Python：每日一题 190,Python交流,编程语言专区,鱼C论坛

冬雪雪冬 发表于 2018-7-24 15:23:28

Python：每日一题 190

本帖最后由冬雪雪冬于 2018-7-28 20:30 编辑

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题目：
23是质数，它的十位和个位2和3也是质数，求出100以内所有这样的两位数质数。
注：最近几期的题目可能有点难度，参与的鱼油不够踊跃，这回来个难度低的题，希望更多的人参加。

小木子c 发表于 2018-7-24 16:21:23

list1 =
list0 = []
for each in range(10,100):
for i in range(1,each//2+1):
   for j in range(1,i+1):
         if i * j == each:
            if each in list0:
               pass
            else:
               list0.append(each)
list2 =
for each in list0:
list1.remove(each)
for each in list1:
if (each//10 in list2) and (each%10 in list2):
   print(each,end=',')

23,37,53,73,

graceasyi 发表于 2018-7-24 16:53:03

本帖最后由 graceasyi 于 2018-7-24 17:06 编辑

def is_prime(n):
if n not in :
   for i in range(2, n):
         if n % i == 0:
            return False
   else:
         return True
else:
   return False

for n in range(11, 100):
if is_prime(n):
   if is_prime(n % 10) and is_prime(n // 10):
         print(n)

另一种思路：由于要求每一位都是质数，因此先找出10以内的质数，2，3，5，7，再组合后判断符合条件的，这样一共只需要判断25次质数就可以了。
p =
result =
print(result)

结果：
23
37
53
73

augur 发表于 2018-7-24 16:56:28

def isPrime(num):
if num == 2:
   return True
elif num == 0 or num == 1:
   return False
else:
   for i in range(2, int(num ** 0.5) + 1):
         if num % i == 0:
            return False
   return True

print()

拉了盏灯 发表于 2018-7-24 17:22:37

明天要去军训了，一个星期都不能摸可爱的python了，{:10_266:}{:10_266:}{:10_266:}{:10_266:}

def fun190(num=100):
for each in range(10,num+1):
   if each % 2 != 0 and each % 3 != 0 and each % 5 != 0 and each % 7 != 0:
         print(each)
         count = 0
         for i in str(each):
            if i in '2357':
               count += 1
            else:
               pass
         if count == len(str(each)):
            print(each)

JessiFly 发表于 2018-7-24 17:55:11

本帖最后由 JessiFly 于 2018-7-24 17:56 编辑

def isprime(n):
if n == 1:
   return False
elif n == 2:
   return True
elif n % 2:
   n_sqrt = n ** 0.5
   i = 3
   while i <= n_sqrt:
         if n % i == 0:
            return False
         i += 1
   return True
else:
   return False

for n in range(10,100):
a = n // 10
b = n % 10
if isprime(n) and isprime(a) and isprime(b):
   print(n)

结果
23
37
53
73

zhou995287902 发表于 2018-7-24 18:05:11

本帖最后由 zhou995287902 于 2018-7-24 21:26 编辑

def isprime(number):
if number==0 or number==1:
   return False
else:
   for i in range(2,number//2+1):
         if number%i==0 :
            break
   else:
         return True

for each in range(10,100):
if isprime(int(str(each))) and isprime(int(str(each))) and isprime(each):
   print(each)

晓屁屁 发表于 2018-7-24 18:07:30

本帖最后由晓屁屁于 2018-7-25 09:40 编辑

class Prime:
def search(self,num):
   for i in range(2, num):
         if num%i == 0:
            return False
   return True
def pos(self,num):
   if len(str(num)) == 2:
         self.gew = num%10
         self.shw = num//10%10
         return (self.shw,self.gew)

p = Prime()
a = []
for i in range(2,100):
if p.search(i) and p.pos(i) != None:
   a.append(p.pos(i))
else:
   pass
for j in a:
if p.search(j) and p.search(j) and j != 1:
   print(''.join( ))
else:
   pass

为了练习类，写的很low

塔利班 发表于 2018-7-24 18:48:58

def isp(x):
if x>1:
   if x==2:
         return True
   if x%2==0:
         return False
   for i in range(3, int(x**0.5 + 1), 2):
         if x % i == 0:
            return False
   return True
return False
for i in range(2,100):
if isp(i) and isp(i//10) and isp(i%10):
   print(i)

