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Antti H S Laaksonen 2017-05-17 21:23:33 +03:00
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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. about inventing an efficient algorithm.
Theoretical knowledge of algorithms Theoretical knowledge of algorithms
is very important to competitive programmers. is important to competitive programmers.
Typically, a solution to a problem is Typically, a solution to a problem is
a combination of well-known techniques and a combination of well-known techniques and
new insights. new insights.
@ -117,11 +117,11 @@ but now it suffices to write \texttt{cout}.
The code can be compiled using the following command: The code can be compiled using the following command:
\begin{lstlisting} \begin{lstlisting}
g++ -std=c++11 -O2 -Wall code.cpp -o bin g++ -std=c++11 -O2 -Wall test.cpp -o test
\end{lstlisting} \end{lstlisting}
This command produces a binary file \texttt{bin} This command produces a binary file \texttt{test}
from the source code \texttt{code.cpp}. from the source code \texttt{test.cpp}.
The compiler follows the C++11 standard The compiler follows the C++11 standard
(\texttt{-std=c++11}), (\texttt{-std=c++11}),
optimizes the code (\texttt{-O2}) optimizes the code (\texttt{-O2})
@ -176,7 +176,7 @@ The following lines at the beginning of the code
make input and output more efficient: make input and output more efficient:
\begin{lstlisting} \begin{lstlisting}
ios_base::sync_with_stdio(0); ios::sync_with_stdio(0);
cin.tie(0); cin.tie(0);
\end{lstlisting} \end{lstlisting}
@ -334,7 +334,7 @@ for (int i = 2; i <= n i++) {
cout << x%m << "\n"; cout << x%m << "\n";
\end{lstlisting} \end{lstlisting}
Usually the remainder should always Usually we want the remainder to always
be between $0\ldots m-1$. be between $0\ldots m-1$.
However, in C++ and other languages, However, in C++ and other languages,
the remainder of a negative number the remainder of a negative number
@ -571,7 +571,7 @@ For example,
is an arithmetic progression with constant 4. is an arithmetic progression with constant 4.
The sum of an arithmetic progression can be calculated The sum of an arithmetic progression can be calculated
using the formula 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, where $a$ is the first number,
$b$ is the last number and $b$ is the last number and
$n$ is the amount of numbers. $n$ is the amount of numbers.
@ -591,7 +591,7 @@ For example,
is a geometric progression with constant 2. is a geometric progression with constant 2.
The sum of a geometric progression can be calculated The sum of a geometric progression can be calculated
using the formula 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, where $a$ is the first number,
$b$ is the last number and the $b$ is the last number and the
ratio between consecutive numbers is $x$. ratio between consecutive numbers is $x$.
@ -599,11 +599,11 @@ For example,
\[3+6+12+24=\frac{24 \cdot 2 - 3}{2-1} = 45.\] \[3+6+12+24=\frac{24 \cdot 2 - 3}{2-1} = 45.\]
This formula can be derived as follows. Let 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 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 and solving the equation
\[ xS-S = bx-a\] \[ kS-S = bk-a\]
yields the formula. yields the formula.
A special case of a sum of a geometric progression is 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, Using the notation described above,
we can express many kinds of logical propositions. we can express many kinds of logical propositions.
For example, 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 means that if a number $x$ is larger than 1
and not a prime number, and not a prime number,
then there are numbers $a$ and $b$ then there are numbers $a$ and $b$
@ -824,8 +824,8 @@ f(n) & = & f(n-1)+f(n-2) \\
The first Fibonacci numbers are The first Fibonacci numbers are
\[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\] \[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\]
There is also a closed-form formula There is also a closed-form formula
for calculating Fibonacci numbers\footnote{This formula is sometimes called for calculating Fibonacci numbers, which is sometimes called
\index{Binet's formula} \key{Binet's formula}.}: \index{Binet's formula} \key{Binet's formula}:
\[f(n)=\frac{(1 + \sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}.\] \[f(n)=\frac{(1 + \sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}.\]
\subsubsection{Logarithms} \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 we have to divide $x$ by $k$ before we reach
the number 1. the number 1.
For example, $\log_2(32)=5$ For example, $\log_2(32)=5$
because 5 divisions are needed: because 5 divisions by 2 are needed:
\[32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 \] \[32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 \]
@ -869,7 +869,6 @@ calculate logarithms to some fixed base.
The \key{natural logarithm} $\ln(x)$ of a number $x$ The \key{natural logarithm} $\ln(x)$ of a number $x$
is a logarithm whose base is $e \approx 2.71828$. is a logarithm whose base is $e \approx 2.71828$.
Another property of logarithms is that Another property of logarithms is that
the number of digits of an integer $x$ in base $b$ is the number of digits of an integer $x$ in base $b$ is
$\lfloor \log_b(x)+1 \rfloor$. $\lfloor \log_b(x)+1 \rfloor$.
@ -913,9 +912,8 @@ Some countries organize online practice contests
for future IOI participants, for future IOI participants,
such as the Croatian Open Competition in Informatics \cite{coci} such as the Croatian Open Competition in Informatics \cite{coci}
and the USA Computing Olympiad \cite{usaco}. and the USA Computing Olympiad \cite{usaco}.
In addition, In addition, a large collection of problems from Polish contests
many problems from Polish contests is available online \cite{main}.
are available online \cite{main}.
\subsubsection{ICPC} \subsubsection{ICPC}
@ -964,13 +962,13 @@ performing well in a contest is a good way to prove one's skills.
\subsubsection{Books} \subsubsection{Books}
There are already some books (besides this book) that 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} \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: \item S. S. Skiena and M. A. Revilla:
\emph{Programming Challenges: The Programming Contest Training Manual} \cite{ski03} \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 \item K. Diks et al.: \emph{Looking for a Challenge? The Ultimate Problem Set from
the University of Warsaw Programming Competitions} \cite{dik12} the University of Warsaw Programming Competitions} \cite{dik12}
\end{itemize} \end{itemize}