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% Document Info
\author{malandro123}
\pdfinfo{
  /Title (computacao-2.pdf)
  /Creator (Cheatography)
  /Author (malandro123)
  /Subject (Computação 2 Cheat Sheet)
}

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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Computação 2 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{malandro123} via \textcolor{DarkBackground}{\uline{cheatography.com/23628/cs/5543/}}}
\end{tabulary}
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\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}malandro123 \\
  \uline{cheatography.com/malandro123} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 27th October, 2015.\\
   Updated 9th May, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
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% www.emerson.emory.edu/services/latex/latex_169.html
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Tipos de Busca}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Busca Sequencial} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Busca Binária} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Arvore de Busca Binária} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Hash} \tn 
% Row Count 4 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Busca Sequencial}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Compara a chave com cada item na array ou lista, até encontrar um item de dado cujo valor é igual o valor da chave.% Row Count 3 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Busca Sequencial - Code}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{for (i=0; i\textless{}n; i++) \newline       if (A{[}i{]}==x) \newline           return(i); /{\emph{chave encontrada}}/ \newline return(-1); /{\emph{chave não encontrada}}/} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Algoritmo de busca seqüencial em um vetor A, com N posições (0 até N-1), sendo x a chave procurada}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Busca Sequencial - Code c/ Sentinela}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{A{[}N{]}=x; \newline for(i=0; x!=A{[}i{]}; i++); \newline if (i\textless{}n) return(i); /{\emph{chave encontrada}}/ \newline else return(-1); /{\emph{sentinela encontrado}}/} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Busca Binária - Code}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Bin-Search(collection c, low, high, k) \newline int mid; \newline if low \textgreater{} high \newline      then return NIL; \newline mid = (high+low)/2; \newline if k = key{[}mid{]} \newline      then return key{[}mid{]}; \newline      else if k \textless{} key{[}mid{]} \newline         then return Bin\_search(c, low, mid-1, k); \newline         else return Bin\_search(c, mid+1, high, k);} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Busca Binária - Complexidade}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{O(log(n)), pois cada comparação reduz o número de possíveis candidatos por um fator de 2.% Row Count 2 (+ 2)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Árvore Binária de Busca - Busca Geral Recursivo}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Search(x, k) \newline if x = NIL or k = key{[}x{]} \newline     then return x \newline if k \textless{} key{[}x{]} \newline     then return Tree-Search(left{[}x{]}, k) \newline     else return Tree-Search(right{[}x{]}, k)} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Árvore Binária de Busca - Busca Geral Iterativo}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Iterative-Tree-Search(x, k) \newline while x ≠ NIL and k ≠ key{[}x{]} \newline     do if k \textless{} key{[}x{]} \newline         then x ← left{[}x{]} \newline         else x ← right{[}x{]} \newline return x} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Árvore Binária de Busca - Busca do Valor Minimo}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Minimum(x) \newline while left{[}x{]} ≠ NIL \newline     do x ← left{[}x{]} \newline return x} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Árvore Binária de Busca - Busca do Valor Maximo}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Maximum(x) \newline while right{[}x{]} ≠ NIL \newline     do x ← right{[}x{]} \newline return x} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Algoritmo de Busca do Valor Sucessor}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{O sucessor do nó x é o nó com o menor chave maior que key{[}x{]}. \newline % Row Count 2 (+ 2)
Case 1: Se a subarvore direita do nó x não for vazio, então, o sucessor do x é o nó mais esquerdo na subarvore direita; \newline % Row Count 5 (+ 3)
Case 2: Se a subarvore direita do nó x for vazio, o sucessor do x (se x é um filho esquerdo) é o antecessor de nível mais baixa ou é o antecessor de nível mais baixa , cujo filho esquerdo também é antecessor do x (se x é um filho \newline % Row Count 10 (+ 5)
