通过 Singular/Plural 差异化做出改变

Question

我正在编写脚本初学者 class，我们需要编写一些代码来根据整数输入进行更改。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时停止，我不确定如何让它继续。我猜这与我如何布置所有这些 elif 语句有关，我不确定正确的做法是让代码继续，如果它仍然有剩余的变化：

# 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

谁能给我个主意？谢谢你的时间。

Answer 1

始终从其余更改中减去您已经匹配的内容。然后继续使用 IF 而不是 ELIF，因为 elif 将永远不会再次匹配之前的 if 匹配。

# 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

如果我答对了你的问题，那就是根据 'change' 打印 dollars/dollar [和子值]，它会一直减少直到为零。

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 = 10 美分
1 夸特 = 25 美分
1 美元 = 100 美分

因此，如果输入是 1451 美分，则预期输出是：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()

输出：

➜ 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

通过 Singular/Plural 差异化做出改变

Making Change with Singular/Plural Differentiation

python

plural