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sergejj [24]
3 years ago
6

Write a program that inputs a five-digit integer, spearates the integer into its digits and prints them seperated by three space

s each. [Hint: Use the ineger division and remainder operators.]
Computers and Technology
1 answer:
FromTheMoon [43]3 years ago
8 0
<h2>Explanation:</h2>

The source code and a sample output have been attached to this response.

The code has been written in Java and it contains comments explaining important parts of the code.

A few things that are worth noting are in the for loop used in the code;

<em><u>The loop goes from i = 4 to i = 0</u></em>

<em>When i = 4;</em>

=> (int) Math.pow(10, i) = (int) Math.pow(10, 4) = 10000

Then the <em>fiveDigit</em> is divided by 10000. Since this is an integer division, the first digit of the 5-digit number will be stored in <em>digit </em>which is then printed to the console.

Next, the remainder is calculated and stored in <em>fiveDigit</em>

<em>When i = 3;</em>

=> (int) Math.pow(10, i) = (int) Math.pow(10, 3) = 1000

Then the <em>fiveDigit</em> (which is the remainder when i = 4) is divided by 1000. Since this is an integer division, the second digit of the 5-digit number will be stored in <em>digit </em>which is then printed to the console.

Next, the remainder is calculated and stored in <em>fiveDigit</em>

<em>When i = 2;</em>

(int) Math.pow(10, i) = (int) Math.pow(10, 2) = 100

Then the <em>fiveDigit</em> (which is the remainder when i = 3) is divided by 100. Since this is an integer division, the third digit of the 5-digit number will be stored in <em>digit </em>which is then printed to the console.

Next, the remainder is calculated and stored in <em>fiveDigit</em>

<em>When i = 1;</em>

(int) Math.pow(10, i) = (int) Math.pow(10, 1) = 10

Then the <em>fiveDigit</em> (which is the remainder when i = 2) is divided by 100. Since this is an integer division, the fourth digit of the 5-digit number will be stored in <em>digit </em>which is then printed to the console.

Next, the remainder is calculated and stored in <em>fiveDigit</em>

<em>When i = 0;</em>

(int) Math.pow(10, i) = (int) Math.pow(10, 0) = 1

Then the <em>fiveDigit</em> (which is the remainder when i = 1) is divided by 1000. Since this is an integer division, the fifth digit of the 5-digit number will be stored in <em>digit </em>which is then printed to the console.

Next, the remainder is calculated and stored in <em>fiveDigit </em>and then the program ends.

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Program:
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Answer:

See explaination

Explanation:

#method to print menu & handle user choice

def print_menu(input_str):

#printing menu

print('MENU')

print('c - Number of non-whitespace characters')

print('w - Number of words')

print('f - Fix capitalization')

print('r - Replace punctuation')

print('s - Shorten spaces')

print('q - Quit\n')

#getting choice

choice=input('Choose an option: ').lower()

#identifying choice

if choice=='c':

#displaying number of non white space chars in input_str

print('Number of non-whitespace characters:',get_num_of_non_WS_characters(input_str))

elif choice=='w':

#displaying number of words in input_str

print('Number of words:',get_num_of_words(input_str))

elif choice=='f':

#fixing capitalization and getting updated string and count of values capitalized

input_str,count=fix_capilization(input_str)

#displaying results

print('Number of letters capitalized:',count)

print('Edited text:',input_str)

elif choice=='r':

#replacing punctuation, displaying updated text

input_str=replace_punctuation(input_str)

print('Edited text:', input_str)

elif choice=='s':

#shortening spaces and displaying updated text

input_str = shorten_space(input_str)

print('Edited text:', input_str)

#returning choice and input_str

return choice,input_str

#returns the number of non white space chars in input_str

def get_num_of_non_WS_characters(input_str):

count=0

#looping through each character in input_str

for i in input_str:

if not i.isspace():

#i is not a space

count+=1

return count

#returns the number of words in input_str

def get_num_of_words(input_str):

