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IRISSAK
2 months ago
6

Some numbers are formed with closed paths. the digits 0, 4, 6 and 9 each have 1 closed path and 8 has 2. None of the other numbe

rs is formed with a closed path. Given a number determine the total number of closed paths in each of its digits combined.
Computers and Technology
1 answer:
Natasha_Volkova [1K]2 months ago
5 0

Answer:

def cal(n):

s=0

while n>0:

r=n%10

if(r==0 or r==4 or r==6 or r==9):

s+=1

elif r==8:

s+=2

n=n//10

print(s)

n=int(input("enter number:"))  

print(n)

cal(n)

Explanation:

  • A function is defined to calculate the number of closed paths.
  • A variable is initialized to hold the count of these paths.
  • As long as the number remains positive , retrieve the last digit of n.
  • Trim the last digit to shorten the number.
  • Invoke the function to find the path count.
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Answer:

input price

set change to one hundred minus price

set quarter to change divided by 25

set change to change mod 25

set dime to change divided by 10

set change to change mod 10

set nickel to change divided by 5

set change to change mod 5

set penny to change

print "You acquired an item for price cents and paid me a dollar. Your change is: "

if quarter is equal to 1

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else if quarter is greater than 1

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if dime is equal to 1

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else if dime is greater than 1

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if nickel is equal to 1

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if penny is equal to 1

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else if penny is greater than 1

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Explanation:

A clearer way to express the pseudocode above could involve using alternative names for change as:

input price

set change to one hundred minus price

set quarter to change divided by 25

set quarter_remainder to change mod 25

set dime to quarter_remainder divided by 10

set dimes_remainder to quarter_remainder mod 10

set nickel to dimes_remainder divided by 5

set nickel_remainder to dimes_remainder mod 5

set penny to nickel_remainder

print "You purchased an item for" price"cents and provided a dollar. Your change is: "

if quarter is equal to 1

print quarter "quarter"

else if quarter is greater than 1

print quarter "quarters"

if dime is equal to 1

print dime "dime"

else if dime is greater than 1

print dime "dimes"

if nickel is equal to 1

print nickel "nickel"

else if nickel is greater than 1

print nickel "nickels"

if penny is equal to 1

print penny "penny"

else if penny is greater than 1

print penny "pennies"

Next, I will illustrate the pseudocode through an example:

Let's assume price = 75

Thus, change = 100 - price = 100 - 75 = 25

change = 25

quarter = change/25 = 25/25 = 1

quarter = 1

quarter_remainder = change % 25 = 25%25 = 0

dime = 0/10 = 0

At this point, all additional values are 0.

The following message will be displayed on the screen:

You acquired an item for 75 cents and paid me a dollar. Your change is:

Now the program proceeds to the conditional statement

if quarter == 1

This condition holds true because the value of quarter is 1

print quarter "quarter"

Subsequently, this line will appear on the screen:

1 quarter

Thus, the full output of the pseudocode is:

You acquired an item for 75 cents and paid me a dollar. Your change is:

1 quarter

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