n /= k;
res += n;
}
return res;
}
Composite $k$¶
The same idea can't be applied directly. Instead we can factor $k$, representing it as $k = k_1^{p_1} \cdot \ldots \cdot k_m^{p_m}$. For each $k_i$, we find the number of times it is present in $n!$ using the algorithm described above - let's call this value $a_i$. The answer for composite $k$ will be