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Write a program in pascal to solve a quadratic equation

GalinKa [24]

Program p1;
var a,b,c,d : integer; {i presume you give integer numbers for the values of a, b, c }
     x1, x2 : real;
begin
write('a='); readln(a);
write('b='); readln(b);
write('c=');readln(c);
d:=b*b - 4*a*c
if a=0 then x1=x2= - c/b
          else
if d>0 then begin
                 x1:=(-b+sqrt(d)) / (2*a);
                 x2:=(-b - sqrt(d))/(2*a);
                end;
          else if d=0 then x1=x2= - b /(2*a)
                           else write ("no specific solution because d<0");
writeln('x1=', x1);
writeln('x2=',x2);
readln;
end.

5 0

3 years ago

I have no idea, this has two parts. Any help I would love.

lana [24]

Answer:

2that is the question??? then many I could help

7 0

2 years ago

A. Assume a computer has a physical memory organized into 64-bit words. Give the word address and offset within the word for eac

Galina-37 [17]

Answer:

Check the explanation

Explanation:

For 9th byte , it is from 8*8 bit to 9"8th bit so each word consists of 64 bits , to find word address u have to divide 8*8 by 64.

Offset within word = 9*8modulo 64.

For 27th byte , word address = 8*27/64.

Offset within word = 27* 8 modulo 64

For 21th byte , word address = 8*31/64

Offset within the word = 31*8 modulo 64

For 120 , word address = 8*120/64

Offset within the word = 120*8 modulo 64.

7 0

3 years ago

There are n poor college students who are renting two houses. For every pair of students pi and pj , the function d(pi , pj ) ou

Nuetrik [128]

Answer:

Here the given problem is modeled as a Graph problem.

Explanation:

Input:- n, k and the function d(pi,pj) which outputs an integer between 1 and n2

Algorithm:-We model each student as a node. So, there would be n nodes. We make a foothold between two nodes u and v (where u and v denote the scholars pu and pv respectively) iff d(pu,pv) > k. Now, Let's call the graph G(V, E) where V is that the vertex set of the graph ( total vertices = n which is that the number of students), and E is that the edge set of the graph ( where two nodes have edges between them if and only the drama between them is bigger than k).

We now need to partition the nodes of the graph into two sets S1 and S2 such each node belongs to precisely one set and there's no edge between the nodes within the same set (if there's a foothold between any two nodes within the same set then meaning that the drama between them exceeds k which isn't allowed). S1 and S2 correspond to the partition of scholars into two buses.

The above formulation is akin to finding out if the graph G(V,E) is a bipartite graph. If the Graph G(V, E) is bipartite then we have a partition of the students into sets such that the total drama <= k else such a partition doesn't exist.

Now, finding whether a graph is bipartite or not is often found using BFS (Breadth First algorithm) in O(V+E) time. Since V = n and E = O(n2) , the worst-case time complexity of the BFS algorithm is O(n2). The pseudo-code is given as

PseudoCode:

// Input = n,k and a function d(pi,pj)

// Edges of a graph are represented as an adjacency list

1. Make V as a vertex set of n nodes.

2. for each vertex u ∈ V

$\rightarrow$ for each vertex v ∈ V

$\rightarrow\rightarrow$ if( d(pu, pj) > k )

$\rightarrow\rightarrow\rightarrow$ add vertex u to Adj[v] // Adj[v] represents adjacency list of v

$\rightarrow\rightarrow\rightarrow$ add vertex v to Adj[u] // Adj[u] represents adjacency list of u

3. bool visited[n] // visited[i] = true if the vertex i has been visited during BFS else false

4. for each vertex u ∈ V

$\rightarrow$ visited[u] = false

5. color[n] // color[i] is binary number used for 2-coloring the graph

6. for each vertex u ∈ V

$\rightarrow$ if ( visited[u] == false)

$\rightarrow\rightarrow$ color[u] = 0;

$\rightarrow\rightarrow$ isbipartite = BFS(G,u,color,visited) // if the vertices reachable from u form a bipartite graph, it returns true

$\rightarrow\rightarrow$ if (isbipartite == false)

$\rightarrow\rightarrow\rightarrow$ print " No solution exists "

$\rightarrow\rightarrow\rightarrow$ exit(0)

7. for each vertex u ∈V

$\rightarrow$ if (color[u] == 0 )

$\rightarrow\rightarrow$ print " Student u is assigned Bus 1"

$\rightarrow$ else

$\rightarrow\rightarrow$ print " Student v is assigned Bus 2"

BFS(G,s,color,visited)

1. color[s] = 0

2. visited[s] = true

3. Q = Ф // Q is a priority Queue

4. Q.push(s)

5. while Q != Ф {

$\rightarrow$ u = Q.pop()

$\rightarrow$ for each vertex v ∈ Adj[u]

$\rightarrow\rightarrow$ if (visited[v] == false)

$\rightarrow\rightarrow\rightarrow$ color[v] = (color[u] + 1) % 2

$\rightarrow\rightarrow\rightarrow$ visited[v] = true

$\rightarrow\rightarrow\rightarrow$ Q.push(v)

$\rightarrow\rightarrow$ else

$\rightarrow\rightarrow\rightarrow$ if (color[u] == color[v])

$\rightarrow\rightarrow\rightarrow\rightarrow$ return false // vertex u and v had been assigned the same color so the graph is not bipartite

}

6. return true

3 0

3 years ago

Create a method called nicknames that passes an array as a parameter. Inside the method initialize it to hold 5 names of your ch

Gre4nikov [31]

<h2>Answer:</h2>

//======METHOD DECLARATION=========//

//method name: nicknames

//method return type: void

//method parameter: an array reference

public static void nicknames(String [] names){

//initialize the array with 5 random names

names = new String[] {"John", "Doe", "Brian", "Loveth", "Chris"};

//using an enhanced for loop, print out the elements in the array

for(String n: names){

System.out.print(n + " ");

}

<h2>Explanation:</h2>

The program is written in Java. It contains comments explaining important parts of the code. Kindly go through these comments.

A few things to note.

i. Since the method does not return any value, its return type is <em>void</em>

ii. The method is made public so that it can be accessible anywhere in and out of the class the uses it.

iii. The method is made static since it will most probably be called in the static main method (at least for testing in this case)

iv. The method receives an array of type <em>String </em>as parameter since the names to be stored are of type <em>String</em>.

v. The format of initializing an array manually should follow as shown on line 7. The <em>new</em> keyword followed by the array type (String), followed by the square brackets ([]) are all important.

vi. An enhanced for loop (lines 9 - 11) is a shorthand way of writing a for loop. The format is as follows;

=> The keyword <em>for</em>

=> followed by an opening parenthesis

=> followed by the type of each of the elements in the array. Type String in this case.

=> followed by a variable name. This holds an element per cycle of the loop.

=> followed by a column

=> followed by the array

=> followed by the closing parenthesis.

=> followed by a pair of curly parentheses serving as a block containing the code to be executed in every cycle of the loop. In this case, the array elements, each held in turn by variable n, will be printed followed by a space.

A complete code and sample output for testing purposes are shown as follows:

==================================================

public class Tester{

//The main method

public static void main(String []args){

String [] names = new String[5];

nicknames(names);

}

//The nicknames method

public static void nicknames(String [] names){

names = new String[] {"John", "Doe", "Brian", "Loveth", "Chris"};

for(String n: names){

System.out.print(n + " ");

}

==================================================

<h2>Output:</h2>

John Doe Brian Loveth Chris

<em>NB: To run this program, copy the complete code, paste in an IDE or editor and save as Tester.java</em>

6 0

2 years ago