Question: Party budget planning - complexPrompt the user to enter their current budget and the number of people who need to share a meal.You are ordering food from a restaurant that has two menu items:Tacos $4Empanadas $3Calculate and print out all the options of meals where the budget is used entirely(where possible) and each person has an equal number of items to eat. They may eat different things, but everyone gets the same number of food items, e.g. 2 tacos or 1 empanada and 1 taco or 2 empanadas.Hints: Here are some things that might help you in your solution.If total items purchased does not equally divide by the number of diners, then abandon that combination.It may help to calculate the max number of empanadas that the budget can buy and also the max number of tacos that the budget could buy. These can be thought of as upper bounds in your loops.In your solution, keep track of items per personAlso, keep track of total number of viable solutions, as that might help you determine when there is no solution.My solution is printing "You cannot buy a meal with these options." multiple times even when it is possible to buy a meal. Also my solution does show "Meal options:" before printing all of the alternatives. I've attached the outcome that the program should be displaying and my current code. Enter budget for meal: $100How many people are eating: 6Meal options:$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.$99 buys 21 empanadas and 9 tacos with $1 change.5 items per person.$100 buys 20 empanadas and 10 tacos without change. 5 items per person.orEnter budget for meal: $33How many people are eating: 15You cannot buy a meal with these options. def get_possible_meals ( budget, num_persons):tacos_cost = 4empanandas_cost= 3budget // tacos_costbudget // empanandas_costmax tacos =max_empanadasfeasible = Falsefor num_tacos in range (max_empanadas + 1):num_empanadasif num_empanadas > 0 and not (num_empanadas + num_tacos) % num_persons:feasible = True(budgetnum_tacos * tacos_cost) // empanandas_costtotal_costnum_tacos * tacos_cost + num_empanadas * empanandas_cost%3Dchangebudgettotal_cost%3Dprint(f'${total_cost} buys {num_empanadas} empanadas and {num_tacos} tacos with ${change} change.')print(f'{(num_empanadas + num_tacos) // num_persons} items per person.')if not feasible:print(f'You cannot buy a meal with these options.')def main() :budgetint(input('Enter budget for meal: $'))num_personsint(input('How many people are eating: '))get_possible_meals(budget, num_persons)main__':ifname==main()→ Enter budget for meal: $100How many people are eating: 6You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.You cannot buy a meal with these options.$98 buys 22 empanadas and 8 tacos with $2 change.5 items per person.

Question: Party budget planning - complex Prompt the user to enter their current budget and the number of people who need to share a meal. You are ordering food from a restaurant that has two menu items: Tacos $4 Empanadas $3 Calculate and print out all the options of meals where the budget is used entirely(where possible) and each person has an equal number of items to eat. They may eat different things, but everyone gets the same number of food items, e.g. 2 tacos or 1 empanada and 1 taco or 2 empanadas. Hints: Here are some things that might help you in your solution. If total items purchased does not equally divide by the number of diners, then abandon that combination. It may help to calculate the max number of empanadas that the budget can buy and also the max number of tacos that the budget could buy. These can be thought of as upper bounds in your loops. In your solution, keep track of items per person Also, keep track of total number of viable solutions, as that might help you determine when there is no solution. My solution is printing "You cannot buy a meal with these options." multiple times even when it is possible to buy a meal. Also my solution does show "Meal options:" before printing all of the alternatives. I've attached the outcome that the program should be displaying and my current code.

Programming Logic & Design Comprehensive

9th Edition

ISBN:9781337669405

Author:FARRELL

Publisher:FARRELL

Chapter4: Making Decisions

Section: Chapter Questions

Problem 18RQ

See similar textbooks

Related questions

Question

Question: Party budget planning - complex

Prompt the user to enter their current budget and the number of people who need to share a meal.

You are ordering food from a restaurant that has two menu items:

Tacos $4
Empanadas $3

Calculate and print out all the options of meals where the budget is used entirely(where possible) and each person has an equal number of items to eat. They may eat different things, but everyone gets the same number of food items, e.g. 2 tacos or 1 empanada and 1 taco or 2 empanadas.

Hints: Here are some things that might help you in your solution.

If total items purchased does not equally divide by the number of diners, then abandon that combination.
It may help to calculate the max number of empanadas that the budget can buy and also the max number of tacos that the budget could buy. These can be thought of as upper bounds in your loops.
In your solution, keep track of items per person
Also, keep track of total number of viable solutions, as that might help you determine when there is no solution.

My solution is printing "You cannot buy a meal with these options." multiple times even when it is possible to buy a meal. Also my solution does show "Meal options:" before printing all of the alternatives.

I've attached the outcome that the program should be displaying and my current code.

Enter budget for meal: $100
How many people are eating: 6
Meal options:
$98 buys 22 empanadas and 8 tacos with $2 change.
5 items per person.
$99 buys 21 empanadas and 9 tacos with $1 change.
5 items per person.
$100 buys 20 empanadas and 10 tacos without change. 5 items per person.
or
Enter budget for meal: $33
How many people are eating: 15
You cannot buy a meal with these options.

def get_possible_meals ( budget, num_persons):
tacos_cost = 4
empanandas_cost
= 3
budget // tacos_cost
budget // empanandas_cost
max tacos =
max_empanadas
feasible = False
for num_tacos in range (max_empanadas + 1):
num_empanadas
if num_empanadas > 0 and not (num_empanadas + num_tacos) % num_persons:
feasible = True
(budget
num_tacos * tacos_cost) // empanandas_cost
total_cost
num_tacos * tacos_cost + num_empanadas * empanandas_cost
%3D
change
budget
total_cost
%3D
print(f'${total_cost} buys {num_empanadas} empanadas and {num_tacos} tacos with ${change} change.')
print(f'{(num_empanadas + num_tacos) // num_persons} items per person.')
if not feasible:
print(f'You cannot buy a meal with these options.')
def main() :
budget
int(input('Enter budget for meal: $'))
num_persons
int(input('How many people are eating: '))
get_possible_meals(budget, num_persons)
main__':
if
name
==
main()
→ Enter budget for meal: $100
How many people are eating: 6
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
You cannot buy a meal with these options.
$98 buys 22 empanadas and 8 tacos with $2 change.
5 items per person.

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 1 images

SEE SOLUTION Check out a sample Q&A here