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tensa zangetsu [6.8K]
3 years ago
5

Write a program that can print a banner with rotated letters, such that they can be read from top to bottom. Specifically, each

letter in the message should be printed using multiple lines composed of * symbols. The code to print each letter should go into a separate method just for that letter (no parameters, no return value). Don’t worry, you don’t need to make 26 different methods, just 8: the seven letters in the message "Hello World" (D, E, H, L, O, R, and W) and a space. In your main method, you should ask the user to type a message. Then, loop through each character in the message and, for each character, if you have a corresponding method, call the method to print out the letter to the screen, ignoring any other characters. To help get you started, here is an example method for the letter D:
Computers and Technology
1 answer:
My name is Ann [436]3 years ago
4 0

Answer:

package Lab6Problema;

import java.util.Scanner;

public class Lab6Problema{

public static void H() // prints H

{

System.out.println();

System.out.println("** **");

System.out.println("** **");

System.out.println("******");

System.out.println("** **");

System.out.println("** **");

}

 

public static void E()//prints E

{

System.out.println();

System.out.println("******");

System.out.println("** ");

System.out.println("******");

System.out.println("** ");

System.out.println("******");

}

 

public static void L()//prints L

{

System.out.println();

System.out.println("* ");

System.out.println("* ");

System.out.println("* ");

System.out.println("* ");

System.out.println("******");

}

 

public static void O()//prints O

{

System.out.println();

System.out.println("******");

System.out.println("* *");

System.out.println("* *");

System.out.println("* *");

System.out.println("******");

}

 

public static void W()//prints W

{

System.out.println();

System.out.println("* *");

System.out.println("* *");

System.out.println("* ** *");

System.out.println("* ** *");

System.out.println("** **");

}

 

public static void R()//prints R

{

System.out.println();

System.out.println("***** ");

System.out.println("** **");

System.out.println("***** ");

System.out.println("** **");

System.out.println("** **");

}

 

public static void D()//prints D

{

System.out.println();

System.out.println("**** ");

System.out.println("* * ");

System.out.println("* *");

System.out.println("** * ");

System.out.println("**** ");

}

public static void blank() //prints blank space

{

System.out.println();

System.out.println();

System.out.println();

}

public static void main(String[] args) {

//just enter "Hello World" as input as specified

System.out.println("Enter a Message");

Scanner scan=new Scanner(System.in);

String myMessage=scan.nextLine(); // read msg using scanner

int numOfCharacters=myMessage.length();// calculate message lenth

for(int i=0;i<numOfCharacters;i++) // loop through each chracter in msg

{

// gets character at position i, and switch accordingly

switch(myMessage.charAt(i))

{

// if character is 'H' or 'h', call method H(). Similarly for other

// Hecharacters in message.

case 'H':

case 'h':H();break;

case 'E':

case 'e':E();break;

case 'L':

case 'l':L();break;

case 'O':

case 'o':O();break;

case 'W':

case 'w':W();break;

case 'R':

case 'r':R();break;

case 'D':

case 'd':D();break;

case ' ':blank();break;

}

}

}

}

Explanation:

You might be interested in
The compare_strings function is supposed to compare just the alphanumeric content of two strings, ignoring upper vs lower case a
Korolek [52]

Answer:

There is a problem in the given code in the following statement:

Problem:

punctuation = r"[.?!,;:-']"

This produces the following error:

Error:

bad character range

Fix:

The hyphen - should be placed at the start or end of punctuation characters. Here the role of hyphen is to determine the range of characters. Another way is to escape the hyphen - using using backslash \ symbol.

