Python：每日一题 206,Python交流,编程语言专区,鱼C论坛

wyp02033 发表于 2018-9-11 17:12:07

def fab():
a, b = 5, 8
while True:
   a, b = b, a + b
   yield b

def main():
num = fab()
count = 0
while count < 10:
   next_num = next(num)
   next_num_sum = sum()

   if next_num % next_num_sum == 0:
         print(next_num)
         count += 1

if __name__ == '__main__':
main()

龙战yl天下 发表于 2018-9-11 18:41:29

本帖最后由龙战yl天下于 2018-9-11 18:43 编辑

import copy
i = 1
n=0
num1 = 5
num2 = 8
while n < 10:
num3 = num2 + num1
#print ('num3=',num3) #斐波那契数
x=copy.copy(num3) #求各位之和
s=len(str(x))
h=0
sum=0 #sum表示各位之和
sz=[]
while h<s:
   sz = sz +
   x=x-sz*(10**(s-h-1))
   sum+=sz
   h+=1
e = num3 % sum    #判断能否被整除
if e==0:
   print (num3)
   n=n+1
   print('n=',n)
i += 1
num1 = num2
num2 = num3

答案：
21
n= 1
144
n= 2
2584
n= 3
14930352
n= 4
86267571272
n= 5
498454011879264
n= 6
160500643816367088
n= 7
114059301025943970552219
n= 8
5358359254990966640871840
n= 9
555565404224292694404015791808
n= 10

土匪加步枪 发表于 2018-9-11 21:08:54

def fib(value=1):
if value % sum_iter(value)==0 :
   print(value)

def sum_iter(value):
   iters=[]
   n=len(str(value))
   print(value)
   for each in range(len(str(value))):
         step=n-each-1
         iters.append(int(value/(10**step)))
         value=value-(int(value/(10**step))*(10**step))
   print(iters)
   return sum(iters)
def fibs(value):
if value % sum_iter(value)==0:
   return value
else:
   False

value=9
n=10
fibss=[]
while True:
value+=1
print(value)
if fibs(value) :
   fibss.append(value)
   n-=1
elif n==0:
   break

BlueSand 发表于 2018-9-11 21:20:06

# encoding: utf-8

#斐波那契数中有些数字是可以被其各位数字之和整除的，例如：
# 144可以被1+4+4整除。
# 要求求出前10个这样的斐波那契数（不包括一位数1, 2, 3, 5, 8）
def getSum(number):
sum = 0
while number >= 1:
sum = sum + number % 10
number = number // 10
return sum

def isGood(n1, n2):
if n1 % n2 == 0:
return True
else:
return False
i = 0
f1 = 1
f2 = 1
result= []
while i < 10:
f3 = f1 + f2
f1, f2 = f2, f3
if f3 > 10:
t = getSum(f3)
if isGood(f3, t):
result.append(f3)
i = i + 1
print(result)
答案:
[21,
144,
2584,
14930352,
86267571272,
498454011879264,
160500643816367088,
114059301025943970552219,
5358359254990966640871840,
555565404224292694404015791808]
为什么你们的代码都这么简洁的,好羡慕

archlzy 发表于 2018-9-11 21:32:41

def f206(n):
def creat_num(m):
   a, b = 1, 0
   for i in range(m):
         a ,b = a+b ,a
   return a


result = []
m = 0
while len(result) < n:
   m += 1
   temp = creat_num(m)
   if temp >10:
         temp_a = 0
         for i in str(temp):
            temp_a += int(i)
         if temp % temp_a == 0:
            result.append(temp)
return result

子沙发表于 2018-9-11 21:47:44

本帖最后由子沙于 2018-9-11 21:54 编辑

def fun_206(num):
x1,x2=1,1
x1,x2=x2,x1+x2
num1=0
while num1<num:
   x1,x2=x2,x1+x2
   if len(str(x2))>=2:
         if x2%sum(list(map(lambda x:int(x),list(str(x2)))))==0:
            print(x2)
            num1+=1
fun_206(10)

运行结果如下：

21
144
2584
14930352
86267571272
498454011879264
160500643816367088
114059301025943970552219
5358359254990966640871840
555565404224292694404015791808

CoderWGB 发表于 2018-9-11 22:26:40

#斐波那契数中有些数字是可以被其各位数字之和整除的，例如：
# 144可以被1+4+4整除。
# 要求求出前10个这样的斐波那契数（不包括一位数1, 2, 3, 5, 8）。

# 666 -> 6+6+6 = 18获取各位之和
def aliquot(num):
   temp = 0
   for i in range(len(str(num))):
            temp += int(str(num))
   return temp

# 1,1,2,3,5,8,13,21...
def fibsNum():
   a = 0
   b = 1
   num_list = []
   while True:
            a,b = b,a+b
            if a > 10:
                  div = aliquot(a)
                  if a%div == 0:
                           num_list.append(a)
                           if len(num_list) == 10:
                                    return num_list

print(fibsNum())

真是没想到,这后面的几个数这么大!!!

