Python：每日一题 207

冬雪雪冬 · 发表于 2018-9-11 11:00:42

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本帖最后由冬雪雪冬于 2018-9-12 19:45 编辑

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题目：
本题与206题类似，斐波那契数中有些数字是可以分别被其每一位数字整除的（忽略数字0），例如：
46368可以被3，4，6，8整除。
主要看大家的程序效率，比如能被6整数就不用考虑2和3了。
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）。

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凌九霄 · 发表于 2018-9-11 11:51:35

本帖最后由凌九霄于 2018-9-11 15:17 编辑

import timeit
P = '''
def fibonacci(n):
if n > 0:
i, a, b = 0, 0, 1
while i < n:
a, b = b, a + b
i += 1
return a
def fnum(n):
t = set(str(n)) - {'0', '1'}
if '6' in t:
t = t - {'2', '3'}
if '8' in t:
t = t - {'2', '4'}
return list(map(int, t))
i, j = 7, 0
while j < 10:
f = fibonacci(i)
t = fnum(f)
for k in t:
if f % k:
break
elif k == t[-1]:
print(f)
j += 1
i += 1
'''
print('运行用时：',timeit.timeit(stmt=P, number=1))

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拉了盏灯 · 发表于 2018-9-11 13:35:56

本帖最后由拉了盏灯于 2018-9-12 00:00 编辑

def feb(n):
a,b = 0,1
while b<n:
a,b=b,a+b
if b > 9:
l=list(map(int,str(b)))
if len(l)-l.count(0) == len([i for i in l if i != 0 and b%i ==0]):
yield b
for i in feb(10**200):
print(i,'\n')

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塔利班 · 发表于 2018-9-11 13:36:57

本帖最后由塔利班于 2018-9-12 10:37 编辑

感觉找去因数的方法很蠢，回头再写- -

def check(x):
p=sorted([x for x in set(map(int,str(x))) if x])
q=p[:]
l=len(p)
for i in range(l):
for j in range(i+1,l):
if q[j]%q[i]==0:
try:
p.remove(q[i])
except:
pass
for each in p:
if x%each:
return False
return True
def fib():
a,b=8,13
while True:
a,b=b,a+b
if check(a):
yield a
a=fib()
for i in range(10):
print(next(a))

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这个没筛，有时候感觉处理了也没提升运行速度，应该有更好的算法和内置吧

from functools import reduce
from math import gcd
a,b,c=8,13,0
while c<10:
a,b=b,a+b
if not a%reduce(lambda x,y:x*y//gcd(x,y),[x for x in set(map(int,str(a))) if x],1):
c+=1
print(a)

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graceasyi · 发表于 2018-9-11 14:13:17

本帖最后由 graceasyi 于 2018-9-11 14:20 编辑

def fib_itr(n):
if n < 0:
return None
if n <= 1:
return (0, 1)[n]
a, b = 0, 1
for i in range(n-1):
a, b = b, a+b
return b
i = 7
n = 0
while n < 10:
fbi = fib_itr(i)
sm = set([int(d) for d in str(fbi)])
if 0 in sm: sm.remove(0)
if 1 in sm: sm.remove(1)
if 9 in sm:
if 3 in sm: sm.remove(3)
if 8 in sm:
if 2 in sm: sm.remove(2)
if 4 in sm: sm.remove(4)
if 6 in sm:
if 2 in sm: sm.remove(2)
if 3 in sm: sm.remove(3)
if 4 in sm:
if 2 in sm: sm.remove(2)
qt = [fbi % q for q in sm]
if qt.count(0) == len(qt):
print(fbi)
n += 1
i += 1

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结果：
55
144
46368
5358359254990966640871840
64202014863723094126901777428873111802307548623680
769246427201094785080787978422393713094534885688979999504447628313150135520
9216845717656874712980450562726202415567360565980794777111390850331644813674856981646960226192287360
110433070572952242346432246767718285942590237357555606380008891875277701705731473925618404421867819924194229142447517901959200
1323171012053243520828784042795469593341319770463238313551473338336502410952765153371119398122747569819754164672344667591018783803781288766524146031040
15853779289614481577191691577533518174485949824765525957623037737532160833631593681512535189759649213804709691705593604227037581022758330489263385511067756715301085561591382880

