Python：每日一题 134,Python交流,编程语言专区,鱼C论坛

冬雪雪冬 发表于 2018-1-8 10:50:44

Python：每日一题 134

本帖最后由冬雪雪冬于 2018-1-11 19:08 编辑

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题目：
我们知道质数是2, 3, 5, 7, 11, 13, 17, 19......
首先生成1000以内的质数列表，临近的质数相减都是偶数（除了3-2），我们把结果为质数或除2后为质数生成一个新的质数列表（例如差值为6，新列表中加入的是3），并去除重复的元素。
要求给出新列表，并由小到大排列。

紫诺发表于 2018-1-8 14:25:46

# -*- coding:utf-8 -*-

zs =
for i in xrange(2, 1000):
for temp in xrange(2, i):
   if i % temp == 0:
         break
   if temp == i - 1:
         zs.append(i)

nzs = []
for i in range(1, len(zs)):
tmp = zs - zs
if tmp / 2 in zs:
   if tmp / 2 not in nzs:
         nzs.append(tmp / 2)

print sorted(nzs)

shigure_takimi 发表于 2018-1-8 16:06:24

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

primeInThousend =
for i in range(3, 1000, 2):
if isPrime(i):
   primeInThousend.append(i)

length = len(primeInThousend)

x = set( - primeInThousend[:length-1] for i in range(length-1)])
result = []
for i in x:
if isPrime(i):
   result.append(i)
elif isPrime(i//2):
   result.append(i//2)

print(sorted(list(set(result))))

#result =
#不知道是不是这个意思。

浪子无名1517 发表于 2018-1-8 17:15:52

from math import sqrt
def is_prime(n):
if n<=1:
   return 0
for i in range(2,int(sqrt(n))+1):
   if n%i==0:
         return 0
return 1
def gg():
li = []
for n in range(1,1001):
   if is_prime(n):
         li.append(n)
re = set()
for i in range(1,len(li)):
   if is_prime(li-li):
         re.add(int(li-li))
   elif is_prime((li-li)/2):
         re.add(int((li-li)/2))
print(re)

coopr 发表于 2018-1-8 18:56:45

本帖最后由 coopr 于 2018-1-8 18:59 编辑

#每日一题134
temp = 0
zhishu = []
for i in range (1,1000):
for j in range (2,i):
sum = i % j
if sum == 0:
temp = 1
if temp == 0:
zhishu.append(i)
temp = 0
print(zhishu) #质数列表
lengths = len(zhishu)
list = []
for k in range(1,lengths):
cha = zhishu - zhishu
if cha %2 == 0:
shang = cha//2
if cha in zhishu and cha not in list:
list.append(cha)
if shang in zhishu and shang not in list:
list.append(shang)
list.sort()
print(list)

bush牛 发表于 2018-1-8 19:58:36

from math import sqrt
def is_prime(n:int):
for i in range(2, int(sqrt(n)) + 1):
   if n % i == 0:
         return False
return True
s = list(filter(is_prime, range(2, 1001)))
a = set()
for i in range(1, len(s)):
tmp = s - s
if tmp in s:
   a.add(tmp)
elif (tmp / 2).is_integer() and int(tmp / 2) in s:
   a.add(int(tmp / 2))
print(sorted(list(a)))

solomonxian 发表于 2018-1-8 20:35:44

差值都是偶数，里面只有2一个质数吧
def fun(n):
prime_nums = [i for i in range(2, n+1)
            if 0 not in {i%j for j in range(2, int(i**0.5)+1)}]
lst = {2 if j-i == 2 else (j-i)/2
      for i,j in zip(prime_nums, prime_nums)}
return sorted(num for num in lst if num in prime_nums)
之前的题目有鱼油介绍过gmpy2，模仿写一个
from gmpy2 import next_prime

def fun2(n):
prms =
while prms[-1] <= n:
   prms.append(next_prime(prms[-1]))
lst = {2 if j-i == 2 else (j-i)/2
      for i,j in zip(prms[:-1], prms)}
return sorted(num for num in lst if num in prms)

流月飞星 发表于 2018-1-8 20:58:17

kankan

wyp02033 发表于 2018-1-8 21:38:58

def generate_primes(n):
primes = * n
primes, primes = False, False
for (i, prime) in enumerate(primes):
   if prime:
         for j in range(i*i, n, i):
            primes = False
primes =
return primes

def main():
primes = generate_primes(1000)
new_primes = []
for i in range(1,len(primes)):
   difference = primes - primes
   if difference in primes and difference not in new_primes:
         new_primes.append(difference)
   elif difference % 2 == 0 and difference // 2 in primes and difference // 2 not in new_primes:
         new_primes.append(difference//2)
new_primes.sort()
print(new_primes)

if __name__ == '__main__':
main()
结果：

lihw 发表于 2018-1-8 21:41:15

import math
lsit1=[]
lsit2=[]
sr_num=0

for i in range (2,1000):
fg=0
for n in range(2,int(math.sqrt(i))+1):
   if i % n ==0:
         fg=1
         break
if fg ==0:
   lsit1.append(i)

print(lsit1)
for i in range (len(lsit1)-1):
nu= lsit1-lsit1
if nu in lsit1:
   lsit2.append(nu)
if nu//2 in lsit1:
   lsit2.append(nu//2)
print(set(lsit2))

编程新血 发表于 2018-1-9 19:47:05

求素数函数来自网络，是个“埃氏筛法”什么的，反正不是我写的。

def numsu(n):
'''返回n以内的素数列表'''
listdef=[]
flag = *(n+2)
p=2
while(p<=n):
   listdef.append(p)
   for i in range(2*p,n+1,p):
         flag = False
   while 1:
         p += 1
         if flag:
            break
return listdef

list1=numsu(1000)
list2=[]
'''取素数列表两元素差的一半组成list2'''
for i in range(1,len(list1)):
temp=int((list1-list1)/2)
list2.append(temp)

