961 lines
28 KiB
TeX
961 lines
28 KiB
TeX
\chapter{Introduction}
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Competitive programming combines two topics:
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(1) the design of algorithms and (2) the implementation of algorithms.
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The \key{design of algorithms} consists of problem solving
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and mathematical thinking.
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Skills for analyzing problems and solving them
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creatively are needed.
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An algorithm for solving a problem
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has to be both correct and efficient,
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and the core of the problem is often
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how to invent an efficient algorithm.
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Theoretical knowledge of algorithms
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is very important to competitive programmers.
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Typically, a solution to a problem is
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a combination of well-known techniques and
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new insights.
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The techniques that appear in competitive programming
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also form the basis for the scientific research
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of algorithms.
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The \key{implementation of algorithms} requires good
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programming skills.
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In competitive programming, the solutions
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are graded by testing an implemented algorithm
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using a set of test cases.
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Thus, it is not enough that the idea of the
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algorithm is correct, but the implementation has
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also to be correct.
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Good coding style in contests is
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straightforward and concise.
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The programs should be written quickly,
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because there is not much time available.
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Unlike in traditional software engineering,
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the programs are short (usually at most some
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hundreds of lines) and it is not needed to
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maintain them after the contest.
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\section{Programming languages}
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\index{programming language}
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At the moment, the most popular programming
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languages in contests are C++, Python and Java.
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For example, in Google Code Jam 2016,
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among the best 3,000 participants,
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73 \% used C++,
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15 \% used Python and
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10 \% used Java \cite{goo16}.
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Some participants also used several languages.
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Many people think that C++ is the best choice
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for a competitive programmer,
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and C++ is nearly always available in
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contest systems.
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The benefits in using C++ are that
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it is a very efficient language and
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its standard library contains a
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large collection
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of data structures and algorithms.
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On the other hand, it is good to
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master several languages and know
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their strengths.
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For example, if large integers are needed
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in the problem,
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Python can be a good choice, because it
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contains built-in operations for
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calculating with large integers.
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Still, most problems in programming contests
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are set so that
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using a specific programming language
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is not an unfair advantage.
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All example programs in this book are written in C++,
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and the standard library's
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data structures and algorithms are often used.
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The programs follow the C++11 standard,
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that can be used in most contests nowadays.
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If you cannot program in C++ yet,
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now it is a good time to start learning.
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\subsubsection{C++ template}
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A typical C++ template for competitive programming
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looks like this:
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\begin{lstlisting}
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#include <bits/stdc++.h>
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using namespace std;
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int main() {
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// solution comes here
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}
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\end{lstlisting}
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The \texttt{\#include} line at the beginning
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of the code is a feature of the \texttt{g++} compiler
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that allows us to include the entire standard library.
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Thus, it is not needed to separately include
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libraries such as \texttt{iostream},
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\texttt{vector} and \texttt{algorithm},
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but they are available automatically.
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The \texttt{using} line determines
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that the classes and functions
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of the standard library can be used directly
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in the code.
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Without the \texttt{using} line we should write,
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for example, \texttt{std::cout},
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but now it suffices to write \texttt{cout}.
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The code can be compiled using the following command:
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\begin{lstlisting}
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g++ -std=c++11 -O2 -Wall code.cpp -o code
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\end{lstlisting}
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This command produces a binary file \texttt{code}
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from the source code \texttt{code.cpp}.
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The compiler follows the C++11 standard
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(\texttt{-std=c++11}),
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optimizes the code (\texttt{-O2})
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and shows warnings about possible errors (\texttt{-Wall}).
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\section{Input and output}
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\index{input and output}
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In most contests, standard streams are used for
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reading input and writing output.
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In C++, the standard streams are
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\texttt{cin} for input and \texttt{cout} for output.
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In addition, the C functions
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\texttt{scanf} and \texttt{printf} can be used.
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The input for the program usually consists of
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numbers and strings that are separated with
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spaces and newlines.
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They can be read from the \texttt{cin} stream
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as follows:
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\begin{lstlisting}
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int a, b;
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string x;
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cin >> a >> b >> x;
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\end{lstlisting}
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This kind of code always works,
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assuming that there is at least one space
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or newline between each element in the input.