HML · 发表于 2018-9-11 14:35:41

Hitdongfeng · 发表于 2018-9-11 14:57:23

JessiFly · 发表于 2018-9-11 15:17:59

def fib():
a,b = 0,1
while True:
a,b = b,a+b
yield a
def my_filter(num):
each_num = []
org_num = num
while num:
temp = num%10
if temp not in each_num:
if temp != 0 and temp != 1:
each_num.append(temp)
num //= 10
#如果有9不考虑3；如果有8不考虑4和2；如果有6不考虑3和2；如果有4不考虑2。
if 9 in each_num:
if 3 in each_num:
each_num.remove(3)
if 8 in each_num:
if 4 in each_num:
each_num.remove(4)
if 2 in each_num:
each_num.remove(2)
if 6 in each_num:
if 3 in each_num:
each_num.remove(3)
if 2 in each_num:
each_num.remove(2)
if 4 in each_num:
if 2 in each_num:
each_num.remove(2)
for each in each_num:
if org_num%each:
return False
return True
def fun207():
count = 0
for each_fib in fib():
if each_fib > 10 and my_filter(each_fib):
print(each_fib)
count += 1
if count == 10:
break
if __name__ == '__main__':
fun207()

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骨傲天 · 发表于 2018-9-11 15:23:38

本帖最后由骨傲天于 2018-9-11 16:17 编辑

def b():
a=1
b=0
while True:
yield a+b
a,b=a+b,a
x=1
for i in b():
if i<10:
continue
elif x<11:
r=list(map(int,str(i)))
re=[]
d,s=0,0
for y in r:
if y in re:
continue
if y!=0 and i%y:
break
elif y%2 and not s:
re.extend([2,4,6,8])
s=1
elif y%3 and not d:
re.extend([3,9])
d=1
else:
re.append(y)
else:
print(i)
x+=1
else:
break

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老猫NAKE · 发表于 2018-9-11 15:51:50

本帖最后由老猫NAKE 于 2018-9-11 15:54 编辑

走错片场了。。

天圆突破 · 发表于 2018-9-11 17:01:30

本帖最后由天圆突破于 2018-9-11 22:22 编辑

class Feb:
def __init__(self):
self.a = self.b = 1
def __iter__(self):
return self
@staticmethod
def filter_num(n):
# 取集合，筛去0和1
lst = list(filter(lambda x:x!=0 and x!=1,set(map(lambda x:int(x), str(n)))))
# 倒序排序，早碰上早滚蛋，提高效率
lst.sort(reverse=True)
for each_n in lst:
if n%each_n != 0:
return False
return True
def __next__(self):
while True:
self.a, self.b = self.b, self.a+self.b
if self.filter_num(self.a) and self.a > 10:
return self.a
if __name__ == '__main__':
for i in Feb():
input(i)

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我觉得考虑效率的话,去重去01外加逆排序就足够了
如果为了去掉2,3,4(只能去这3个数字),需要遍历4次才能去干净（从集合里找9864）,不去的话只会多遍历3次（带着234去除）.
所以还是不去了吧.

胡尔汉 · 发表于 2018-9-11 19:16:17

本帖最后由胡尔汉于 2018-9-11 19:40 编辑

实在没有好的办法按照题目的方法优化效率，能不能给点建议或其他思路？

def confirm_num(num):
l1 = [2,3,4,5,6,7,8,9]
num1 = str(num)
for i in num1:
i = int(i)
if i in l1:
if num%i != 0:
return False
else:
l1.remove(i)
return True
x,y,i = 5,8,0
list1 = [] #存放符合要求的数字
while i <10:
x,y = y,x+y
if confirm_num(y):
list1.append(y)
i+=1
print(list1)

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龙战yl天下 · 发表于 2018-9-11 19:58:55

