题解 | #自动售货系统# 找零的地方直接循环处理简单点

s = input().split(";")
 
 
# 商品
key = s[0].split(' ')
t1 = ['A1', 'A2', 'A3', 'A4', 'A5', 'A6']
t2 = map(int, key[1].split('-'))
goods_ct = dict(zip(t1, t2))
goods_value = {'A1': 2, 'A2': 3, 'A3': 4, 'A4': 5, 'A5': 8, 'A6': 6}
 
# 钱
t1 = [1, 2, 5, 10]
t2 = map(int, key[2].split('-'))
money_ct = dict(zip(t1, t2))
 
print("S001:Initialization is successful")
 
# 投币额
global cost
cost = 0
 
 
def compute(order):
    global cost
    if order[0] == 'q':
        if order == 'q 0':
            for good in goods_value.keys():
                print("{} {} {}".format(
                    good, goods_value[good], goods_ct[good]))
        elif order == 'q 1':
            for m in money_ct.keys():
                print("{} yuan coin number={}".format(m, money_ct[m]))
 
        else:
            print("E010:Parameter error")
 
    if order[0] == 'p':
        cost_temp = int(order[2:])
        if cost_temp not in [1, 2, 5, 10]:
            print("E002:Denomination error")
        elif cost_temp in [5, 10] and money_ct[1]*1 + money_ct[2]*2 < cost_temp:
            print("E003:Change is not enough, pay fail")
        elif sum(goods_ct.values()) == 0:
            print("E005:All the goods sold out")
        else:
            cost += cost_temp
            # 增加一张库存
            money_ct[cost_temp] += 1
            print(f"S002:Pay success,balance={cost}")
 
    if order[0] == 'b':
        name = order[2:]
        if name not in goods_value.keys():
            print("E006:Goods does not exist")
        elif goods_ct[name] == 0:
            print("E007:The goods sold out")
        elif cost < goods_value[name]:
            print("E008:Lack of balance")
        else:
            goods_ct[name] -= 1
            cost -= goods_value[name]
            print(f"S003:Buy success,balance={cost}")
 
    if order[0] == 'c':
        if cost == 0:
            print("E009:Work failure")
        else:
            re_ct = dict(zip([1, 2, 5, 10], [0, 0, 0, 0]))
            
            for n in [10, 5, 2, 1]:
                while money_ct[n] > 0 and cost - n >= 0:
                    re_ct[n] += 1
                    money_ct[n] -= 1
                    cost -= n
            for m in re_ct.keys():
                print("{} yuan coin number={}".format(m, re_ct[m]))
         
 
 
# 命令
orders = s[1:-1]
for order in orders:
    compute(order)

针对找零模块原来会遇到 8 = 5+2 而不是 8 = 2 + 2 +2 +2的情形增加了dfs回溯 这也太复杂了

r 22-18-21-21-7-20 0-23-10-6;p 2;p 2;p 2;p 2;c;

用这个自测看看你的是返回 2 5 还是 2 2 2 2 大部分人都不对 而且没有设计这样的用例 不是考点 只能说题目出的太难或者有点问题

s = input().split(";")


# 商品
key = s[0].split(' ')
t1 = ['A1', 'A2', 'A3', 'A4', 'A5', 'A6']
t2 = map(int, key[1].split('-'))
goods_ct = dict(zip(t1, t2))
goods_value = {'A1': 2, 'A2': 3, 'A3': 4, 'A4': 5, 'A5': 8, 'A6': 6}

# 钱
t1 = [1, 2, 5, 10]
t2 = map(int, key[2].split('-'))
money_ct = dict(zip(t1, t2))

print("S001:Initialization is successful")

# 投币额
global cost
cost = 0


def compute(order):
    global cost
    global money_ct
    global goods_ct
    global res
    if order[0] == 'q':
        if order == 'q 0':
            for good in goods_value.keys():
                print("{} {} {}".format(
                    good, goods_value[good], goods_ct[good]))
        elif order == 'q 1':
            for m in money_ct.keys():
                print("{} yuan coin number={}".format(m, money_ct[m]))

        else:
            print("E010:Parameter error")

    if order[0] == 'p':
        cost_temp = int(order[2:])
        if cost_temp not in [1, 2, 5, 10]:
            print("E002:Denomination error")
        elif cost_temp in [5, 10] and money_ct[1]*1 + money_ct[2]*2 < cost_temp:
            print("E003:Change is not enough, pay fail")
        elif sum(goods_ct.values()) == 0:
            print("E005:All the goods sold out")
        else:
            cost += cost_temp
            # 增加一张库存
            money_ct[cost_temp] += 1
            print(f"S002:Pay success,balance={cost}")

    if order[0] == 'b':
        name = order[2:]
        if name not in goods_value.keys():
            print("E006:Goods does not exist")
        elif goods_ct[name] == 0:
            print("E007:The goods sold out")
        elif cost < goods_value[name]:
            print("E008:Lack of balance")
        else:
            goods_ct[name] -= 1
            cost -= goods_value[name]
            print(f"S003:Buy success,balance={cost}")

    if order[0] == 'c':
        if cost == 0:
            print("E009:Work failure")
        else:
            re_ct = dict(zip([1, 2, 5, 10], [0, 0, 0, 0]))
            # for n in [10, 5, 2, 1]:
            #     while money_ct[n] > 0 and cost - n >= 0:
            #         re_ct[n] += 1
            #         money_ct[n] -= 1
            #         cost -= n
            # for m in re_ct.keys():
            #     print("{} yuan coin number={}".format(m, re_ct[m]))
            res = []
            tuibi(cost, money_ct, re_ct)
            res.sort(key=lambda x: (x[0], sum(x[1].values())))
            cost = res[0][0]
            solution = res[0][1]
            money_ct = res[0][2]
            for m in solution.keys():
                print("{} yuan coin number={}".format(m, solution[m]))


def tuibi(target, money_ct, solution):
    for n in [10, 5, 2, 1]:
        if money_ct[n] and target-n >= 0:
            money_ct[n] -= 1
            solution[n] += 1
            tuibi(target-n, money_ct, solution)
            # 还原
            money_ct[n] += 1
            solution[n] -= 1

    # 结束标志
    # 无钱可找
    res.append([target, solution.copy(), money_ct.copy()])


# 命令
orders = s[1:-1]
for order in orders:
    compute(order)