我的任务是
'编写一个函数 selectCoins 要求用户输入金额
(以便士为单位)然后输出每种面额的硬币数量(从 2 英镑起)
到 1p) 应该用来准确地弥补那个数量(使用尽可能少的
硬币数量)。例如，如果输入是 292，那么函数应该报告:
1 × £2、0 × £1、1 × 50p、2 × 20p、0 × 10p、0 × 5p、1 × 2p、0 × 1p。 (提示:使用整数
除法和余数)。

def selectCoins():
    twopound = 200
    onepound = 100
    fiftyp = 50
    twentyp = 20
    tenp = 10
    fivep = 5
    twop = 2
    onep = 1
    a = 0
    b = 0
    c = 0
    d = 0
    e = 0
    f = 0
    g = 0
    h = 0
    money = int(input('Enter how much money you have in pence'))

    while True:
        if money >= twopound:
            money = money - twopound
            a = a + 1
        elif money >= onepound:
            money = money - onepound
            b = b + 1
        elif money >= fiftyp:
            money = money - fiftyp
            c = c + 1
        elif money >= twentyp:
            money = money - twentyp
            d = d + 1
        elif money >= tenp:
            money = money - tenp
            e = e + 1
        elif money >= fivep:
            money = money - fivep
            f = f + 1
        elif money >= twop:
            money = money - twop
            g = g + 1
        elif money >= onep:
            money = money - onep
            h = h + 1
        else:
            money = 0
        break
    print(a,b,c,d,e,f,g,h)

最佳答案

由于您是编码新手，您应该开始在纸上编写您要遵循的过程，然后找出可以使用哪些工具来自动化此过程。

Important

Read the full answer in order!
Don't fall for the temptation of reading the code right away.

The solutions I provide are hidden, but you can read them hovering your mouse over them or clicking on them (if you're using StackExchange mobile app, touch the "spoiler" link in each block).

|  200 |  100 |   50 |   20 |   10 |    5 |    2 |    1 |
+------+------+------+------+------+------+------+------+
|   I  |   II |  III |   IV |    V |   VI |  VII | VIII |

Write the value:   292
Start with the first bin (denomination: 200)
Pick 1 coin from the bin
    The total amount picked from the bin is 200
    The remainder is 92
Strike the previous value
    The new value is 92
Move to the next bin (denomination: 100)
Pick 0 coins from the bin
    The total amount picked from the bin is 0
    The remainder is 92
Strike the previous value
    The new value is 92
Move to the next bin (denomination: 50)
Pick 1 coin from the bin
    The total amount picked from the bin is 50
    The remainder is 42
Move to the next bin (denomination: 20)
    Pick 2 coins from the bin
    The total amount picked from the bin is 20
    The remainder is 2
Move to the next bin (denomination: 10)
    Pick 0 coins from the bin
    The total amount picked from the bin is 0
    The remainder is 2
Move to the next bin (denomination: 10)
    Pick 0 coin from the bin
    The total amount picked from the bin is 0
    The remainder is 2
Move to the next bin (denomination: 5)
    Pick 0 coin from the bin
    The total amount picked from the bin is 0
    The remainder is 2
Move to the next bin (denomination: 2)
    Pick 1 coin from the bin
    The total amount picked from the bin is 2
    The remainder is 0
Done

292 // 200 == 2
292 % 200 == 92

92 // 100 == 0
92 % 100 == 92

...

# Initialize the value:
value = 292
# Initialize the denomination:
denomination = 200
# Calculate the amount of coins needed for the specified denomination
# and get the remainder (overwriting the value), in one single step:
coins, value = value // denomination, value % denomination
#              ^^^^^^^^^^^^^^^^^^^^^  ^^^^^^^^^^^^^^^^^^^^
#              |                      The remainder
#              The number of coins
#              (using integer division)

def selectCoins():
    twopound = 200
    onepound = 100
    fiftyp = 50
    twentyp = 20
    tenp = 10
    fivep = 5
    twop = 2
    onep = 1
    a = 0
    b = 0
    c = 0
    d = 0
    e = 0
    f = 0
    g = 0
    h = 0
    money = int(input('Enter how much money you have in pence')) # Example: 292
    # Calculate the number of coins needed and the remainder
    # The remainder will "overwrite" the value previously held in the "money" variable
    a, money = money // twopound, money % twopound # a = 1, money = 92
    b, money = money // onepound, money % onepound # b = 0, money = 92
    c, money = money // fiftyp,   money % fiftyp   # c = 1, money = 42
    d, money = money // twentyp,  money % twentyp  # d = 2, money = 2
    e, money = money // tenp,     money % tenp     # e = 0, money = 2
    f, money = money // fivep,    money % fivep    # f = 0, money = 2
    g, money = money // twop,     money % twop     # g = 1, money = 0
    e, money = money // onep,     money % onep     # e = 0, money = 0
    print(a,b,c,d,e,f,g,h)

denominations = [200, 100, 50, 20, 10, 5, 2, 1]

coins = [] # We'll use the '.append()' method to add elements to this list

n, money = money // denominations[0] , money % denominations[0]
    coins.append(n)

def select_coins_v2():
        denominations = [200, 100, 50, 20, 10, 5, 2, 1]
        coins = []
        money = int(input('Enter how much money you have in pence'))
        for i in range(len(denominations)):
            n, money = money // denominations[i], money % denominations[i]
            coins.append(n)
        print(coins)

def select_coins_v3():
        denominations = [200, 100, 50, 20, 10, 5, 2, 1]
        coins = []
        money = int(input('Enter how much money you have in pence'))
        for d in denominations:  # 'd' will hold the value of the denomination
            n, money = money // d, money % d
            coins.append(n)
        print(coins)

def select_coins_v4():
        denominations = [200, 100, 50, 20, 10, 5, 2, 1]
        coins = []
        money = int(input('Enter how much money you have in pence'))
        for d in denominations:  # 'd' will hold the value of the denomination
            n, money = money // d, money % d
            coins.append(n)
        number_of_coins = dict(zip(denominations, coins))
        print(number_of_coins)

关于python - 如何将钱(以便士计)转换为单独的硬币？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/52706139/