我正在嘗試構(gòu)建一個(gè)功能，當(dāng)人們輸入金額后，它會(huì)顯示他們需要的最小硬幣或紙幣數(shù)量。但這有什么方法可以讓我改變它，這樣它就不會(huì)打印未使用的硬幣的名稱和數(shù)量嗎？（作為初學(xué)者）感謝您的幫助?。梢杂胒or循環(huán)來處理嗎？）
3 回答
慕的地10843
TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個(gè)贊
循環(huán)可用于遍歷筆記列表，并且在循環(huán)內(nèi)，如果發(fā)現(xiàn)有任何筆記被計(jì)數(shù)，則可以打印該筆記。
notes=[1000,500,100,50,20,10,5,2,1]
amount=int(input('Enter an amount: '))
print('Total number of notes/coins=')
for notesAmount in notes:
if amount>=notesAmount:
notesCount=amount//notesAmount
amount%=notesAmount
if notesCount>0:
print(notesAmount, ":", notesCount)
慕少森
TA貢獻(xiàn)2019條經(jīng)驗(yàn) 獲得超9個(gè)贊
不要為每個(gè)面額保留一個(gè)變量，而是保留一個(gè)字典并根據(jù)所使用的面額更新 key: val 。查看代碼
amount=int(input('Enter an amount: '))
denominations = dict()
print('Total number of notes/coins=')
if amount>=1000:
denominations['1000'] = amount//1000
amount%=1000
if amount>=500:
denominations['500'] = amount//500
amount= amount%500
if amount>=100:
denominations['100'] = amount//100
amount= amount%100
if amount>=50:
denominations['50'] = amount//50
amount= amount%50
if amount>=20:
denominations['20'] = amount//20
amount= amount%20
if amount>=10:
denominations['10'] = amount//10
amount= amount%10
if amount>=5:
denominations['5'] = amount//5
amount= amount%5
if amount>=2:
denominations['2'] = amount//2
amount= amount%2
if amount>=1:
denominations['1'] = amount//1
for key, val in denominations.items():
print(f"{key}: {val}")
Enter an amount: 523
Total number of notes/coins=
500: 1
20: 1
2: 1
1: 1
如果使用如下所示的簡單邏輯，則可以減少代碼行數(shù)，
def find_denominations():
amount=int(input('Enter an amount: '))
denominations = dict()
DENOMINATIONS = [1000, 500, 100, 50, 20, 10, 5, 2, 1]
print('Total number of notes/coins=')
for d in DENOMINATIONS:
if amount >= d:
denominations[d] = amount // d
amount %= d
for key, val in denominations.items():
print(f"{key}: {val}")
倚天杖
TA貢獻(xiàn)1828條經(jīng)驗(yàn) 獲得超3個(gè)贊
Sreerams 的類似實(shí)現(xiàn)使用 while 循環(huán)而不是 for 循環(huán)：
amount = int(input("Enter an amount: "))
counter = amount
pos = 0
notes = [1000, 500, 100, 50, 20, 10, 5, 2, 1]
output = []
while counter > 0:
remainder = counter % notes[pos]
sub = counter - remainder
num = int(sub / notes[pos])
counter -= sub
output.append({notes[pos]: num})
pos += 1
print("Total number of notes/coins=")
for r in output:
for k,v in r.items():
if v > 0:
print("{}: {}".format(k, v))
請(qǐng)注意，Sreerams 代碼優(yōu)于我的代碼，它更易于閱讀，并且在規(guī)模上性能更高。
添加回答
舉報(bào)
0/150
提交
取消