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N76 [4]
3 years ago
14

How do graphic designers showcase their work?

Computers and Technology
2 answers:
LekaFEV [45]3 years ago
8 0

on their own webistes, you can make one on weebly.com for free


Luda [366]3 years ago
6 0

They showcase it digitally since their work is done on computers.

You might be interested in
You are an interior decorator, confronted with a dark living room. To lighten the room up, you have n candles and want to build
user100 [1]

Answer:

a)

Algorithm to find a solution of min. cost

Function:cost(Graph G,Graph G1);

GC --> empty graph

for i in edges E:

if E(i,j) in G:

c(i,j)=c(i,j)

else if E(i,j) in G1:

c(i,j)=-c(i,j)

Function:Mincost(Graph G):

GC=Cost(G,G1)

while(negativecycle(GC)):

Update residal graph(G1)

GC=Cost(G,G1)

mincost=sum of Cij*F(i,j)

return mincost;

Explanation:

a)

1) Start the program

2) Read the no. of edges and vertices and also read the cost of the two nodes.

3) Find the min cost by travelling to the destination i.e.. finding all possible min. cost values.

4) Compare the all possible min.cost values

5) And display the least min. cost

6) Stop the program

b)

<u>Correctness of algorithm</u>

1)Here in these algorithm we are calculating all the possible cases.

2)These algorithm also supports the negative cost.

3)These algorithm occupies more space.

4)Takes less time

5)so,these algorithm provides the cost efficient solution very effectively.

c)

<u>Run Time Analysis</u>

1) While reading the values during the run time the program execution will stop until user provides the values.

2) Based on the User input Output vary.

3) Time consumption and space consumption is depends on the no. of inputs the user is given.

6 0
3 years ago
Consider a company that needs to sort an array of structures of type Customer by balance, with the largest balance first. Here i
8090 [49]

Answer:

#include <iostream>

#include <string>

#include <cstring>

#include <cstdlib>

using namespace std;

struct Person

{

string name;

int age;

};

int main()

{

struct Person data[10];

struct Person *pData[10],*temp;

string names[] = {"a","b","c","g","z","l","p","q","r","w"};

int num[] = {4,6,34,8,13,90,33,22,18,23};

for(int i=0;i<9;i++)

{

data[i].name = names[i];

data[i].age = num[i];

pData[i] = &data[i];

}

for(int i=0;i<9;i++)

{

for(int j=i+1;j<9;j++)

{

if(pData[i]->name.compare(pData[j]->name)>0)

{

temp = pData[i];

pData[i] = pData[j];

pData[j] = temp;

}

}

}

for(int i=0;i<9;i++)

{

cout<<pData[i]->name<<" "<<pData[i]->age<<endl;

}

}

Explanation:

The line #include <iostream> initializes the program.

This program created an array of only 10 structures, Customer data. Also defined an auxiliary array Customer *pData.

The program uses these parameters to sorts the array of pointers so that when you go through pData in increasing order of index k, the entries pData[k] point to Customer objects in decreasing order by balance, meaning that pdata[0] now points to the customer with the highest balance, and pData[9] points to the customer with the smallest balance.

4 0
2 years ago
Which of the following should NOT be done to keep people from slipping on floors? A) Have the proper absorbents in an easy-to-fi
RoseWind [281]

The following options helps keep people from slipping on floors

A) Have the proper absorbents in an easy-to-find place to quickly put on spills : So that accidental little spills can be cleaned right away.

C) Wear slip-resistant footwear : It is a precautionary step to follow, no matter where we go.

D) Mark a slippery area with an easy-to-use tent sign that says " Caution, Slippery floor" : This allows people to be aware of the wet area and cross it cautiously.

Leaving oil and fluids to air dry thoroughly, takes a long time and there are chances that people step over it and slip.

So, the answer is

(B) Anytime oil or fluids are spilled on the floor, leave them to thoroughly air dry.

8 0
3 years ago
Read 2 more answers
Need Help!
Aleksandr-060686 [28]

Answer:

A and C are the only <u>legal</u> but unethical options

4 0
3 years ago
Write a function that returns a chessboard pattern ("B" for black squares, "W" for white squares). The function takes a number N
Kryger [21]

Answer:

Here is the C++ program.