学习有限，目前也就仅到能做出来的程度了，后续改进

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) #
s=len(str(x))
h=0
sum=0
sz=[]
while h<s:
      sz = sz + [x//(10**(s-h-1))]
      x=x-sz[h]*(10**(s-h-1))
      if sz[h] != 0:
         e = num3 % sz[h]
         sum+=e
      h+=1

''' h=h+1
sz.sort()
rz=sz
m=0
while m<len(rz)-1:
      if rz[m]==rz[m+1]:
         rz.remove(rz[m])
      else:
         m=m+1
p=0
while p<len(rz)-1:
   if rz[p] != 0:
      e = num3 % rz[p]
      sum+=e
      p+=1
'''
if sum==0:
      print (num3)
      n=n+1
      print('n=',n)
i += 1
num1 = num2
num2 = num3

答案：
55
n= 1
144
n= 2
46368
n= 3
5358359254990966640871840
n= 4
64202014863723094126901777428873111802307548623680
n= 5
769246427201094785080787978422393713094534885688979999504447628313150135520
n= 6
9216845717656874712980450562726202415567360565980794777111390850331644813674856981646960226192287360
n= 7
110433070572952242346432246767718285942590237357555606380008891875277701705731473925618404421867819924194229142447517901959200
n= 8
1323171012053243520828784042795469593341319770463238313551473338336502410952765153371119398122747569819754164672344667591018783803781288766524146031040
n= 9
15853779289614481577191691577533518174485949824765525957623037737532160833631593681512535189759649213804709691705593604227037581022758330489263385511067756715301085561591382880
n= 10

瓦蓝 · 发表于 2018-9-11 20:13:45

zhou995287902 · 发表于 2018-9-11 20:45:58

def feibonacci(num):
a,b = 0,1
count = 0
while count <num:
yield a
a,b = b,a+b
count +=1
def fun207(n):
result = list()
for i in feibonacci(5000):
if i in [0,1,2,3,5,8]:
pass
else:
for each in str(i):
if int(each) == 0:
pass
elif i % int(each) == 0:
a=True
else:
a=False
break
if a:
result.append(i)
if len(result)==n:
break
print(result)
if __name__ == '__main__':
fun207(10)

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hzmmz · 发表于 2018-9-11 20:58:52

谁与争锋 · 发表于 2018-9-11 21:02:10

本帖最后由谁与争锋于 2018-9-11 21:26 编辑

list1 = []
list2 = [5,8]

while len(list1)<10:
list2[-2],list2[-1]=list2[-1],list2[-1]+list2[-2]
a=list2[-1]
#print(a)
num1=1
b=str(a)
c=list(b)
c=list(set(c))
c.sort(reverse=True)
#print('c=',c)
list3 = []

if '0' in c:
      c.remove('0')
while len(c)>0:
      d=int(c.pop(0))
      list3.append(d)
      if num1%d!=0:
         num1*=d
      list4 = c.copy()
      for e in list4:
         if d%(int(e))==0:
            c.remove(e)

if a%num1==0:
      list1.append(a)

print(list1)

archlzy · 发表于 2018-9-11 21:34:42

本帖最后由 archlzy 于 2018-9-11 22:34 编辑

def f207(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_set = set([float(i) for i in list(set(str(temp))) if i not in ["0","1"]])
temp_set -= set([j for i in temp_set for j in temp_set if i/j in temp_set])
temp_a = 0
for i in temp_set:
temp_a += temp % int(i)
if temp_a == 0:
result.append(temp)
return result

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没抢到 206的楼好遗憾~

在东边 · 发表于 2018-9-11 21:48:08

def fun207(n):
def fib():
a, b = 0, 1
while 1:
a, b = b, a + b
yield a
g = fib()
flt = filter(lambda x: len(str(x)) > 1 and not sum(map(lambda y, x=x: x % y, {int(i) for i in str(x)} - {0, 1})), g)
return [next(flt) for _ in range(n)]
print(fun207(10))

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对韭当割 · 发表于 2018-9-11 22:01:46

hao

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