Python练习题答案: 圆素数【难度：1级】

Python练习题答案: 圆素数【难度：1级】

圆素数【难度:1级】:

答案1:

def circular_permutations(n):n = str(n)return [int(n[i:] + n[:i]) for i in range(len(n))]def is_prime(n):return n > 1 and all(n % i != 0 for i in range(2, int(n**0.5)+1))def circular_prime(n):return all(is_prime(x) for x in circular_permutations(n))
​

答案2:

def circular_prime(n):return n in [2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 113, 131, 197, 199, 311, 337, 373, 719, 733, 919, 971, 991, 1193, 1931, 3119, 3779, 7793, 7937, 9311, 9377, 11939]​

答案3:

def is_prime(n):return n > 1 and all(n % i for i in xrange(2, int(n ** 0.5) + 1))def circular_prime(n):num = str(n) * 2length = len(num) / 2for a in xrange(length):if not is_prime(int(num[a:a + length])):return Falsereturn True
​

答案4:

from gmpy2 import is_primedef circular_prime(n):n=str(n)return all(is_prime(int(n[i:]+n[:i])) for i in range(len(n)))​

答案5:

def circular_prime(n):ns = str(n)lp = []for i in ns:ns = ns[1:] + ns[0]lp.append(ns)for i in lp:for j in range(2,int(i)):if int(i)%j == 0:return Falsereturn True and n != 1​

答案6:

from collections import dequedef is_prime(n):return n > 1 and all(n % i != 0 for i in range(2, int(n**0.5) + 1))def circular_prime(n):items = deque(str(n))for _ in range(len(items)):ate(1)if not is_prime(int("".join(items))):return Falsereturn True​

答案7:

def circular_permutations(n):s = str(n)return [int(s[i:] + s[:i]) for i in range(len(s))]def is_prime(n):return n == 2 or pow(2, n - 1, n) == 1def circular_prime(n):return all(is_prime(c) for c in circular_permutations(n))​

答案8:

def is_prime(n):d = 2while d * d <= n:if n % d == 0:return Falsed += 1return n > 1def rotate(l, n):return l[-n:] + l[:-n]def circular_prime(number):number_list = [int(x) for x in str(number)]if is_prime(number):check_list = [True]for index in range(1,len(number_list)):number_rotated = rotate(number_list,index)number_join = int(''.join(map(str, number_rotated)))if is_prime(number_join):check_list.append(True)if (len(check_list) == len(number_list)): return Trueelse: return Falseelse: return False​

答案9:

def circular_prime(n):return chr(n + 0x4E00) in '丑冗凋僝七乏併乡粣仅助佑亃丂丅峃乇义像严洁仇凟不下丟娯喋汱犡伷牟万乱'
​

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