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For example, the above code can read
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both the following inputs:
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\begin{lstlisting}
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123 456 monkey
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\end{lstlisting}
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\begin{lstlisting}
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123 456
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monkey
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\end{lstlisting}
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The \texttt{cout} stream is used for output
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as follows:
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\begin{lstlisting}
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int a = 123, b = 456;
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string x = "monkey";
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cout << a << " " << b << " " << x << "\n";
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\end{lstlisting}
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Input and output is sometimes
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a bottleneck in the program.
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The following lines at the beginning of the code
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make input and output more efficient:
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\begin{lstlisting}
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ios_base::sync_with_stdio(0);
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cin.tie(0);
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\end{lstlisting}
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Note that the newline \texttt{"\textbackslash n"}
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works faster than \texttt{endl},
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because \texttt{endl} always causes
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a flush operation.
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The C functions \texttt{scanf}
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and \texttt{printf} are an alternative
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to the C++ standard streams.
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They are usually a bit faster,
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but they are also more difficult to use.
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The following code reads two integers from the input:
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\begin{lstlisting}
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int a, b;
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scanf("%d %d", &a, &b);
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\end{lstlisting}
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The following code prints two integers:
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\begin{lstlisting}
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int a = 123, b = 456;
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printf("%d %d\n", a, b);
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\end{lstlisting}
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Sometimes the program should read a whole line
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from the input, possibly with spaces.
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This can be accomplished by using the
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\texttt{getline} function:
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\begin{lstlisting}
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string s;
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getline(cin, s);
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\end{lstlisting}
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If the amount of data is unknown, the following
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loop is useful:
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\begin{lstlisting}
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while (cin >> x) {
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// code
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}
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\end{lstlisting}
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This loop reads elements from the input
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one after another, until there is no
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more data available in the input.
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In some contest systems, files are used for
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input and output.
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An easy solution for this is to write
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the code as usual using standard streams,
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but add the following lines to the beginning of the code:
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\begin{lstlisting}
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freopen("input.txt", "r", stdin);
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freopen("output.txt", "w", stdout);
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\end{lstlisting}
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After this, the program reads the input from the file
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''input.txt'' and writes the output to the file
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''output.txt''.
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\section{Working with numbers}
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\index{integer}
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\subsubsection{Integers}
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The most used integer type in competitive programming
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is \texttt{int}, that is a 32-bit type with
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value range $-2^{31} \ldots 2^{31}-1$
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or about $-2 \cdot 10^9 \ldots 2 \cdot 10^9$.
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If the type \texttt{int} is not enough,
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the 64-bit type \texttt{long long} can be used,
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with value range $-2^{63} \ldots 2^{63}-1$
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or about $-9 \cdot 10^{18} \ldots 9 \cdot 10^{18}$.
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The following code defines a
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\texttt{long long} variable:
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\begin{lstlisting}
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long long x = 123456789123456789LL;
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\end{lstlisting}
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The suffix \texttt{LL} means that the
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type of the number is \texttt{long long}.
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A common error when using the type \texttt{long long}
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is that the type \texttt{int} is still used somewhere
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in the code.
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For example, the following code contains
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a subtle error:
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\begin{lstlisting}
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int a = 123456789;
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long long b = a*a;
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cout << b << "\n"; // -1757895751
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\end{lstlisting}
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Even though the variable \texttt{b} is of type \texttt{long long},
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both numbers in the expression \texttt{a*a}
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are of type \texttt{int} and the result is
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also of type \texttt{int}.
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Because of this, the variable \texttt{b} will
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contain a wrong result.
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The problem can be solved by changing the type
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of \texttt{a} to \texttt{long long} or
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by changing the expression to \texttt{(long long)a*a}.
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Usually contest problems are set so that the
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type \texttt{long long} is enough.
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Still, it is good to know that
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the \texttt{g++} compiler also provides
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an 128-bit type \texttt{\_\_int128\_t}
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with value range
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$-2^{127} \ldots 2^{127}-1$ or $-10^{38} \ldots 10^{38}$.
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However, this type is not available in all contest systems.
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\subsubsection{Modular arithmetic}
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\index{remainder}
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\index{modular arithmetic}
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We denote by $x \bmod m$ the remainder
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when $x$ is divided by $m$.
