請(qǐng)您幫忙更正此代碼！這是為了找到低于 10,000 的設(shè)定限制的所有完美數(shù)字，我用評(píng)論來(lái)解釋我在做什么。謝謝！#Stores list of factors and first perfect numberfacs = []x = 1#Defines function for finding factorsdef findfactors(num): #Creates for loop to find all factors of num for i in range(1, num + 1): if num % i == 0: #Stores factor in facs facs.append(i)#Activates loopwhile x < 10000: #Finds factors of x and appends them to list findfactors(x) #Finds sum of list facsum = sum(facs) #Makes decision based on sum of factors and original number if facsum == x: #Ouputs and increases x print(x) x += 1 else: #Increases x x += 1
3 回答
守著星空守著你
TA貢獻(xiàn)1799條經(jīng)驗(yàn) 獲得超8個(gè)贊
在 def 中初始化列表并返回，并且范圍不應(yīng)包括原始num范圍，因此范圍將從 1 到 num，其中包括 1 但不包括原始num范圍，因此它將生成范圍從1到num-1
#Stores list of factors and first perfect number
x = 1
#Defines function for finding factors
def findfactors(num):
facs = []
for i in range(1, num):
if num % i == 0:
#Stores factor in facs
facs.append(i)
return facs
#Activates loop
while x < 1000:
#Finds factors of x and appends them to list
facs=findfactors(x)
#Finds sum of list
facsum = sum(facs)
#Makes decision based on sum of factors and original number
if facsum == x:
#Ouputs and increases x
print(x)
x+= 1
比從Python 中查找數(shù)字的所有因子的最有效方法是什么生成列表方法要快得多？
#Stores list of factors and first perfect number
x = 1
#Defines function for finding factors
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
#Activates loop
while x < 10000:
#Finds factors of x and appends them to list
facs=factors(x)
facs.remove(x) # remove original number as it is not required further
#Finds sum of list
facsum = sum(facs)
#Makes decision based on sum of factors and original number
if facsum == x:
#Ouputs and increases x
print(x)
x+= 1
長(zhǎng)風(fēng)秋雁
TA貢獻(xiàn)1757條經(jīng)驗(yàn) 獲得超7個(gè)贊
這是您的邏輯的更簡(jiǎn)單實(shí)現(xiàn)。
x = 1
#Defines function for finding factors
def isperfectnum(num):
sum=0
#Creates for loop to find all factors of num
for i in range(1, num ):
if num % i == 0:
sum=sum+i
if sum==num:
return TRUE
else
return FALSE
#Activates loop
while x < 10000:
#Finds perfect numbers
if isperfectnum(x):
print(x)
x=x+1
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