'''set排重'''
list2=list(set(list2))

'''是素数放入list3'''
list3=[]
for i in list2:
if i in list1:
   list3.append(i)
print(list3)

achencan 发表于 2018-1-9 20:12:31

def is_sushu(num):
res=True
for x in range(2,num-1):
if num%x==0:
res=False
return res
return res
list1=
list2=[]
for x in list1:
list2.append(x-list1)
list2.pop(0)
list3=set(list2)
list4=[]
for x in list3:
if is_sushu(int(x/2)) and int(x/2)>0:
list4.append(int(x/2))
print(list4)

h20060304 发表于 2018-1-14 16:56:33

本帖最后由 h20060304 于 2018-1-14 17:17 编辑

在下不才，献丑了。

# 在质数列表中，除了第一个质数2为偶数外，均为奇数，故先生成一个无限的奇数序列，从而为后续可以生成无限素数，打下基础，摆脱局限性。

def _odd_iter():
n = 1
while True:
   n += 2
   yield n

# 根据“埃氏筛选法”，逐步筛出所有的素数。

def _not_divisible(n):
return lambda x: x % n > 0

def primes():
yield 2
it = _odd_iter()
while True:
   n = next(it)
   yield n
   it = filter(_not_divisible(n), it)

# 主程序，这里用first，和two，依次从素数序列中，取出两个元素，按照题目设定条件，进行筛查，并最终总返回结果。

def check_prime():
prime = primes()
first = next(prime)
two = next(prime)
lst = []
while two < 1000:
   comp =
   for prime1 in primes():
         if comp >= prime1:
            if prime1 in comp:
               lst.append(prime1)
               continue
         else:
            break
   first, two = two, next(prime)
lst = set(lst)
print('运行结果：{}'.format(lst))

check_prime()

wojiaodabai 发表于 2018-1-16 22:34:40

def primeNum(low,upper):
num=[]
for i in range(low,upper+1):
   fg=0

   for j in range(2,i):
         if i%j==0:
            fg=1
            break
   if fg==0:
         num.append(i)
return num

def fun(list1):
new=[]
for i in range(1,len(list1)):
   x=list1-list1

   if x in list1:
         new.append(int(x))


   elif x/2 in list1:
         new.append(int(x/2))
return new



list1=primeNum(1,1001)
list3=list(set(fun(list1)))

print(list3)

YBKathus 发表于 2018-2-5 22:04:17

萌新{:10_254:}
import math
def prime(n):
if n == 2 or n == 3:
      return True
i = int(math.sqrt(abs(n)))
while i >1:
   if n % i == 0:
         return False
         break
   else:
         i -= 1
         if i < 2:
            return True
def main():
a = []
b = []
for i in range(1, 1000):
   if prime(i):
         a.append(i)
for p in range(2, len(a)):
   m = (a - a)/2
   if prime(m) == True :
         if m not in b:
            b.append(m)
b.sort()
print(b)
main()

dunhee 发表于 2018-2-8 16:44:55

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

def f(n):
S = set()
L =
for i in range(len(L)-1):
   if isPrime(L - L):
         S.add(L - L)
   elif isPrime((L - L)/2):
         S.add(int((L - L)/2))
return sorted(list(S))

print(f(1000))

fan1993423 发表于 2018-3-26 18:38:59

本帖最后由 fan1993423 于 2018-3-26 18:40 编辑

m=[]
for a in range(2,1001):
if a==2:
   m.append(a)
else:
   if 0 not in :
            m.append(a)
i=1
l=[]
while i<=len(m)-2:
if m-m in m:
   l.append(m-m)
elif (m-m)//2 in m:
            l.append((m-m)//2)
i+=1
print(list(set(l)))

fan1993423 发表于 2018-3-26 18:41:04

<m=[]
for a in range(2,1001):
if a==2:
   m.append(a)
else:
   if 0 not in :
            m.append(a)
i=1
l=[]
while i<=len(m)-2:
if m-m in m:
   l.append(m-m)
elif (m-m)//2 in m:
            l.append((m-m)//2)
i+=1
l=set(l)
print(list(l))>

foxiangzun 发表于 2019-1-3 23:10:32

#coding=utf-8

primeList = []
primeNew = []

#生成质数
def getPrime() :
   for i in range(2, 1001) :
            flag = 0
            for j in range(2, i + 1) :
                     if i % j == 0 :
                           flag += 1
            if flag < 2 :
                     primeList.append(i)

#质数判定并返回
def primeConfirm(n) :
   flag = 0
   for i in range(2, n + 1) :
            if n % i == 0 :
                     flag += 1
   if flag < 2 :
            primeNew.append(n)

#质数差值计算
def start() :
   getPrime()
   tempList = []
   tempIndex = 0
   for i in primeList :
            tempIndex = primeList.index(i)
            if tempIndex > 1 :
                     tempList.append(i - primeList)
   for i in tempList :
            if i % 2 != 0 :
                     primeConfirm(i)
            else :
                     primeConfirm(i // 2)
   print(set(primeNew))

start()

kinkon 发表于 2022-3-9 13:58:53

def fun134(n):
p = []
for i in range(2, n + 1):
   for j in range(2, int(i ** 0.5) + 1):
         if i % j == 0:
            break
   else:
         p.append(i)
#print(p)

s = set(p)
ans = set()
for i in range(1, len(p)):
   cur = p - p
   if cur in s:
         ans.add(cur)
   if cur / 2 in s:
         ans.add(cur // 2)

return sorted(ans)

print(fun134(1000))

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Python：每日一题 134