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	Add exercise: [M20] Reverse factorisation of factorials - libzahl - big integer…
	git clone git://git.suckless.org/libzahl
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	---
	commit 555b57b3190c2ed6f73970c0515ac77dc4087220
	parent 67ebaf88644f0bf47103af79fee76d015d43ce00
	Author: Mattias Andrée <[email protected]>
	Date: Sat, 23 Jul 2016 20:07:26 +0200

	Add exercise: [M20] Reverse factorisation of factorials

	Signed-off-by: Mattias Andrée <[email protected]>

	Diffstat:
	M doc/exercises.tex \| 49 +++++++++++++++++++++++++++++…

	1 file changed, 48 insertions(+), 1 deletion(-)
	---
	diff --git a/doc/exercises.tex b/doc/exercises.tex
	@@ -52,6 +52,27 @@ The function shall be efficient for all $n$ where all primes…
	be found efficiently. You can assume that $n \ge 2$. You should not evaluate $…


	+\item {[\textit{M20}]} \textbf{Reverse factorisation of factorials}
	+
	+You should already have solved ``Factorisation of factorials''
	+before you solve this problem.
	+
	+Implement the function
	+
	+\vspace{-1em}
	+\begin{alltt}
	+ void unfactor_fact(z_t x, z_t *P,
	+ unsigned long long int *K, size_t n);
	+\end{alltt}
	+\vspace{-1em}
	+
	+\noindent
	+which given the factorsation of $x!$ determines $x$.
	+The factorisation of $x!$ is
	+$\displaystyle{\prod_{i = 1}^{n} P_i^{K_i}}$, where
	+$P_i$ is \texttt{P[i - 1]} and $K_i$ is \texttt{K[i - 1]}.
	+
	+
	\end{enumerate}


	@@ -77,7 +98,7 @@ $$ 1 + \frac{L_{n - 2}}{L_{n - 1}} = \frac{L_{n - 1}}{L_{n - …

	$$ 1 + \varphi = \frac{1}{\varphi} $$

	-So the ratio tends toward the golden ration.
	+So the ratio tends toward the golden ratio.


	\item \textbf{Factorisation of factorials}
	@@ -93,4 +114,30 @@ There is no need to calculate $\lfloor \log_p n \rfloor$,
	you will see when $p^a > n$.


	+\item \textbf{Reverse factorisation of factorials}
	+
	+$\displaystyle{x = \max_{p ~\in~ P} ~ p \cdot f(p, k_p)}$,
	+where $k_p$ is the power of $p$ in the factorisation
	+of $x!$. $f(p, k)$ is defined as:
	+
	+\vspace{1em}
	+\hspace{-2.8ex}
	+\begin{minipage}{\linewidth}
	+\begin{algorithmic}
	+ \STATE $k^\prime \gets 0$
	+ \WHILE{$k > 0$}
	+ \STATE $a \gets 0$
	+ \WHILE{$p^a \le k$}
	+ \STATE $k \gets k - p^a$
	+ \STATE $a \gets a + 1$
	+ \ENDWHILE
	+ \STATE $k^\prime \gets k^\prime + p^{a - 1}$
	+ \ENDWHILE
	+ \RETURN $k^\prime$
	+\end{algorithmic}
	+\end{minipage}
	+\vspace{1em}
	+
	+
	+
	\end{enumerate}