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chapter01.tex
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@ -13,7 +13,7 @@ and the core of the problem is often
about inventing an efficient algorithm.
Theoretical knowledge of algorithms
is very important to competitive programmers.
is important to competitive programmers.
Typically, a solution to a problem is
a combination of well-known techniques and
new insights.
@ -117,11 +117,11 @@ but now it suffices to write \texttt{cout}.
The code can be compiled using the following command:
\begin{lstlisting}
g++ -std=c++11 -O2 -Wall code.cpp -o bin
g++ -std=c++11 -O2 -Wall test.cpp -o test
\end{lstlisting}
This command produces a binary file \texttt{bin}
from the source code \texttt{code.cpp}.
This command produces a binary file \texttt{test}
from the source code \texttt{test.cpp}.
The compiler follows the C++11 standard
(\texttt{-std=c++11}),
optimizes the code (\texttt{-O2})
@ -176,7 +176,7 @@ The following lines at the beginning of the code
make input and output more efficient:
\begin{lstlisting}
ios_base::sync_with_stdio(0);
ios::sync_with_stdio(0);
cin.tie(0);
\end{lstlisting}
@ -334,7 +334,7 @@ for (int i = 2; i <= n i++) {
cout << x%m << "\n";
\end{lstlisting}
Usually the remainder should always
Usually we want the remainder to always
be between $0\ldots m-1$.
However, in C++ and other languages,
the remainder of a negative number
@ -571,7 +571,7 @@ For example,
is an arithmetic progression with constant 4.
The sum of an arithmetic progression can be calculated
using the formula
\[\frac{n(a+b)}{2}\]
\[\underbrace{a + \cdots + b}_{n \,\, \textrm{numbers}} = \frac{n(a+b)}{2}\]
where $a$ is the first number,
$b$ is the last number and
$n$ is the amount of numbers.
@ -591,7 +591,7 @@ For example,
is a geometric progression with constant 2.
The sum of a geometric progression can be calculated
using the formula
\[\frac{bx-a}{x-1}\]
\[a + ak + ak^2 + \cdots + b = \frac{bk-a}{k-1}\]
where $a$ is the first number,
$b$ is the last number and the
ratio between consecutive numbers is $x$.
@ -599,11 +599,11 @@ For example,
\[3+6+12+24=\frac{24 \cdot 2 - 3}{2-1} = 45.\]
This formula can be derived as follows. Let
\[ S = a + ax + ax^2 + \cdots + b .\]
\[ S = a + ak + ak^2 + \cdots + b .\]
By multiplying both sides by $x$, we get
\[ xS = ax + ax^2 + ax^3 + \cdots + bx,\]
\[ kS = ak + ak^2 + ak^3 + \cdots + bk,\]
and solving the equation
\[ xS-S = bx-a\]
\[ kS-S = bk-a\]
yields the formula.
A special case of a sum of a geometric progression is the formula
@ -774,7 +774,7 @@ but false in the set of natural numbers.
Using the notation described above,
we can express many kinds of logical propositions.
For example,
\[\forall x ((x>1 \land \lnot P(x)) \Rightarrow (\exists a (\exists b (x = ab \land a > 1 \land b > 1))))\]
\[\forall x ((x>1 \land \lnot P(x)) \Rightarrow (\exists a (\exists b (a > 1 \land b > 1 \land x = ab))))\]
means that if a number $x$ is larger than 1
and not a prime number,
then there are numbers $a$ and $b$
@ -824,8 +824,8 @@ f(n) & = & f(n-1)+f(n-2) \\
The first Fibonacci numbers are
\[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\]
There is also a closed-form formula
for calculating Fibonacci numbers\footnote{This formula is sometimes called
\index{Binet's formula} \key{Binet's formula}.}:
for calculating Fibonacci numbers, which is sometimes called
\index{Binet's formula} \key{Binet's formula}:
\[f(n)=\frac{(1 + \sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}.\]
\subsubsection{Logarithms}
@ -843,7 +843,7 @@ that $\log_k(x)$ equals the number of times
we have to divide $x$ by $k$ before we reach
the number 1.
For example, $\log_2(32)=5$
because 5 divisions are needed:
because 5 divisions by 2 are needed:
\[32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 \]
@ -869,7 +869,6 @@ calculate logarithms to some fixed base.
The \key{natural logarithm} $\ln(x)$ of a number $x$
is a logarithm whose base is $e \approx 2.71828$.
Another property of logarithms is that
the number of digits of an integer $x$ in base $b$ is
$\lfloor \log_b(x)+1 \rfloor$.
@ -913,9 +912,8 @@ Some countries organize online practice contests
for future IOI participants,
such as the Croatian Open Competition in Informatics \cite{coci}
and the USA Computing Olympiad \cite{usaco}.
In addition,
many problems from Polish contests
are available online \cite{main}.
In addition, a large collection of problems from Polish contests
is available online \cite{main}.
\subsubsection{ICPC}
@ -964,13 +962,13 @@ performing well in a contest is a good way to prove one's skills.
\subsubsection{Books}
There are already some books (besides this book) that
concentrate on competitive programming and algorithmic problem solving:
focus on competitive programming and algorithmic problem solving:
\begin{itemize}
\item S. Halim and F. Halim:
\emph{Competitive Programming 3: The New Lower Bound of Programming Contests} \cite{hal13}
\item S. S. Skiena and M. A. Revilla:
\emph{Programming Challenges: The Programming Contest Training Manual} \cite{ski03}
\item S. Halim and F. Halim:
\emph{Competitive Programming 3: The New Lower Bound of Programming Contests} \cite{hal13}
\item K. Diks et al.: \emph{Looking for a Challenge? The Ultimate Problem Set from
the University of Warsaw Programming Competitions} \cite{dik12}
\end{itemize}