最小硬币更换问题-回溯

Question

我正试图用最少的硬币数量来解决硬币更换问题使用回溯方法。

我实际上已经完成了，但我想添加一些选项，按单位打印硬币数量，而不仅仅是总金额。

下面是我的python代码。

def minimum_coins(coin_list, change):
min_coins = change 
if change in coin_list: 
return 1
else:
for coin in coin_list:
if coin < change: 
num_coins = 1 + minimum_coins(coin_list, change - coin)
if num_coins < min_coins:
min_coins = num_coins
return min_coins
coin_list = []
unit = input("Enter the Coin Unitn")
#1 10 15 20
change = int(input("Enter the Total Changen"))
#73
unit = unit.split()
for i in unit:
coin_list.append(int(i))
min = minimum_coins(coin_list, change)
#print("You need 3 1's, 1 10's, 0 15's, 3 20's") <- I want to implement this option.
print("Total number of coins required is %s." % min) #7

我正在努力在代码中实现这一点。

由于回溯很复杂，我不知道如何按单位检查硬币数量。

Answer 1

可能的方式：

def minimum_coins(coin_list, change):
min_coins = change
if change in coin_list:
return 1, [change]
else:
cl = []
for coin in coin_list:
if coin < change:
mt, t = minimum_coins(coin_list, change - coin)
num_coins = 1 + mt
if num_coins < min_coins:
min_coins = num_coins
cl = t + [coin]
return min_coins, cl
change = 73
coin_list = [1, 10, 15, 20]
min, c = minimum_coins(coin_list, change)
#print("You need 3 1's, 1 10's, 0 15's, 3 20's") <- I want to implement this option.
print("Total number of coins required is %s." % min, c) #7
>>>Total number of coins required is 7. [20, 20, 20, 10, 1, 1, 1]

Answer 2

# create dictionary from list of notes

#i am newbie in python
#the code is not mine, i just found it on the net
notes = [ 200, 100, 50, 20, 10, 5, 1]
notesCount = {}
x=122

用于纸币中的纸币：

if x >= note:
notesCount[note] = x//note
x=x%note
print(notesCount)
values = notesCount.values()
print(values)
total = sum(values)
print(total)

Answer 3

give_back = 122
return_bills = []
bill_stock = [ 
{ "bill_type":200,    "instock":1 },
{ "bill_type":100,    "instock":1 },
{ "bill_type":50,     "instock":1 },
{ "bill_type":20,     "instock":1 },
{ "bill_type":10,     "instock":1 },
{ "bill_type":5,      "instock":1 },
{ "bill_type":1,      "instock":1 },
]
for item in bill_stock:
# return when give back is 0
if give_back == 0: break
# get maximum amount to give back on the highest bill type
qty = give_back // item["bill_type"]
qty = min(qty, item["instock"])
if qty > 0:
give_back -= (item["bill_type"] * qty)
return_bills.append({
"bill_type": item["bill_type"],
"qty": qty
})
# display what to give back
for item in return_bills:
quantity = item["qty"]
bill_type = item["bill_type"]
print(f"{quantity}X ---- {bill_type}")
# output:
# > py .calc.py
# 1X ---- 100
# 1X ---- 20
# 1X ---- 1