通过单数/复数差异做出改变

Question

我在编写 class 的初学者脚本中，我们需要编写一些代码来根据 integer 输入进行更改。 它还需要为每种硬币类型打印正确的复数。 我在让我的代码工作时遇到了一些麻烦，因为它在它执行的第一个操作处停止，我不知道如何让它继续。 我猜这与我如何布置所有这些 elif 语句有关，我不确定如果代码仍有剩余部分可以进行更改，那么让代码继续执行的正确做法是：

# Get input for amount of change
change = int(input())

# If change is less than 1, print 'No change'
if change < 1:
    print('No change')

# If change is over 199, subtract and print 'X Dollars'
elif change > 199:
    print(change // 100, 'Dollars\n')
    change = change % 100

# If change is 100-199, subtract and print '1 Dollar'
elif 99 < change <= 199:
    print('1 Dollar\n')
    change = change % 100

# If remaining change is 50-99, subtract and print 'X Quarters'
elif 49 < change <= 99:
    print(change // 25, 'Quarters\n')
    change = change % 25

# If remaining change is 25-49, subtract and print '1 Quarter'
elif 24 < change <= 49:
    print('1 Quarter\n')
    change = change % 25

# If remaining change is 25, print '1 Quarter' and set change to 0
elif change == 25:
    print('1 Quarter')
    change = 0

# If remaining change is 20-24, subtract and print '2 Dimes'
elif 19 < change <= 24:
    print('2 Dimes\n')
    change = change % 10

# If remaining change is 10, print '1 Dime' and set change to 0
elif change == 10:
    print('1 Dime')
    change = 0

# If remaining change is 5-9, subtract and print '1 Nickel'
elif 4 < change <= 9:
    print('1 Nickel\n')
    change = change % 5

# If remaining change is 2-4, subtract and print 'X Pennies'
elif 1 < change <= 4:
    print(change // 1, 'Pennies')
    change = change % 1

# If remaining change is 1, print '1 Penny' and set change to 0
elif change == 1:
    print('1 Penny')
    change = 0

谁能给我一个想法？ 感谢您的时间。

Answer 1

始终从更改的 rest 中减去您已经匹配的内容。 然后继续使用 IF 而不是 ELIF，因为如果匹配，elif 将永远不会再匹配一次。

# Get input for amount of change
change = int(input())

# If change is less than 1, print 'No change'
if change < 1:
    print('No change')

# If change is over 199, subtract and print 'X Dollars'
if change > 199:
    print(change // 100, 'Dollars\n')
    change = change - change // 100 * 100

# If change is 100-199, subtract and print '1 Dollar'
if 99 < change <= 199:
    print('1 Dollar\n')
    change = change - 100

# If remaining change is 50-99, subtract and print 'X Quarters'
if 49 < change <= 99:
    print(change // 25, 'Quarters\n')
    change = change - change // 25 * 25
# ....

等等...

Answer 2

如果我的问题是正确的，那就是根据不断减少直到为零的“变化”打印dollars / dollar [和子价值]。

让我澄清一下（不是美元用户：）：

• 1 便士 = 1 美分
• 1 镍 = 5 美分
• 1 角钱 = 10 美分
• 1 季度 = 25 美分
• 1 美元 = 100 美分

因此，如果输入是1451美分，则预期的 output 是： 1451 = 14 Dollars 2 Quarters 0 Dime 0 Nickel 1 Penny

你可以这样做：

change = int(input())
wt = {'Dollar': 0, 'Quarter': 0, 'Dime': 0, 'Nickel': 0, 'Penny': 0} # weight

wt['Dollar'] = change // 100
change = change % 100
wt['Quarter'] = change // 25
change = change % 25
wt['Dime'] = change // 10
change = change % 10
wt['Nickel'] = change // 5
change = change % 5
wt['Penny'] = change // 1
change = change % 1

for key, value in wt.items():
    print(f'{value} {key}s' if value > 1 else f'{value} {key}', end=' ')
print()

Output：

➜ python file.py 
112341
1123 Dollars 1 Quarter 1 Dime 1 Nickel 1 Penny 


➜ python file.py 
24
0 Dollar 0 Quarter 2 Dimes 0 Nickel 4 Pennys # you might wanna add some small condition to rectify this :)


➜ python file.py 
521
5 Dollars 0 Quarter 2 Dimes 0 Nickel 1 Penny