学学看看 发表于 2018-9-12 11:30:53

小沈同学 发表于 2018-9-13 20:51:54

jj482846 发表于 2018-9-13 22:34:04

n：每日一题 206 [修改]

Frankly 发表于 2018-9-14 07:45:42

看看

Y2M5210 发表于 2018-9-14 11:03:43

"""
斐波那契数列中有些数字可以被其他各位数字之和整除的，求前十个，除（1,2,3,5,8）,用三重双引号
"""
def snum(n):
if n < 10:
   return n
else:
   return n%10 + snum(n//10)

fa = 5
fb = 8
count = 0
while count < 10:
fc = fa + fb
fa = fb
fb = fc
if fc%snum(fc) == 0:
   count += 1
   print('第{0}个是{1}'.format(count,fc),end='\t')
   if count%4 == 0:
         print()

Chysial 发表于 2018-9-14 14:30:52

# 斐波那切数列 c=a+b
def fibonacci():
a = 1
b = 0
ailquot = 0
n = 0
fibo_sequence = []
fibo_divisible_number = []
while n<10:
   c = a+b
   b = a
   a = c
   fibo_sequence.append(c)
   if len(str(c)) > 1:
         t = sum()
         if c%t == 0:
            fibo_divisible_number.append(c)
            n += 1
return fibo_sequence,fibo_divisible_number
print(fibonacci())
结果：
fibo_divisible_number:

tkxlyytqy 发表于 2018-10-12 11:11:41

# -*- coding:utf-8 -*-
list_fib=[]
def fib_fun(n):
i,a,b=0,1,1
list_fib.append(a)
while True:
   a,b=b,a+b
   i+=1
   list_fib.append(a)
   if i==n-1:
         break
return list_fib
fib_fun(200)
i_str =""
sum=0
list_divd=[]
num=0
for i in list_fib:
if i>10:
   i_str=str(i)
   for j in i_str:
         j=int(j)
         sum+=j
   if (i%sum)==0:
         list_divd.append(i)
         num += 1
         if num == 10:
            break
sum=0
for i in list_divd:
print(i)

抬起头~ 发表于 2018-10-15 21:42:10

加油

喜欢吃菠菜 发表于 2018-10-17 09:44:03

塔利班发表于 2018-9-10 10:43

这么多答案中，我看到的最pythonnic的，特别是map这个函数的写法。👍

sunrise085 发表于 2019-1-21 16:35:24

为了节省程序运行时间，斐波那契数就别再调用函数了，那样会多出很多次重复运算，直接在主程序中这样写，相当于只调用了一次斐波那契，而不是每进行一次while循环，就调用一次。
def mysum(num):
result=0
while num:
result+=num%10
num=num//10
return result

i,x1,x2=1,8,13
while i<=10:
if x2%mysum(x2)==0:
print('第%2d个数是%d' %(i,x2))
i+=1
x1,x2=x2,x1+x2
结果：
第 1个数是21
第 2个数是144
第 3个数是2584
第 4个数是14930352
第 5个数是86267571272
第 6个数是498454011879264
第 7个数是160500643816367088
第 8个数是114059301025943970552219
第 9个数是5358359254990966640871840
第10个数是555565404224292694404015791808

永恒的蓝色梦想 发表于 2019-8-17 22:04:57

本帖最后由永恒的蓝色梦想于 2019-8-17 22:05 编辑

不加限制算出来好多……
（部分结果）21
144
2584
14930352
86267571272
498454011879264
160500643816367088
114059301025943970552219
5358359254990966640871840
555565404224292694404015791808
1226132595394188293000174702095995
18547707689471986212190138521399707760
107168651819712326877926895128666735145224
280571172992510140037611932413038677189525
619220451666590135228675387863297874269396512
64202014863723094126901777428873111802307548623680
271964099255182923543922814194423915162591622175362
6656593304481317393598839952151746590023553382130993248
263621064469290555679241849789653324393054271110084140201023
690168906931029935139391829792095612517948949963798093315456
137347080577163115432025771710279131845700275212767467264610201
23041483585524168262220906489642018075101617466780496790573690289968
293825989466396564333419951255644330166833468672422805842178911936214659279
769246427201094785080787978422393713094534885688979999504447628313150135520
28624020537229717283244863695841661789309459371299941322398704536965075971940465647510004531000816
955620997609204752896986850849030784038174487916669186294134525152075026461787231598163278910835948084128
99080696264900721472208515956748759187919804840608734007367661059706052043279378932885908102386332543066271936
31903676490304597847185685736169548906931858202641803975680844768196795244879691985828019808017263472521442003280
1065113236465588309403889415460645093083860991848425732542338227915288346612042420944981983005010603735148681490199640832代码a,b=1,1
while 1:
a,b=b,a+b
if a%sum((int(i)for i in str(a)))==0 and len(str(a))!=1:print(a)

小陨aoq 发表于 2020-8-15 21:37:12

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19971023 发表于 2020-12-14 10:18:06

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