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For example, $17 \bmod 5 = 2$,
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because $17 = 3 \cdot 5 + 2$.
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Sometimes, the answer to a problem is a
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very large number but it is enough to
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output it ''modulo $m$'', i.e.,
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the remainder when the answer is divided by $m$
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(for example, ''modulo $10^9+7$'').
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The idea is that even if the actual answer
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may be very large,
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it suffices to use the types
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\texttt{int} and \texttt{long long}.
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An important property of the remainder is that
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in addition, subtraction and multiplication,
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the remainder can be taken before the operation:
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\[
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\begin{array}{rcr}
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(a+b) \bmod m & = & (a \bmod m + b \bmod m) \bmod m \\
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(a-b) \bmod m & = & (a \bmod m - b \bmod m) \bmod m \\
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(a \cdot b) \bmod m & = & (a \bmod m \cdot b \bmod m) \bmod m
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\end{array}
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\]
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Thus, we can take the remainder after every operation
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and the numbers will never become too large.
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For example, the following code calculates $n!$,
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the factorial of $n$, modulo $m$:
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\begin{lstlisting}
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long long x = 1;
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for (int i = 2; i <= n i++) {
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x = (x*i)%m;
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}
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cout << x << "\n";
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\end{lstlisting}
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Usually the remainder should be always
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be between $0\ldots m-1$.
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However, in C++ and other languages,
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the remainder of a negative number
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is either zero or negative.
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An easy way to make sure there
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are no negative remainders is to first calculate
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the remainder as usual and then add $m$
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if the result is negative:
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\begin{lstlisting}
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x = x%m;
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if (x < 0) x += m;
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\end{lstlisting}
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However, this is only needed when there
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are subtractions in the code and the
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remainder may become negative.
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\subsubsection{Floating point numbers}
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\index{floating point number}
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The usual floating point types in
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competitive programming are
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the 64-bit \texttt{double}
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and, as an extension in the \texttt{g++} compiler,
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the 80-bit \texttt{long double}.
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In most cases, \texttt{double} is enough,
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but \texttt{long double} is more accurate.
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The required precision of the answer
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is usually given in the problem statement.
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An easy way to output the answer is to use
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the \texttt{printf} function
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and give the number of decimal places
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in the formatting string.
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For example, the following code prints
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the value of $x$ with 9 decimal places:
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\begin{lstlisting}
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printf("%.9f\n", x);
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\end{lstlisting}
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A difficulty when using floating point numbers
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is that some numbers cannot be represented
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accurately as floating point numbers,
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but there will be rounding errors.
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For example, the result of the following code
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is surprising:
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\begin{lstlisting}
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double x = 0.3*3+0.1;
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printf("%.20f\n", x); // 0.99999999999999988898
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\end{lstlisting}
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Due to a rounding error,
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the value of \texttt{x} is a bit smaller than 1,
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while the correct value would be 1.
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It is risky to compare floating point numbers
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with the \texttt{==} operator,
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because it is possible that the values should
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be equal but they are not because of rounding.
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A better way to compare floating point numbers
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is to assume that two numbers are equal
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if the difference between them is $\varepsilon$,
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where $\varepsilon$ is a small number.
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In practice, the numbers can be compared
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as follows ($\varepsilon=10^{-9}$):
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\begin{lstlisting}
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if (abs(a-b) < 1e-9) {
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// a and b are equal
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}
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\end{lstlisting}
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Note that while floating point numbers are inaccurate,
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integers up to a certain limit can be still
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represented accurately.
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For example, using \texttt{double},
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it is possible to accurately represent all
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integers whose absolute value is at most $2^{53}$.
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\section{Shortening code}
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Short code is ideal in competitive programming,
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because programs should be written
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as fast as possible.
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Because of this, competitive programmers often define
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shorter names for datatypes and other parts of code.
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\subsubsection{Type names}
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\index{tuppdef@\texttt{typedef}}
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Using the command \texttt{typedef}
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it is possible to give a shorter name
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to a datatype.
