Python：你如何找到最简单的可分性形式

Question

I am working on a program where the user inputs the number of people going on a trip and the program determines the required number of each vehicle type to be used. Saying that: Car - 4 passengers Minivan - 7 passengers Small Bus - 15 passengers Big Bus - 60 passengers

This is what I have so far:

car_capacity = int(4)
minivan_capacity = int(7)
short_bus_capacity = int(15)
full_size_bus_capacity = int(60)

cars_needed = 0
minivan_needed = 0
short_buses_needed = 0
full_size_buses_needed = 0

passengers = int(input("How many people will be going?\n"))
if passengers < 1:
    print("Nobody is going, therefore no vehicles will be needed.")

while passengers > 0:
    print(passengers % 60)

Basically I'm trying to find out the fewest Vehicles needed. Ex (Input = 63. Output 1 Big Bus, 1 Small)

Please don't think I'm asking for answers, I just need a clue or a hint of what to research or how to approach this

Answer 1

passengers = int(input("How many people will be going?\n"))
if passengers < 1:
    print("Nobody is going, therefore no vehicles will be needed.")

# make it a dictionary, so we can loop through
vehicles = {
    4: 'car',
    7: 'minivan',
    15: 'short_bus',
    60: 'full_size_bus'
} 
vehicles_needed = {}

# a helper to check one type of vehicle
def check_vehicle(passengers, capacity, name):
    if passengers >= capacity:
        amount = passengers / capacity
        # substract the passengers 
        passengers -= (amount * capacity)
        return {name: amount}, passengers
    return {}, passengers

# loop through all capacities from the biggest one
for k, v in sorted(vehicles.items(), reverse=True):
    # if any vehicle needed, it will be put in the vehicles_needed
    vehicle_update, passengers = check_vehicle(passengers, k, v)
    vehicles_needed.update(vehicle_update)
# if there is a leftover, add one car
if passengers > 0:
    vehicles_needed.update({'car':  vehicles_needed.get('car', 0) + 1})

print vehicles_needed # 67 -> {'minivan': 1, 'full_size_bus': 1}