设若我们不考虑效率问题，如何编写一个代码量最小的求素数的代码
def issu(num):flag=1
for j in range(2,num):
if num % j ==0:
return 0
return flag flag=1去掉一行，加入import math，如此行数不变，效率高了
import math
def issu(num):
for j in range(2,int(math.sqrt(num)):
if num % j ==0:
return 0
return 1 def issu(num):
for i in range(2,num):
if num % i == 0:return False
return True qiuyouzhi 发表于 2020-6-24 09:18
还敢再精简一下吗
和菠菜的差不多哈哈，麻烦了点
def issu(num):
for j in range(2,num):
if num % j ==0:break
return bool(num%j) wp231957 发表于 2020-6-24 09:30
还敢再精简一下吗
当然
issu = lambda num:all(map(lambda p:num % p != 0,range(2,num)))
这样也可行：
f = lambda num : [ num for i in range(2,num) if (num % i == 0)]
print(bool(f(3)))
嘿嘿 继续来个简单 :
f = lambda num : 1 if [ num for i in range(2,num) if (num % i == 0)] == [] else 0
页:
[1]