這是我正在處理的代碼。它應該從用戶那里獲取一個數字，如果它是一個完美的數字，它會這樣說，但如果不是，它會要求輸入一個新數字。當我到達輸入新數字部分時，它不會記錄我的輸入。有人可以幫我嗎？
def isPerfect(num):
if num <= 0:
return False
total = 0
for i in range(1,num):
if num%i== 0:
total = total i
if total == num:
return True
else:
return False
def main():
num = int(input("Enter a perfect integer: "))
if isPerfect(num) == False:
op = int(input(f"{num} is not a perfect number. Re-enter:"))
isPerfect(op)
elif isPerfect(num) == True:
print("Congratulations!",num, 'is a perfect number.')
if __name__ == '__main__':
main()
uj5u.com熱心網友回復：
你可以像這樣將 while True 回圈放入你的 main 函式中：
def main():
first_run = True
perfect_num_received = False
while not perfect_num_received:
if first_run:
num = int(input("Enter a perfect integer: "))
first_run = False
if isPerfect(num) == False:
num = int(input(f"{num} is not a perfect number. Re-enter:"))
elif isPerfect(num) == True:
perfect_num_received = True
print("Congratulations!",num, 'is a perfect number.')
但也已經有一個內置函式來檢查變數的資料型別，所以你可以做這樣的事情：
if type(num) == int:
...
uj5u.com熱心網友回復：
根據問題描述，我假設程式應該在找到完全數后終止。可以添加兩件事，一個是每次用戶輸入數字時檢查輸入提示的回圈，另一個是找到完美數字后的中斷條件。這是一個有效的實作。
def main():
while True:
num = int(input("Enter a perfect integer: "))
if isPerfect(num) == False:
num = int(input(f"{num} is not a perfect number. Re-enter: "))
isPerfect(num)
elif isPerfect(num) == True:
print("Congratulations! ",num, ' is a perfect number.')
break
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