Write C program, vending machine C program generation, Generation C language Program work

Problem description

Assuming that there are multiple currencies [1,x1,x2,x3,..., xn] in a country, the country's vending machine will follow a principle of finding zero-"The total number of change is the least". So, how to write programs that help the vending machine automatically
Where's the change?

Problem analysis

The most direct way to solve this problem is to be exhaustive. Suppose you need change Y, then all the options for change are listed through all the currencies less than Y, and then the least total number is compared. This kind of thinking needs to be calculated
There is a great deal of repetition and time complexity.
An effective solution to a similar problem is to use the idea of dynamic programming to deal with it.

The minimum number of currencies needed to solve the change

For example, a currency with a 1,5,10,25 value:

The key to solving this problem is to use similar equations to shrink the scale of the problem. When the problem narrowed to "how many currencies are needed to find $0", the answer to the question is clearly 0.
What we need to do extra is to create a list of requirements that are smaller than the change needs to be calculated.

# Need_change for the amount that needs change,
# currency_list is a list of the denominations of the country's currency,
# num_list is the minimum amount of money needed for change, num_list length is at least (need_change+1)
def givechange (Need_change, Currency_list, num_list):
For change in range (need_change+1): #从0开始计算最少需要的货币数
For currency in Currency_list: #遍历每一种货币
if (change-currency >= 0) and (Num_list[change-currency]+1<num_list[change]): #计算最少货币需求数
Num_list[change] = num_list[change-currency] + 1
Return

def main ():
Need_change = 63
Currency_list = [1,5,10,21,25]
Num_list = List (range (need_change+1)) #初始化num_list为0到need_change, Total (need_change+1)
Givechange (Need_change, Currency_list, Num_list)
Print ("%d requires%d currency for change"% (Need_change, Num_list[need_change]))
if __name__ = = "__main__":
Main ()
The result of the operation is:

63 need 3 Currency for change
Just outputting the correct number of currencies is not enough, and we also need to export the specific denominations of the currency.

Solve the problem of automatic change

As is common, the shortest path is the same as the record. In order to output the change of the currency in which the face value is to be found, we need to record the minimum used currency for each step based on the previous step.
To do this, we need to add a list to record this value.

# Need_change for the amount that needs change,
# currency_list is a list of the denominations of the country's currency,
# num_list is the minimum amount of money needed for change, num_list length is at least (need_change+1)
# used_list is the minimum number of currencies that need change, the same length as Num_list
def givechange (Need_change, Currency_list, Num_list, used_list):
For change in range (need_change+1): #从0开始计算最少需要的货币数
For currency in Currency_list: #遍历每一种货币
if (change-currency >= 0) and (Num_list[change-currency]+1<=num_list[change]): #计算最少货币需求数
Num_list[change] = num_list[change-currency] + 1
Used_list[change] = Currency #记录消耗的货币
Return

# Return the required currency
def showchange (Need_change, used_list):
Give_list = []
While Need_change > 0:
Give_list.append (Used_list[need_change])
Need_change-= Used_list[need_change]
Give_list.sort () #排序
Return give_list

def main ():
Need_change = #需要找零的钱数
Currency_list = [1,5,10,21,25] # List of currency denominations of the country
Num_list = List (range (need_change+1)) #初始化num_list为0到need_change, Total (need_change+1)
Used_list = List (range (need_change+1)) #初始化used_list为0到need_change, Total (need_change+1)
Givechange (Need_change, Currency_list, Num_list, Used_list)
Print ("%d requires%d currency for change"% (Need_change, Num_list[need_change]))
Give_list = Showchange (Need_change, Used_list)
Print (give_list)

if __name__ = = "__main__":
Main ()
The result of the calculation is:

64 need 4 Currency for change
[1, 21, 21, 21]

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Write C program, vending machine C program generation, Generation C language Program work