SA=1000.00AI=0.12MP=100.00def remainb(x):if x==0: return 0else: return x=(SA+(AI/12)*SA)-MP for i in range(x,1000000): x=(x+(AI/12)*x)-MP CIwoP=(x+(AI/12)*x)-x #interest every month ptd=MP*i#payment to date #ptdreal=(ptd-CIwoP) #rbal=(CIwP-ptd) print(i)#payment no. print(ptd)#amount paid to date print(CIwoP)#interest for that month print(x)#balance for each month after payment #if rbal==0: return 0 #return 多次嘗試對此進行調試，但數小時內反復失敗。坦率地說，我被困住了。如果有人能就如何解決這個問題給我建議（例如，運行循環直到 SA==0），我將永遠感激不已。先感謝您。
1 回答
繁華開滿天機
TA貢獻1816條經驗 獲得超4個贊
我認為這就是您要實現的目標。切換到 while 循環更合適，因為您不確定需要循環多少次。只要 x 大于 0，這種情況就會繼續。
循環方法
x = (SA + (AI / 12) * SA) - MP
payment_number = 0
while x > 0:
x = (x + (AI / 12) * x) - MP
CIwoP = (x + (AI / 12) * x) - x # interest every month
ptd = MP * payment_number # payment to date
payment_number += 1
print(payment_number) # payment no.
print(ptd) # amount paid to date
print(CIwoP) # interest for that month
print(x) # balance for each month after payment
遞歸方法
def remainb(x, payment_number=0):
if x < 0: return
x = (x + (AI / 12) * x) - MP
CIwoP = (x + (AI / 12) * x) - x # interest every month
ptd = MP * payment_number # payment to date
payment_number += 1
print(payment_number) # payment no.
print(ptd) # amount paid to date
print(CIwoP) # interest for that month
print(x) # balance for each month after payment
remainb(x, payment_number)
順便提一下，使用良好的描述性變量名稱而不是 x 是一個好習慣，我 :)
添加回答
舉報
0/150
提交
取消