Python：每日一题 106,Python交流,技术交流,鱼C论坛

冬雪雪冬 发表于 2017-9-29 10:15:00

Python：每日一题 106

本帖最后由冬雪雪冬于 2017-9-30 21:46 编辑

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题目：
我们知道数字2、3既是质数又属于斐波那契数列，要求求出1000以内所有既是质数又属于斐波那契数列的数字。

堕落之翼 发表于 2017-9-29 12:28:52

import math

def fun(num):
if num == 1 :
   return 1
elif num == 2:
   return 1
else:
   return fun(num-1) + fun(num - 2)

fun_num = []
n = 1
while True:
prime = True
# 获得斐波那契数
totle = fun(n)

if totle >= 1000:
   break
else:
   if totle != 1:
         # 判断斐波那契数是不是质数
         num =math.floor(math.sqrt(totle))
         for j in range(2,num+1):
            if totle % j == 0 :
               prime = False
               break
         if prime:
            fun_num.append(totle)
n += 1
print(fun_num)

jerryxjr1220 发表于 2017-9-29 13:15:29

本帖最后由 jerryxjr1220 于 2017-9-29 13:58 编辑

根据Sixpy大大的建议改进：
f1,f2=1,2
while f2<1000:
is_prime = lambda n: True if n in {2,3,5,7,11,13,17,19,23,29,31,37} else all(False for p in {2,3,5,7,11,13,17,19,23,29} if pow(p,n-1,n) != 1)
if is_prime(f2): print(f2, end=' ')
f1,f2 = f2,f1+f2
2 3 5 13 89 233

bush牛 发表于 2017-9-29 21:07:50

def fun(n):
a, b = 2, 3
while b < n:
   if is_prime(a):
         yield a
   a, b = b, a+b

def is_prime(n):
for i in range(2,n):
   if n % i == 0:
         return False
return True

for i in fun(1000):
print(i)

solomonxian 发表于 2017-9-29 21:20:28

取n以内质数集合，与100项以内斐波那契集合取交集
第100项斐波那契已经有10的20次方了，基本保证 >n
def fun(n):
prime = {i for i in range(2,n+1) if 0 not in (i%j for j in range(2,int(i**0.5)+1))}
fib = {k for k in [(a[-1],a.append(a[-1]+a[-2])) for a in (,) for j in range(100)]}
return prime & fib

sige 发表于 2017-9-30 08:35:37

感觉自己的代码写的可丑……
#求出斐波拉契数列
m = 0
n = 1
total = 1
list1 = []
while True:
total = m + n
if total < 1000:
   list1.append(total)
   m = n
   n = total
else:
   break

#求出质数
list2 =
for each in list1:
t = 0
each = int(each)

for i in range(2,each):
   if (each % i) != 0:
         t = 1
   else:
         t = 0
         break
if t != 0:
   list2.append(each)

print(list2)
输入结果：

sige 发表于 2017-9-30 09:03:25

递归写法#求出斐波拉契数列
def fun(n):

if n <= 1 :
   return n
else:
   result = int(fun(n-1))+ int(fun(n-2))
   returnresult
list1 = []
for i in range(17):
list1.append(fun(i))

#求出质数
list2 =
for each in list1:
t = 0
each = int(each)

for i in range(2,each//2 + 2):
   if (each % i) != 0:
         t = 1
   else:
         t = 0
         break
if t != 0:
   list2.append(each)

print(list2)

776667 发表于 2017-10-6 22:05:20

import math

def double_check(x):
def fibonacci(n):
   if n == 0:
         return 0
   if n == 1:
         return 1
   return fibonacci(n-1) + fibonacci(n-2)
def prime_check(y):
   list_a =
   list_prime = []
   for m in list_a:
         for k in range(2,m):
            if m%k == 0 and k != 1:
               break
         else:
            list_prime.append(m)
   for l in list_prime:
         if y%l == 0:
            return False
   return True
n = 3
result_list = []
while fibonacci(n) <= x:
   if prime_check(fibonacci(n)):
         result_list.append(fibonacci(n))
   n += 1
return result_list

if __name__ == '__main__':

wyp02033 发表于 2017-10-6 23:03:47

def isprime(n):
if n == 1:
   return False
if n == 2 :
   return True

for i in range(2,n//2 + 1):
   if n % i == 0:
         return False

return True

def fab():
n1, n2 = 1, 1

while True:
   n1, n2 = n2,n1+n2
   yield n1

a = fab()
while True:
b = next(a)
if b <= 1000 and isprime(b):
   print(b)

fan1993423 发表于 2018-3-14 16:44:03

x=1
y=1
z=1
m=0
q=[]
while m<=20:
z=x+y
x=y
y=z
q.append(z)
m+=1

p=[]
for a in range(2,101):
if a==2:
   p.append(a)
else:
   if 0 not in :
         p.append(a)
for h in q:
if h in p:
   print(h)

新手潘包邮 发表于 2018-4-28 00:02:35

solomonxian 发表于 2017-9-29 21:20
取n以内质数集合，与100项以内斐波那契集合取交集
第100项斐波那契已经有10的20次方了，基本保证 >n

怎么您对python的语法，各种库函数这么熟悉啊，

foxiangzun 发表于 2019-2-14 17:13:16

list1 = []

def addNum(num) :
   if num in :
            return 1
   else :
            return addNum(num - 1) + addNum(num - 2)

def getNum() :
   globallist1
   for i in range(20) :
            if addNum(i) <= 1000 :
                     list1.append(addNum(i))
            else :
                     break
   print(list1)

def getPrime() :
   getNum()
   global list1
   list2 = []
   for i in list1 :
            flag = 0
            for j in range(2, i // 2 + 1) :
                     if i == 1 :
                           continue
                     if i % j == 0 :
                           flag += 1
            if flag == 0 :
                     list2.append(i)
   for i in list2[:] :
            if i == 1 :
                     list2.remove(i)
   print(list2)

getPrime()

foxiangzun 发表于 2019-2-15 14:30:59

def fib(num) :
   if num in :
            return 1
   else :
            return fib(num - 1) + fib(num - 2)

def isPrime(num) :
   flag = 0
   for i in range(2, num // 2) :
            if num % i == 0 :
                     flag += 1
   if flag > 0 :
            return False
   else :
            return True

def getNum() :
   list1 = []
   for i in range(1, 20) :
            if fib(i) >= 1000 :
                     break
            else :
                     list1.append(fib(i))
   for i in list1[:] :
            if (i == 1) or (not isPrime(i)) :
                     list1.remove(i)
   print(list1)

getNum()

咕咕鸡鸽鸽 发表于 2019-3-12 10:55:05

def fun106():
a = 1
b = 1
list1 = []
while b < 1000:
   a, b = b, a+b
   prime = True
   if b > 3:
         for i in range(2,b):
            #排除非质数
            if b%i == 0:
               prime = False
               break
   if prime:
         list1.append(b)

return list1[:-1]

print(fun106())

小贤啊哦 发表于 2020-5-6 19:18:59

#判断费那波切数：
def fei():
s=
a=1
b=1
for i in range(1000):
   a,b=b,a+b
   if a>=1000:
         break
   s.append(a)
return s
#判断是否为质数
def zhi():
z=[]
for i in range(2,1000):
   for j in range(2,i):
         if i%j==0:
            break
   else:
         z.append(i)
return z
a=zhi()
b=fei()
for i in a:
for j in b:
   if i==j:
         print(i)

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Python：每日一题 106