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For example, the name \texttt{long long} is long,
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so we can define a shorter name \texttt{ll}:
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\begin{lstlisting}
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typedef long long ll;
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\end{lstlisting}
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After this, the code
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\begin{lstlisting}
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long long a = 123456789;
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long long b = 987654321;
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cout << a*b << "\n";
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\end{lstlisting}
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can be shortened as follows:
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\begin{lstlisting}
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ll a = 123456789;
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ll b = 987654321;
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cout << a*b << "\n";
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\end{lstlisting}
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The command \texttt{typedef}
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can also be used with more complex types.
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For example, the following code gives
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the name \texttt{vi} for a vector of integers
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and the name \texttt{pi} for a pair
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that contains two integers.
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\begin{lstlisting}
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typedef vector<int> vi;
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typedef pair<int,int> pi;
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\end{lstlisting}
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\subsubsection{Macros}
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\index{macro}
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Another way to shorten the code is to define
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\key{macros}.
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A macro means that certain strings in
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the code will be changed before the compilation.
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In C++, macros are defined using the
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command \texttt{\#define}.
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For example, we can define the following macros:
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\begin{lstlisting}
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#define F first
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#define S second
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#define PB push_back
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#define MP make_pair
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\end{lstlisting}
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After this, the code
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\begin{lstlisting}
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v.push_back(make_pair(y1,x1));
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v.push_back(make_pair(y2,x2));
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int d = v[i].first+v[i].second;
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\end{lstlisting}
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can be shortened as follows:
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\begin{lstlisting}
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v.PB(MP(y1,x1));
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v.PB(MP(y2,x2));
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int d = v[i].F+v[i].S;
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\end{lstlisting}
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A macro can also have parameters
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which makes it possible to shorten loops and other
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structures.
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For example, we can define the following macro:
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\begin{lstlisting}
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#define REP(i,a,b) for (int i = a; i <= b; i++)
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\end{lstlisting}
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After this, the code
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\begin{lstlisting}
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for (int i = 1; i <= n; i++) {
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search(i);
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}
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\end{lstlisting}
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can be shortened as follows:
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\begin{lstlisting}
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REP(i,1,n) {
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search(i);
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}
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\end{lstlisting}
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\section{Mathematics}
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Mathematics plays an important role in competitive
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programming, and it is not possible to become
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a successful competitive programmer without
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having good mathematical skills.
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This section discusses some important
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mathematical concepts and formulas that
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are needed later in the book.
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\subsubsection{Sum formulas}
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Each sum of the form
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\[\sum_{x=1}^n x^k = 1^k+2^k+3^k+\ldots+n^k,\]
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where $k$ is a positive integer,
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has a closed-form formula that is a
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polynomial of degree $k+1$.
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For example,
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\[\sum_{x=1}^n x = 1+2+3+\ldots+n = \frac{n(n+1)}{2}\]
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and
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\[\sum_{x=1}^n x^2 = 1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6}.\]
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An \key{arithmetic progression} is a \index{arithmetic progression}
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sequence of numbers
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where the difference between any two consecutive
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numbers is constant.
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For example,
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\[3, 7, 11, 15\]
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is an arithmetic progression with constant 4.
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The sum of an arithmetic progression can be calculated
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using the formula
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\[\frac{n(a+b)}{2}\]
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where $a$ is the first number,
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$b$ is the last number and
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$n$ is the amount of numbers.
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For example,
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\[3+7+11+15=\frac{4 \cdot (3+15)}{2} = 36.\]
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The formula is based on the fact
|
|
that the sum consists of $n$ numbers and
|
|
the value of each number is $(a+b)/2$ on average.
|
|
|
|
\index{geometric progression}
|
|
A \key{geometric progression} is a sequence
|
|
of numbers
|
|
where the ratio between any two consecutive
|
|
numbers is constant.
|
|
For example,
|
|
\[3,6,12,24\]
|
|
is a geometric progression with constant 2.
|
|
The sum of a geometric progression can be calculated
|
|
using the formula
|
|
\[\frac{bx-a}{x-1}\]
|
|
where $a$ is the first number,
|
|
$b$ is the last number and the
|
|
ratio between consecutive numbers is $x$.