So the above statement becomes:

punctuation = r"[-.?!,;:']"  

You can also do this:

punctuation = r"[.?!,;:'-]"  

You can also change this statement as:

punctuation = r"[.?!,;:\-']"

Explanation:

The complete program is as follows. I have added a print statement print('string1:',string1,'\nstring2:',string2) that prints the string1 and string2 followed by return string1 == string2  which either returns true or false. However you can omit this print('string1:',string1,'\nstring2:',string2) statement and the output will just display either true or false

import re  #to use regular expressions

def compare_strings(string1, string2):  #function compare_strings that takes two strings as argument and compares them

   string1 = string1.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   string2 = string2.lower().strip()  # converts the string1 characters to lowercase using lower() method and removes trailing blanks

   punctuation = r"[-.?!,;:']"  #regular expression for punctuation characters

   string1 = re.sub(punctuation, r"", string1)  # specifies RE pattern i.e. punctuation in the 1st argument, new string r in 2nd argument, and a string to be handle i.e. string1 in the 3rd argument

   string2 = re.sub(punctuation, r"", string2)  # same as above statement but works on string2 as 3rd argument

   print('string1:',string1,'\nstring2:',string2)  #prints both the strings separated with a new line

   return string1 == string2  # compares strings and returns true if they matched else false

#function calls to test the working of the above function compare_strings

print(compare_strings("Have a Great Day!","Have a great day?")) # True

print(compare_strings("It's raining again.","its raining, again")) # True

print(compare_strings("Learn to count: 1, 2, 3.","Learn to count: one, two, three.")) # False

print(compare_strings("They found some body.","They found somebody.")) # False

The screenshot of the program along with its output is attached.

4 0
4 years ago
How do you program a computer
Vesna [10]
Here is a website about computer programming. Hope this helps. 
http://www.wikihow.com/Start-Learning-Computer-Programming
8 0
3 years ago
Write a complete function called lowestPosition() which takes as array (of double) and the number of elements used in the array
stiv31 [10]

Answer:

#include <bits/stdc++.h>

using namespace std;

int lowestPosition(double elements[],int n)//function to find position of lowest value element.

{

   int i,pos;

   double min=INT_MAX;

   for(i=0;i<n;i++)

   {

       if(elements[i]<min)

       {

           min=elements[i];

           pos=i;//storing index.

       }

   }

   return pos+1;//returning pos +1 to give the correct position in the array.

}

int main() {

double ele[500];

int n;

cout<<"Enter number of values"<<endl;

cin>>n;

cout<<"Enter elements"<<endl;

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

{

   cin>>ele[i];

}

int pos=lowestPosition(ele,n);//function call.

cout<<"The position of element with minimum value is "<<pos<<endl;

return 0;

}

Output:-

Enter number of values

5

Enter elements

4 6 7 5 1

The position of element with minimum value is 5

Explanation:

I have created a function lowestPosition() with arguments double array and it's size.I have taken a integer variable pos to store the index and after finding minimum and index i am returning pos+1 just to return correct position in the array.

3 0
4 years ago
What is the disk bandwidth? the total number of bytes transferred total time between the first request for service and the compl
Vikki [24]

Answer:

(c) the total number of bytes transferred divided by the total time between the first request for service and the completion on the last transfer

Explanation:

disk bandwidth

  • Disk read and write in bandwidth is strongly dependent on the size of the block.
  • Disk read / write bandwidth on Commodity hardware varies from slower like 10 MB/s  to faster like 500 MB/s.

This difference can be represented by the following rules:

  • Reading / writing big data is faster than reading / writing small blocks of data
  • Writing is faster than reading
  • Solid state drives are faster than spinning disks, especially for very small reads / rights

so we can say Disk bandwidth is the total number of bytes transferred divided by  the total time between the first request for service and the completion of the last transfer.

7 0
3 years ago
PLEASE HELP
lilavasa [31]

This assignment is required to be executed in python programming language. See the code below.

<h3>What is the code that displays word pairs that differ between the two sentences?</h3>

# taking two sentence as input s1 = input() s2 = input()  

# getting the words in both sentences in list w1 = s1.split() w2 = s2.split()  # looping through the word lists and checking if they are equal or not for i in range(len(w1)):  

# printing word pairs if they are not equal

if (w1[i] != w2[i]):

 print(w1[i],w2[i])    

Learn more about python programming language at;
brainly.com/question/26497128
#SPJ1

4 0
2 years ago
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