#include <iostream>  //to use input output functions

using namespace std;  // to access objects like cin cout

 

void chessboard(int N){  // function that takes a number N as parameter generates corresponding board

   int i, j;  

   string black = "B";  // B for black squares

   string white = "W"; // W for white squares

   for(i=1;i<=N;i++){  //iterates through each column of the board

       for(j=1;j<=N;j++){  //iterates through each row of the board

           if((i+j)%2==0)  // if sum of i and j is completely divisible by 2

               cout<<black<<" ";  //displays B when above if condition is true

           else  //if (i+j)%2 is not equal to 0

           cout<<white<<" ";  }  // displays W when above if condition is false

      cout<<endl;    }    }  //prints the new line

       

int main(){    //start of the main function

   int num;  //declares an integer num

   cout << "Enter an integer representing the size of the chessboard: ";  //prompts user to enter size of chess board

   cin >> num;  //reads value of num from user

   chessboard(num); } //calls chessboard function to display N lines consist of N space-separated characters representing the chessboard pattern

Explanation:

The function works as follows:

Lets say that N = 2

two string variables black and white are declared. The value of black is set to "B" and value of white is set to "W" to return a chessboard pattern in B and W squares.

The outer loop for(i=1;i<=N;i++) iterates through each column of the chess board. The inner loop  for(j=1;j<=N;j++) iterates through each row of chess board.

At first iteration of outer loop:

N = 2

outer loop:

i=1

i<=N is true because i=1 and 1<=2

So the program enters the body of the outer for loop. It has an inner for loop:

for(j=1;j<=N;j++)

At first iteration of inner loop:

j = 1

j<=N is true because j=1 and 1<=2

So the program enters the body of the inner for loop. It has an if statement:

if((i+j)%2==0) this statement works as:

if(1+1) % 2 == 0

(1+1 )% 2 takes the mod of 1+1 with 2 which gives the remainder of the division.

2%2 As 2 is completely divisible by 2 so 2%2 == 0

Hence the if condition evaluates to true. So the statement in if part executes:

cout<<black<<" ";

This prints B on the output screen with a space.

B

The value of j is incremented to 1.

j = 2

At second iteration of inner loop:

j = 2

j<=N is true because j=2 and 2=2

So the program enters the body of the inner for loop. It has an if statement:

if((i+j)%2==0) this statement works as:

if(1+2) % 2 == 0

(1+2 )% 2 takes the mod of 1+2 with 2 which gives the remainder of the division.

3%2 As 3 is not completely divisible by 2

Hence the if condition evaluates to false. So the statement in else part executes:

cout<<white<<" ";

This prints W on the output screen with a space.

B W

The value of j is incremented to 1.

j = 3

Now  

j<=N is false because j=3 and 3>2

So the loop breaks and program moves to the outer loop to iterate through the next row.

At second iteration of outer loop:

N = 2

outer loop:

i=2

i<=N is true because i=2 and 2 = 2

So the program enters the body of the outer for loop. It has an inner for loop:

for(j=1;j<=N;j++)

At first iteration of inner loop:

j = 1

j<=N is true because j=1 and 1<=2

So the program enters the body of the inner for loop. It has an if statement:

if((i+j)%2==0) this statement works as:

if(2+1) % 2 == 0

(2+1 )% 2 takes the mod of 2+1 with 2 which gives the remainder of the division.

3%2 As 3 is not completely divisible by 2

Hence the if condition evaluates to false. So the statement in else part executes:

cout<<white<<" ";

This prints W on the output screen with a space.

B W

W

The value of j is incremented to 1.

j = 2

At second iteration of inner loop:

j = 2

j<=N is true because j=2 and 2=2

So the program enters the body of the inner for loop. It has an if statement:

if((i+j)%2==0) this statement works as:

if(2+2) % 2 == 0

(2+2 )% 2 takes the mod of 2+2 with 2 which gives the remainder of the division.

4%2 As 4 is completely divisible by 2 so 4%2 == 0

Hence the if condition evaluates to false. So the statement in if part executes:

cout<<black<<" ";

This prints B on the output screen with a space.

B W

W B

The value of j is incremented to 1.

j = 3

Now  

j<=N is false because j=3 and 3>2

So the loop breaks and program moves to the outer loop. The value of outer loop variable i is incremented to 1 so i = 3

N = 2

outer loop:

i=3

i<=N is false because i=3 and 3>2

So this outer loop ends.

Now the output of this program is:

B W

W B

Screenshot of this program along with its output is attached.

8 0
2 years ago
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