|
|
For example,
|
|
\[3+6+12+24=\frac{24 \cdot 2 - 3}{2-1} = 45.\]
|
|
|
|
This formula can be derived as follows. Let
|
|
\[ S = a + ax + ax^2 + \cdots + b .\]
|
|
By multiplying both sides by $x$, we get
|
|
\[ xS = ax + ax^2 + ax^3 + \cdots + bx,\]
|
|
and solving the equation
|
|
\[ xS-S = bx-a\]
|
|
yields the formula.
|
|
|
|
A special case of a sum of a geometric progression is the formula
|
|
\[1+2+4+8+\ldots+2^{n-1}=2^n-1.\]
|
|
|
|
\index{harmonic sum}
|
|
|
|
A \key{harmonic sum} is a sum of the form
|
|
\[ \sum_{x=1}^n \frac{1}{x} = 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}.\]
|
|
|
|
An upper bound for a harmonic sum is $\log_2(n)+1$.
|
|
Namely, we can
|
|
modify each term $1/k$ so that $k$ becomes
|
|
the nearest power of two that does not exceed $k$.
|
|
For example, when $n=6$, we can estimate
|
|
the sum as follows:
|
|
\[ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \le
|
|
1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}.\]
|
|
This upper bound consists of $\log_2(n)+1$ parts
|
|
($1$, $2 \cdot 1/2$, $4 \cdot 1/4$, etc.),
|
|
and the value of each part is at most 1.
|
|
|
|
\subsubsection{Set theory}
|
|
|
|
\index{set theory}
|
|
\index{set}
|
|
\index{intersection}
|
|
\index{union}
|
|
\index{difference}
|
|
\index{subset}
|
|
\index{universal set}
|
|
\index{complement}
|
|
|
|
A \key{set} is a collection of elements.
|
|
For example, the set
|
|
\[X=\{2,4,7\}\]
|
|
contains elements 2, 4 and 7.
|
|
The symbol $\emptyset$ denotes an empty set,
|
|
and $|S|$ denotes the size of a set $S$,
|
|
i.e., the number of elements in the set.
|
|
For example, in the above set, $|X|=3$.
|
|
|
|
If a set $S$ contains an element $x$,
|
|
we write $x \in S$,
|
|
and otherwise we write $x \notin S$.
|
|
For example, in the above set
|
|
\[4 \in X \hspace{10px}\textrm{and}\hspace{10px} 5 \notin X.\]
|
|
|
|
\begin{samepage}
|
|
New sets can be constructed using set operations:
|
|
\begin{itemize}
|
|
\item The \key{intersection} $A \cap B$ consists of elements
|
|
that are both in $A$ and $B$.
|
|
For example, if $A=\{1,2,5\}$ and $B=\{2,4\}$,
|
|
then $A \cap B = \{2\}$.
|
|
\item The \key{union} $A \cup B$ consists of elements
|
|
that are in $A$ or $B$ or both.
|
|
For example, if $A=\{3,7\}$ and $B=\{2,3,8\}$,
|
|
then $A \cup B = \{2,3,7,8\}$.
|
|
\item The \key{complement} $\bar A$ consists of elements
|
|
that are not in $A$.
|
|
The interpretation of a complement depends on
|
|
the \key{universal set} that contains all possible elements.
|
|
For example, if $A=\{1,2,5,7\}$ and the universal set is
|
|
$\{1,2,\ldots,10\}$, then $\bar A = \{3,4,6,8,9,10\}$.
|
|
\item The \key{difference} $A \setminus B = A \cap \bar B$
|
|
consists of elements that are in $A$ but not in $B$.
|
|
Note that $B$ can contain elements that are not in $A$.
|
|
For example, if $A=\{2,3,7,8\}$ and $B=\{3,5,8\}$,
|
|
then $A \setminus B = \{2,7\}$.
|
|
\end{itemize}
|
|
\end{samepage}
|
|
|
|
If each element of $A$ also belongs to $S$,
|
|
we say that $A$ is a \key{subset} of $S$,
|
|
denoted by $A \subset S$.
|
|
A set $S$ always has $2^{|S|}$ subsets,
|
|
including the empty set.
|
|
For example, the subsets of the set $\{2,4,7\}$ are
|
|
\begin{center}
|
|
$\emptyset$,
|
|
$\{2\}$, $\{4\}$, $\{7\}$, $\{2,4\}$, $\{2,7\}$, $\{4,7\}$ and $\{2,4,7\}$.
