Style fix - libzahl - big integer library | |
git clone git://git.suckless.org/libzahl | |
Log | |
Files | |
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README | |
LICENSE | |
--- | |
commit 4d2e79e7eec793a557c26d1253bcfc13f6b555d6 | |
parent 243a542dce0f8da6fc3ac43d5e5fcb144559b507 | |
Author: Mattias Andrée <[email protected]> | |
Date: Mon, 25 Jul 2016 15:58:29 +0200 | |
Style fix | |
Signed-off-by: Mattias Andrée <[email protected]> | |
Diffstat: | |
M doc/exercises.tex | 8 ++++---- | |
1 file changed, 4 insertions(+), 4 deletions(-) | |
--- | |
diff --git a/doc/exercises.tex b/doc/exercises.tex | |
@@ -168,13 +168,13 @@ For improved performance, instead of using \texttt{zmod}, | |
you can use the recursive function | |
% | |
\( \displaystyle{ | |
- k ~\mbox{Mod}~ 2^n - 1 = | |
+ k \mod (2^n - 1) = | |
\left ( | |
- (k ~\mbox{Mod}~ 2^n) + \lfloor k \div 2^n \rfloor | |
- \right ) ~\mbox{Mod}~ 2^n - 1, | |
+ (k \mod 2^n) + \lfloor k \div 2^n \rfloor | |
+ \right ) \mod (2^n - 1), | |
}\) | |
% | |
-where $k ~\mbox{Mod}~ 2^n$ is efficiently calculated | |
+where $k \mod 2^n$ is efficiently calculated | |
using \texttt{zand($k$, $2^n - 1$)}. (This optimisation | |
is not part of the difficulty rating of this problem.) | |