direito) .% Row Count 11 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Algoritmo de Busca do Valor Sucessor}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Successor(x) \newline if right{[}x{]} ≠ NIL \newline     then return Tree-Minimum(right{[}x{]}) \newline y ← p{[}x{]} \newline while y ≠ NIL and x = right{[}p{[}x{]}{]} \newline     do x ← y \newline         y ← p{[}y{]} \newline return y} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Algoritmo de Inserção}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Insert(T, z) \newline y ← NIL \newline x ← root{[}T{]} \newline while x ≠ NIL \newline     do y ← x \newline         if key{[}z{]} \textless{} key{[}x{]} \newline             then x ← left{[}x{]} \newline             else x ← right{[}x{]} \newline p{[}z{]} ← y \newline if y = NIL \newline     then root{[}T{]} ← z \newline     else if key{[}z{]} \textless{} key{[}x{]} \newline         then left{[}y{]} ← z \newline         else right{[}y{]} ← z} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Algoritmo de Remoção}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Tree-Delete(T, z) \newline if left{[}z{]} = NIL ou right{[}z{]} = NIL \newline     then y ← z \newline     else y ← Tree-Successor(z) \newline if left{[}y{]} ≠ NIL \newline     then x ← left{[}y{]} \newline     else x ← right{[}y{]} \newline if x ≠ NIL \newline     then p{[}y{]} ← p{[}x{]} \newline if p{[}y{]} = NIL \newline     then root{[}T{]} ← x \newline     else if y =left{[}p{[}y{]}{]} \newline         then left{[}p{[}y{]}{]} ← x \newline         else right{[}p{[}y{]}{]} ← x \newline if y ≠ z \newline     then key{[}z{]} ← key{[}y{]} \newline return y} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Folha 1}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{} \tn 
% Row Count 0 (+ 0)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Ver Note}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{\#include \textless{}stdio.h\textgreater{} \newline \#include \textless{}stdlib.h\textgreater{} \newline  \newline int main() \newline \{ \newline     int conj1,conj2,conj3,n1,n2,n3,i,k1,k2,k3,soma1,soma2,soma3,mediageral; \newline     float media1,media2,media3; \newline  \newline     printf("Insira o numero de valores para serem somados no conjunto 1: \textbackslash{}n"); \newline     scanf("\%d", \&n1); \newline     soma1=0; \newline     for(i=0;i\textless{}n1;i++)\{ \newline        printf("insira os valores: \textbackslash{}n"); \newline        scanf("\%d",\&k1); \newline        soma1=k1+soma1; \newline     \} \newline    media1= (float) soma1/n1; \newline  \newline         printf("A média do conjunto 1 é: \%f\textbackslash{}n", media1); \newline    printf("Insira o numero de valores para serem somados no conjunto 2: \textbackslash{}n"); \newline     scanf("\%d", \&n2); \newline     soma2=0; \newline     for(i=0;i\textless{}n2;i++)\{ \newline        printf("insira os valores: \textbackslash{}n"); \newline        scanf("\%d",\&k2); \newline        soma2=k2+soma2; \newline     \} \newline         media2= (float)soma2/n2; \newline  \newline         printf("A média do conjunto 1 é: \%f\textbackslash{}n", media2); \newline  \newline  \newline  \newline   printf("Insira o numero de valores para serem somados no conjunto 3: \textbackslash{}n"); \newline     scanf("\%d", \&n3); \newline     soma3=0; \newline     for(i=0;i\textless{}n3;i++)\{ \newline        printf("insira os valores: \textbackslash{}n"); \newline        scanf("\%d",\&k3); \newline        soma3=k3+soma3; \newline     \} \newline         media3= (float)soma3/n3; \newline  \newline         printf("A média do conjunto 3 é: \%f\textbackslash{}n", media3); \newline  \newline    mediageral= \seqsplit{(float)(media1+media2+media3)/3;} \newline  \newline     printf ("A média final gerada dos três conjuntos é: \%f\textbackslash{}n",mediageral);} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Exercício 2.7.1. Escrever um programa C, sem utilizar funções, que  \newline a) Leia três conjuntos de n números reais digitados pelo usuário (n pode ser diferente para cada conjunto)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Ver Note 2}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{\#include \textless{}stdio.h\textgreater{} \newline \#include \textless{}stdlib.h\textgreater{} \newline  \newline float conjunto(void)\{ \newline     int i,n,k; \newline     float soma,media; \newline  \newline     printf("Insira o numero de valores para serem somados no conjunto: \textbackslash{}n"); \newline     scanf("\%d", \&n); \newline     soma=0; \newline     for(i=0;i\textless{}n;i++)\{ \newline        printf("insira os valores: \textbackslash{}n"); \newline        scanf("\%d",\&k); \newline        soma=k+soma; \newline     \} \newline         media= (float) soma/n; \newline         return (media) \newline  \newline \} \newline  \newline int main()\{ \newline  \newline  \newline     float media1,media2,media3, mediageral; \newline     media1=conjunto (); \newline      media2=conjunto (); \newline       media3=conjunto (); \newline     printf("A média do conjunto 1 é: \%f\textbackslash{}n", media1); \newline  \newline     printf("A média do conjunto 2 é: \%f\textbackslash{}n", media2); \newline  \newline     printf("A média do conjunto 3 é: \%f\textbackslash{}n", media3); \newline  \newline      mediageral= \seqsplit{(media1+media2+media3)/3.0;} \newline  \newline     printf ("A média final gerada dos três conjuntos é: \%f\textbackslash{}n",mediageral);} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{b) Imprima a média e o desvio padrão de cada um dos três conjuntos;}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{8.4cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Questao 4}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{int soma (int valor)\{ \newline     int aux; \newline     if (valor == -1 || valor == 1)\{ \newline         valor = -1; \newline     \} \newline     else\{ \newline         valor = ((-2*valor) +1) + soma(valor-1); \newline      \} \newline  \newline      return valor; \newline \}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{8.4cm}}{Escreve uma função recursiva para calcular a seuinte soma: -1-3-5-7-...-(2N-1)}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}