|
|
\end{center}
|
|
|
|
Often used sets are
|
|
$\mathbb{N}$ (natural numbers),
|
|
$\mathbb{Z}$ (integers),
|
|
$\mathbb{Q}$ (rational numbers) and
|
|
$\mathbb{R}$ (real numbers).
|
|
The set $\mathbb{N}$
|
|
can be defined in two ways, depending
|
|
on the situation:
|
|
either $\mathbb{N}=\{0,1,2,\ldots\}$
|
|
or $\mathbb{N}=\{1,2,3,...\}$.
|
|
|
|
We can also construct a set using a rule of the form
|
|
\[\{f(n) : n \in S\},\]
|
|
where $f(n)$ is some function.
|
|
This set contains all elements of the form $f(n)$,
|
|
where $n$ is an element in $S$.
|
|
For example, the set
|
|
\[X=\{2n : n \in \mathbb{Z}\}\]
|
|
contains all even integers.
|
|
|
|
\subsubsection{Logic}
|
|
|
|
\index{logic}
|
|
\index{negation}
|
|
\index{conjuction}
|
|
\index{disjunction}
|
|
\index{implication}
|
|
\index{equivalence}
|
|
|
|
The value of a logical expression is either
|
|
\key{true} (1) or \key{false} (0).
|
|
The most important logical operators are
|
|
$\lnot$ (\key{negation}),
|
|
$\land$ (\key{conjunction}),
|
|
$\lor$ (\key{disjunction}),
|
|
$\Rightarrow$ (\key{implication}) and
|
|
$\Leftrightarrow$ (\key{equivalence}).
|
|
The following table shows the meaning of these operators:
|
|
|
|
\begin{center}
|
|
\begin{tabular}{rr|rrrrrrr}
|
|
$A$ & $B$ & $\lnot A$ & $\lnot B$ & $A \land B$ & $A \lor B$ & $A \Rightarrow B$ & $A \Leftrightarrow B$ \\
|
|
\hline
|
|
0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\
|
|
0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 \\
|
|
1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\
|
|
1 & 1 & 0 & 0 & 1 & 1 & 1 & 1 \\
|
|
\end{tabular}
|
|
\end{center}
|
|
|
|
The expression $\lnot A$ has the opposite value of $A$.
|
|
The expression $A \land B$ is true if both $A$ and $B$
|
|
are true,
|
|
and the expression $A \lor B$ is true if $A$ or $B$ or both
|
|
are true.
|
|
The expression $A \Rightarrow B$ is true
|
|
if whenever $A$ is true, also $B$ is true.
|
|
The expression $A \Leftrightarrow B$ is true
|
|
if $A$ and $B$ are both true or both false.
|
|
|
|
\index{predicate}
|
|
|
|
A \key{predicate} is an expression that is true or false
|
|
depending on its parameters.
|
|
Predicates are usually denoted by capital letters.
|
|
For example, we can define a predicate $P(x)$
|
|
that is true exactly when $x$ is a prime number.
|
|
Using this definition, $P(7)$ is true but $P(8)$ is false.
|
|
|
|
\index{quantifier}
|
|
|
|
A \key{quantifier} connects a logical expression
|
|
to the elements of a set.
|
|
The most important quantifiers are
|
|
$\forall$ (\key{for all}) and $\exists$ (\key{there is}).
|
|
For example,
|
|
\[\forall x (\exists y (y < x))\]
|
|
means that for each element $x$ in the set,
|
|
there is an element $y$ in the set
|
|
such that $y$ is smaller than $x$.
|
|
This is true in the set of integers,
|
|
but false in the set of natural numbers.
|
|
|
|
Using the notation described above,
|
|
we can express many kinds of logical propositions.
|
|
For example,
|
|
\[\forall x ((x>1 \land \lnot P(x)) \Rightarrow (\exists a (\exists b (x = ab \land a > 1 \land b > 1))))\]
|
|
means that if a number $x$ is larger than 1
|
|
and not a prime number,
|
|
then there are numbers $a$ and $b$
|
|
that are larger than $1$ and whose product is $x$.
