Python 3.x 寻找素数的函数
我有两个相同的函数 该函数查找给定数之前的所有素数Python 3.x 寻找素数的函数,python-3.x,function,Python 3.x,Function,我有两个相同的函数 该函数查找给定数之前的所有素数 def count_primes2(num): primes = [2,3,] x = 5 if num < 2: return 0 while x <= num: for y in primes: # use the primes list! if x%y == 0: x += 2
def count_primes2(num):
primes = [2,3,]
x = 5
if num < 2:
return 0
while x <= num:
for y in primes: # use the primes list!
if x%y == 0:
x += 2
break
else:
primes.append(x)
x += 2
print(primes)
return len(primes)
def count_primes3(num):
primes = [2]
x = 3
if num < 2 :
return 0
while x <= num:
for y in range(3,x,2):
if x%y == 0:
x += 2
break
else:
primes.append(y)
x += 2
print(primes)
def count_primes2(num):
素数=[2,3,]
x=5
如果num<2:
返回0
当x时,您尝试附加y
而不是x
,就像您在count\u primes2
中所做的那样
def count_primes3(num):
primes = [2]
x = 3
if num < 2 :
return 0
while x <= num:
for y in range(3,x,2):
if x%y == 0:
x += 2
break
else:
primes.append(x) ### This shouldn't be a y
x += 2
print(primes)
def count_primes3(num):
素数=[2]
x=3
如果num<2:
返回0
当x时,您尝试附加y
而不是x
,就像您在count\u primes2
中所做的那样
def count_primes3(num):
primes = [2]
x = 3
if num < 2 :
return 0
while x <= num:
for y in range(3,x,2):
if x%y == 0:
x += 2
break
else:
primes.append(x) ### This shouldn't be a y
x += 2
print(primes)
def count_primes3(num):
素数=[2]
x=3
如果num<2:
返回0
而x