题解 | #自动售货系统#
自动售货系统
https://www.nowcoder.com/practice/cd82dc8a4727404ca5d32fcb487c50bf
from sys import flags
dict_goods = {
"A1":0,
"A2":0,
"A3":0,
"A4":0,
"A5":0,
"A6":0
}
dict_prices = {
"A1":2,
"A2":3,
"A3":4,
"A4":5,
"A5":8,
"A6":6
}
dict_money = {
1:0,
2:0,
5:0,
10:0
}
def sold_out(*good):
# 可变参数总是元组形式
if good[0] in dict_goods:
if dict_goods[good[0]] == 0:
return True
for tmp in dict_goods:
if dict_goods[tmp] != 0:
return False
return True
rest_money = 0
class found(Exception):
pass
def refund():
result_10 = 0
result_5 = 0
result_2 = 0
result_1 = 0
flag = False
rest_now = rest_money
# print(f'rest_now:{rest_now}')
try:
for n10 in range(dict_money[10],-1,-1):
# 四个数据位一组 需要每轮循环重置一次
result_10 = n10
result_5 = 0
result_2 = 0
result_1 = 0
rest_now = rest_money
# 从10块钱找起时恢复余额 在第一趟找不到的时候重新找
if rest_now == n10*10:
# rest_now 等于 当前用于找零的10块钱总额 找零成功
rest_now-=n10*10
result_10 = n10
flag = True
raise found
elif rest_now < n10*10:
# 10块找多了 得继续减少
continue
else:
# rest_now 大于 当前用于找零的10块钱总额
rest_now-=n10*10
result_10 = n10
# 检查下一种面值前 记录当前取用的值
for n5 in range(dict_money[5],-1,-1):
if rest_now == n5*5:
rest_now -= n5*5
result_5 = n5
flag = True
raise found
elif rest_now < n5*5:
continue
else:
rest_now -= n5*5
result_5 = n5
for n2 in range(dict_money[2],-1,-1):
if rest_now == n2*2:
rest_now -= n2*2
result_2=n2
flag = True
raise found
elif rest_now < n2*2:
continue
else:
rest_now -= n2*2
result_2 = n2
for n1 in range(dict_money[1],-1,-1):
if rest_now == n1*1:
rest_now -= n1*1
result_1=n1
flag = True
raise found
elif rest_now < n1*1:
continue
else:
# rest_now > n1*1 说明1块钱不够找 break进下一轮2块钱循环
result_1 = n1
break
except found:
print(f'{1} yuan coin number={result_1}')
print(f'{2} yuan coin number={result_2}')
print(f'{5} yuan coin number={result_5}')
print(f'{10} yuan coin number={result_10}')
dict_money[1]-=result_1
dict_money[2]-=result_2
dict_money[5]-=result_5
dict_money[10]-=result_10
while True:
try:
orders = input().split(';')[:-1]
# print(orders)
for order in orders:
if order[0] == 'r':
# 执行初始化
list_g = order.split(' ')[1].split('-')
list_m = order.split(' ')[2].split('-')
# print(f'list_g {list_g} list_m {list_m}')
i = 0
for tmp in dict_goods.keys():
dict_goods[tmp] = int(list_g[i])
i+=1
i = 0
for tmp in dict_money.keys():
dict_money[tmp] = int(list_m[i])
i+=1
# print(dict_goods)
# print(dict_money)
print("S001:Initialization is successful")
elif order[0] == 'p':
# # print(f'order {order}')
# # 检测投币
money = int(order.split(' ')[1])
# # print(f'money {money}')
if money not in dict_money.keys():
# 1.非法输入 报错
print('E002:Denomination error')
elif money != 1 and money != 2 and 1*dict_money[1]+2*dict_money[2] < money:
# ？ 2.零钱不足
print("E003:Change is not enough, pay fail")
elif sold_out(0):
# 3.售罄 报错
print('E005:All the goods sold out')
else:
# 4.成功
rest_money+=money
dict_money[money]+=1
# 存入余额
print(f"S002:Pay success,balance={rest_money}")
pass
elif order[0] == 'b':
# print(f'order :{order}')
# 购买商品
good = order.split(' ')[1]
# print(f'good:{good}')
if good not in dict_goods:
# 1.不在列表
print('E006:Goods does not exist')
elif sold_out(good):
# 2.售罄
print('E007:The goods sold out')
elif rest_money < dict_prices[good]:
# 3.余额不足
print('E008:Lack of balance')
else:
# 4.购买成功
rest_money-=dict_prices[good]
dict_goods[good]-=1
print(f'S003:Buy success,balance={rest_money}')
elif order[0] == 'c':
# 退币
if rest_money == 0:
# 1.余额不足
print('E009:Work failure')
else:
# 2.退币 打印退币信息
refund()
# 重置余额
rest_money = 0
pass
elif order[0] == 'q':
# print(f'order [q] {order}')
if len(order) == 3:
list_sorted = []
for good in dict_goods:
list_sorted.append([good, dict_prices[good],dict_goods[good]])
list_sorted.sort(key=lambda x:x[2],reverse=False)
# print('sorted success')
# 按余量排序 升序 余量一致时保持原商品名顺序
if order[2] == '0':
# print("0 success")
for good in list_sorted:
# print(f'good:{good}')
print(' '.join(list(map(str,good))))
# 0查询商品
elif order[2] == '1':
# print("1 success")
for money in dict_money:
print(f'{money} yuan coin number={dict_money[money]}')
# 1查询零钱
else:
print("E010:Parameter error")
except:
break