|
|
This proposition is true in the set of integers.
|
|
|
|
\subsubsection{Functions}
|
|
|
|
The function $\lfloor x \rfloor$ rounds the number $x$
|
|
down to an integer, and the function
|
|
$\lceil x \rceil$ rounds the number $x$
|
|
up to an integer. For example,
|
|
\[ \lfloor 3/2 \rfloor = 1 \hspace{10px} \textrm{and} \hspace{10px} \lceil 3/2 \rceil = 2.\]
|
|
|
|
The functions $\min(x_1,x_2,\ldots,x_n)$
|
|
and $\max(x_1,x_2,\ldots,x_n)$
|
|
give the smallest and largest of values
|
|
$x_1,x_2,\ldots,x_n$.
|
|
For example,
|
|
\[ \min(1,2,3)=1 \hspace{10px} \textrm{and} \hspace{10px} \max(1,2,3)=3.\]
|
|
|
|
\index{factorial}
|
|
|
|
The \key{factorial} $n!$ can be defined
|
|
\[\prod_{x=1}^n x = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n\]
|
|
or recursively
|
|
\[
|
|
\begin{array}{lcl}
|
|
0! & = & 1 \\
|
|
n! & = & n \cdot (n-1)! \\
|
|
\end{array}
|
|
\]
|
|
|
|
\index{Fibonacci number}
|
|
|
|
The \key{Fibonacci numbers} arise in many situations.
|
|
They can be defined recursively as follows:
|
|
\[
|
|
\begin{array}{lcl}
|
|
f(0) & = & 0 \\
|
|
f(1) & = & 1 \\
|
|
f(n) & = & f(n-1)+f(n-2) \\
|
|
\end{array}
|
|
\]
|
|
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:
|
|
\[f(n)=\frac{(1 + \sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}}.\]
|
|
|
|
\subsubsection{Logarithms}
|
|
|
|
\index{logarithm}
|
|
|
|
The \key{logarithm} of a number $x$
|
|
is denoted $\log_k(x)$, where $k$ is the base
|
|
of the logarithm.
|
|
According to the definition,
|
|
$\log_k(x)=a$ exactly when $k^a=x$.
|
|
|
|
A useful property of logarithms is
|
|
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:
|
|
|
|
\[32 \rightarrow 16 \rightarrow 8 \rightarrow 4 \rightarrow 2 \rightarrow 1 \]
|
|
|
|
Logarithms are often used in the analysis of
|
|
algorithms, because many efficient algorithms
|
|
halve something at each step.
|
|
Hence, we can estimate the efficiency of such algorithms
|
|
using logarithms.
|
|
|
|
The logarithm of a product is
|
|
\[\log_k(ab) = \log_k(a)+\log_k(b),\]
|
|
and consequently,
|
|
\[\log_k(x^n) = n \cdot \log_k(x).\]
|
|
In addition, the logarithm of a quotient is
|
|
\[\log_k\Big(\frac{a}{b}\Big) = \log_k(a)-\log_k(b).\]
|
|
Another useful formula is
|
|
\[\log_u(x) = \frac{\log_k(x)}{\log_k(u)},\]
|
|
and using this, it is possible to calculate
|
|
logarithms to any base if there is a way to
|
|
calculate logarithms to some fixed base.
|
|
|
|
\index{natural logarithm}
|
|
|
|
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
|
|
$\lfloor \log_b(x)+1 \rfloor$.
|
|
For example, the representation of
|
|
$123$ in base $2$ is 1111011 and
|
|
$\lfloor \log_2(123)+1 \rfloor = 7$.
|
|
|
|
\section{Contests}
|
|
|
|
\subsubsection{IOI}
|
|
|
|
The International Olympiad in Informatics (IOI) \cite{ioi}
|
|
is an annual programming contests for
|
|
secondary school students.
|
|
Each country is allowed to send a team of
|
|
four students to the contest.
|
|
There are usually about 300 participants
|
|
from 80 countries \cite{iois}.
|
|
|
|
The IOI consists of two five-hour long contests.
|
|
In both contests, the participants are asked to
|
|
solve three algorithm tasks of various difficulty.
|
|
The tasks are divided into subtasks,
|
|
each of which has an assigned score.