#splitting words into list of tokens by space

words=input_str.split(' ')

count=0

#counting all non empty strings in words list

for i in words:

if len(i)>0:

count+=1

return count

#method to fix capitalization and return updated string and count of letters updated

def fix_capilization(input_str):

count=0

beginning=True #starting letter should be capitalized

result=''

for i in input_str:

if beginning and i.isalpha():

#start of a sentence and i is alphabetic

if i.islower():

#converting i to upper case, incrementing count

i=i.upper()

count+=1

result+=i

#not start of a sentence

beginning=False

elif i in '?.!':

#i is either ? or . or !, next letter should be capitalized

beginning=True

result+=i

else:

#any other character

result+=i

return result,count

#method to replace exclamation and semicolons with period and comma respectively

def replace_punctuation(input_str,exclamationCount =0,semicolonCount=0):

result=''

for i in input_str:

if i=='!':

i='.'

exclamationCount+=1

elif i==';':

i=','

semicolonCount+=1

result+=i

print('Punctuation replaced')

#displaying replaced values counts

print('exclamationCount:',exclamationCount)

print('semicolonCount:',semicolonCount)

return result

#removes all double or more spaces in input_str

def shorten_space(input_str):

input_str=input_str.strip()

result=''

prev=None

for i in input_str:

if prev==None:

result+=i

elif i==' ':

if prev != ' ':

result+=i

else:

result+=i

prev=i

return result

if __name__ == '__main__':

#getting input, printing it

input_str=input('Enter a sample text:\n')

print('\nYou entered:',input_str)

choice=' '

#looping until choice becomes q

while choice!='q':

choice,input_str=print_menu(input_str)

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Answer:

in the paragraph

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The central limit theorem is known to be important to the study of sampling distributions because it enables us to disregard the shape of the population when the value of n is relatively large. Was this answer helpful?

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4.11 lesson practice <br> Need help
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Answer:

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Explanation:

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Through this system it is possible to solve complex problems such as scientific research, product development, analysis and predictions about certain phenomena, development of technology applied to health, etc.

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You are the leader of your school's Go Green Club, which supports environmentally friendly choices and lifestyles. Your club wil
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Answer:

(Answers may vary.)

My Considerations

The Go Green Club has a total of 100 members. I plan to photograph each of them in front of our club's "Go Green" sign. I have a deadline of 30 days to deliver the ID cards. Therefore, I have to ensure that all the cards are printed and ready for delivery in 20 days, with 10 days of buffer time (in case the cards require any edits or changes). Additionally, after I have submitted the draft for review, I will need the coordinator's final approval on the design and layout of the ID card.

Hardware Requirements

  • a desktop computer or laptop with adequate memory to run the software to create the ID cards
  • some external storage devices, such as pen drives and DVD disks, to share the design and layout for the coordinator to review; I will also need these devices to store the final digital file for print
  • a color scanner to scan hand-drawn artwork and signatures, which I will place on the ID card
  • a digital camera to capture photos of members, and to transfer them to the computer/laptop
  • a printer to make hard copies of the cards.
  • Software Requirements

I intend to use Scribus to create the ID cards because it is a free DTP application, and has most of the common features I'd need to create an ID card.

Formatting Techniques

  • I plan to use the following techniques to design the club ID cards.
  • I'll use the landscape or horizontal orientation. This will enable proper spacing for the ID card elements.
  • I'm planning to use a card measuring 3.370 inches × 2.125 inches. A card this size will ensure that all the elements of the card are visible.
  • I am planning to place the club logo on the top left corner, because the eye normally follows the path of the letter Z. This'll help the viewer to see the logo and identify/recognize the club.
  • Next I'll place the club name; I'll use the top-center alignment and also vertically align it to the logo. I'll use the sans serif font for the club name.
  • Following the Z-path rule, I'll place the member's photo to the right and the member's name under the photo. In this position, the photo and the name will be visible, and will not get mixed up with other elements. I'll print the member name with a serif font.
  • Finally, again in line with the Z-path rule, I'll position the club seal and the club director's signature at the right lower bottom. Although these elements are essential on an ID card, they do not require as much viewer attention as the club logo, name, and member photo.

Explanation:

I used Canva for this card.

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