我遇到一个代码挑战,要求我们使用上一个函数创建3个函数。我们正在使用“基本python”,因此没有导入。没有lambda的版本将是理想的,但是都欢迎使用。
find_factors
functionis_prime
function - using thefind_factors
functionhcf
function - using theis_prime
function
The first two functions return the factors and prime numbers and the is_prime
function is using the find_factors
function as our instructor asked.
def find_factors(num):
for value in range(1, num + 1):
if num % value == 0:
print("{0}".format(value))
return num
def is_prime(x):
if x < 2:
return False
else:
for a in range(2,find_factors(x)):
if x % a == 0:
return False
return True
Next we were asked to use the is_prime
function in this hcf
function to find the HCF.
How do I use the second is_prime
function in this third hcf
function?
def hcf(x, y):
if x > y:
small = y
else:
small = x
for i in range(1, small+1):
if((x % i == 0) and (y % i == 0)):
hcf = i
return hcf
Is it even possible to find HCF from normal primes? Maybe our instructor meant prime factorization?