這段代碼應(yīng)該給我最少數(shù)量的硬幣（25 美分硬幣、5 美分硬幣、5 美分硬幣和 1 美分硬幣），這些硬幣加起來等于所欠的金額。當(dāng)我輸入的值是 0.25 的倍數(shù)時(shí)，它可以無縫工作。但是當(dāng)我在終端中輸入其他值時(shí)，它只是插入一個(gè)新行而不做任何事情。我怎么搞砸了？owed = float(input("how much change is owed?"))coins = 0if owed % 0.25 == 0: coins = owed / 0.25 print(int(coins)) exit()elif owed % 0.25 != 0: while owed > 0: if (owed - 0.25) >= 0: coins += 1 owed -= 0.25 elif (owed - 0.10) >= 0: coins += 1 owed -= 0.10 elif (owed - 0.05) >= 0: coins += 1 owed -= 0.05 elif (owed - 0.01) >= 0: coins += 1 owed -= 0.01 print(int(coins)) exit()
2 回答
POPMUISE
TA貢獻(xiàn)1765條經(jīng)驗(yàn) 獲得超5個(gè)贊
while由于浮點(diǎn)數(shù)的內(nèi)部表示引起的與浮點(diǎn)值相關(guān)的一些舍入錯(cuò)誤，您的程序正在運(yùn)行無限循環(huán)。因此，要修復(fù)這些錯(cuò)誤，我們可以使用round()函數(shù)在每次循環(huán)迭代owed結(jié)束時(shí)對(duì)值進(jìn)行四舍五入：while
elif owed % 0.25 != 0:
while owed > 0:
if (owed - 0.25) >= 0:
coins += 1
owed -= 0.25
elif (owed - 0.10) >= 0:
coins += 1
owed -= 0.10
elif (owed - 0.05) >= 0:
coins += 1
owed -= 0.05
elif (owed - 0.01) >= 0:
coins += 1
owed -= 0.01
owed = round(owed, 3) # In this line, we roundoff the value of owed
print(int(coins))
exit()
這很好用。
希望這可以幫助 ：）
浮云間
TA貢獻(xiàn)1829條經(jīng)驗(yàn) 獲得超4個(gè)贊
while由于浮點(diǎn)錯(cuò)誤，您的程序卡在循環(huán)中。嘗試在while循環(huán)中添加以下代碼，您會(huì)看到 whileowed確實(shí)變得無限小，它永遠(yuǎn)不會(huì)變?yōu)榱悖?/p>
...
while owed > 0:
print(owed)
...
輸出：
...
8.326672684688674e-17
8.326672684688674e-17
8.326672684688674e-17
8.326672684688674e-17
...
考慮將輸入乘以100然后將其作為整數(shù)處理：
owed = int(float(input("How much change is owed? $")) * 100)
quarters = int(owed / 25)
dimes = int((owed - quarters * 25) / 10)
nickels = int((owed - quarters * 25 - dimes * 10) / 5)
cents = int((owed - quarters * 25 - dimes * 10 - nickels * 5))
coins = (quarters + dimes + nickels + cents)
print('Quarters (${}): {}'.format(quarters*0.25, quarters))
print('Dimes (${}): {}'.format(dimes*0.1, dimes))
print('Nickels (${}): {}'.format(nickels*0.05, nickels))
print('Cents (${}): {}'.format(cents, cents))
print('Coins:', coins)
或者，如果您想堅(jiān)持使用貪心算法：
owed = int(float(input("How much change is owed? $")) * 100)
while owed > 0:
if (owed - 25) >= 0:
coins += 1
owed -= 25
elif (owed - 10) >= 0:
coins += 1
owed -= 10
elif (owed - 5) >= 0:
coins += 1
owed -= 5
elif (owed - 1) >= 0:
coins += 1
owed -= 1
coins = (quarters + dimes + nickels + cents)
print('Quarters (${}): {}'.format(quarters*0.25, quarters))
print('Dimes (${}): {}'.format(dimes*0.1, dimes))
print('Nickels (${}): {}'.format(nickels*0.05, nickels))
print('Cents (${}): {}'.format(cents, cents))
print('Coins:', coins)
輸出
>>> How much change is owed? $1.42
Quarters ($1.25): 5
Dimes ($0.1): 1
Nickels ($0.05): 1
Cents ($2): 2
Coins: 9
有關(guān)浮點(diǎn)限制的更多信息，請(qǐng)查看以下內(nèi)容：https ://docs.python.org/3.8/tutorial/floatingpoint.html
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