凌九霄 发表于 2018-7-24 19:25:36

本帖最后由凌九霄于 2018-7-26 15:48 编辑

def isPrime(n):
if n <= 1:
   return False
i = 2
while i * i <= n:
   if n % i == 0:
         return False
   i += 1
return True

def outP(num):
bool = True
if isPrime(num):
   strnum = str(num)
   for i in strnum:
         if not isPrime(int(i)):
            bool = False
            break
else:
   bool = False
return bool

X =

print(X)

from itertools import permutations
from math import sqrt

A = * 10 + x, list(permutations(, 2)))) if
0 not in ]

import itertools

nums = list(map(lambda x: x + x, list(itertools.permutations(('2', '3', '5', '7'), 2))))

def checkit(num):
P =
rtn = 1
for i in P:
   result = num % i
   if result == 0:
         rtn = result
         break
return rtn

A =

硫代硫酸钠 发表于 2018-7-24 19:29:13

def func():
def is_prime(x):
   import math
   if x<2:
         return False
   temp = int(math.sqrt(x))
   for i in range(2, temp+1):
         if x%i == 0:
            return False
   return True

for i in range(11, 100):
   if is_prime(i):
         if is_prime(i//10) and is_prime(i%10):
            print(i)

咯洛落 发表于 2018-7-24 20:07:17

for zs in range(10,101):
n = 0
g = 0
s = 0
gw = zs%10
sw = int(zs/10)
for i in range(1,zs+1):
   a=zs%i
   gi=gw%i
   si=sw%i
   if a == 0:
         n +=1
   if gi == 0:
         g +=1
   if si == 0:
         s +=1
if n == 2 and g ==2 and s ==2:
   print(zs)

LeiShen 发表于 2018-7-24 20:31:12

本帖最后由 LeiShen 于 2018-7-24 20:37 编辑

import math

for number in range(10,100):
m = int(number / 10)
n = int(number % 10)
m_f,n_f,number_f = 0,0,0

for i in range(2,int(math.sqrt(m))+1):
if m % i == 0:
m_f = 1
break
if m_f != 1 :
for i in range(2,int(math.sqrt(n))+1):
if n % i == 0:
n_f = 1
break
if m_f != 1 and n_f != 1 :
for i in range(2,int(math.sqrt(number)+1)):
if number % i == 0:
number_f = 1
break
if number_f != 1 and m !=1 and n !=1:
print(number)

LeiShen 发表于 2018-7-24 21:16:56

LeiShen 发表于 2018-7-24 20:31

新手就要勇往直前！！！

久疤K 发表于 2018-7-24 21:44:26

>>> def is_prime( n ):
if n < 2:
return False
for i in range(2,int(n**0.5)+1):
if n % i == 0:
return False
return True

>>> is_prime(4)
False
>>>
>>> def fun():
ls =
r = []
for x in ls:
for y in ls:
t = x*10+y
if is_prime( t ):
r.append( t )
print(r)

>>> fun()

>>>

liuzihe1979 发表于 2018-7-24 22:08:32

def func190():
num = 100
result = []

for i in range(10,100):
   flag = 0
   for j in range(2,i):
         if i%j == 0:
            flag = 1
   if flag == 0 :
         result.append(i)
return result

print(func190())

糊涂windy 发表于 2018-7-24 22:27:05

def Isprime(x):
if x<2:
   return False
for i in range(2,x):
   if x%i==0:
         return False
return True

for i in range(1,101):
c = i
flag =1
while i:
   b = i%10
   if Isprime(b):
         i = int(i/10)
   else:
         flag = 0
         break
if flag == 1:
   print(c)

ErTian 发表于 2018-7-25 01:06:07

答案

grf1973 发表于 2018-7-25 09:26:11

def isprime(n):
if n==2:
return True
if n==1 or n%2==0:
return False
for i in range(3,int(n**0.5)+1,2):
if n%i==0:
return False
return True

print()

喂猪专业户 发表于 2018-7-25 09:54:48

自己写了一个，提示最后一个判断是int和nonetype做与运算，问题出在哪

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Python：每日一题 190