|
|
Even if the contestants are divided into teams,
|
|
they compete as individuals.
|
|
|
|
The IOI syllabus \cite{ioiy} regulates the topics
|
|
that may appear in IOI tasks.
|
|
This book covers almost all topics in the IOI syllabus.
|
|
|
|
The participants for the IOI are selected through
|
|
national contests.
|
|
Before the IOI, many regional contests are organized,
|
|
such as the Baltic Olympiad in Informatics (BOI),
|
|
the Central European Olympiad in Informatics (CEOI)
|
|
and the Asia-Pacific Informatics Olympiad (APIO).
|
|
|
|
Some countries organize online practice contests
|
|
for future IOI participants,
|
|
such as the Croatian Open Competition in Informatics (COCI)
|
|
and the USA Computing Olympiad (USACO).
|
|
In addition,
|
|
many problems from Polish contests
|
|
are available online \cite{main}.
|
|
|
|
\subsubsection{ICPC}
|
|
|
|
The International Collegiate Programming Contest (ICPC) \cite{icpc}
|
|
is an annual programming contest for university students.
|
|
Each team in the contest consists of three students,
|
|
and unlike in the IOI, the students work together;
|
|
there is even only one computer available for each team.
|
|
|
|
The ICPC consists of several stages, and finally the
|
|
best teams are invited to the World Finals.
|
|
While there are tens of thousands of participants
|
|
in the contest, there are only 128 final slots available,
|
|
so even advancing to the finals
|
|
is a great achievement in some regions.
|
|
|
|
In each ICPC contest, the teams have five hours time to solve
|
|
about ten algorithm problems.
|
|
A solution to a problem is accepted only if it solves
|
|
all test cases efficiently.
|
|
During the contest, the teams see the results of the other teams,
|
|
but for the last hour the scoreboard is frozen and it
|
|
is not possible to see the results of the last submissions.
|
|
|
|
The topics that may appear at the ICPC are not so well
|
|
specified as those at the IOI.
|
|
In any case, it is clear that more knowledge is needed
|
|
at the ICPC, especially more mathematical skills.
|
|
|
|
\subsubsection{Online contests}
|
|
|
|
There are also many online contests that are open for everybody.
|
|
At the moment, the most active contest site is Codeforces
|
|
that organizes contests about weekly.
|
|
In Codeforces, participants are divided into two divisions:
|
|
beginners compete in Div2 and more experienced programmers in Div1.
|
|
Other contest sites include AtCoder, CS Academy, HackerRank and Topcoder.
|
|
|
|
Some companies organize online contests with onsite finals.
|
|
Examples of such contests are Facebook Hacker Cup,
|
|
Google Code Jam and Yandex.Algorithm.
|
|
Of course, companies also use those contests for recruiting:
|
|
performing well in a contest is a good way to prove one's skills.
|
|
|
|
\section{Books}
|
|
|
|
\subsubsection{Competitive programming books}
|
|
|
|
There are already some books (besides this book) that
|
|
concentrate 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}, 2013
|
|
\item S. S. Skiena and M. A. Revilla:
|
|
\emph{Programming Challenges: The Programming Contest Training Manual},
|
|
Springer, 2003
|
|
\item \emph{Looking for a Challenge? The Ultimate Problem Set from
|
|
the University of Warsaw Programming Competitions}, 2012
|
|
\end{itemize}
|
|
|
|
The first two books are intended for beginners,
|
|
whereas the last book contains advanced material.
|
|
|
|
\subsubsection{General algorithm books}
|
|
|
|
Of course, general algorithm books are also suitable for
|
|
competitive programmers.
|
|
Some good books are:
|
|
|
|
\begin{itemize}
|
|
\item T. H. Cormen, C. E. Leiserson, R. L. Rivest and C. Stein:
|
|
\emph{Introduction to Algorithms}, MIT Press, 2009 (3rd edition)
|
|
\item J. Kleinberg and É. Tardos:
|
|
\emph{Algorithm Design}, Pearson, 2005
|
|
\item S. S. Skiena:
|
|
\emph{The Algorithm Design Manual}, Springer, 2008 (2nd edition)
|
|
\end{itemize}
|