I have to write a code to get 10 numbers at input. Then, find the number with higher number of prime factors. If two number have the same amount of prime factor, then print the bigger one.
For instance, for sample input of : input values:
123
43
54
12
76
84
98
678
543
231
the Output values should be like this:
678 3
Here is my code, however the output is not correct:
import math
# Below function will print the
# all prime factor of given number
def prime_factors(num):
# Using the while loop, we will print the number of two's that divide n
added1 = 0
added2 = 0
while num % 2 == 0:
added1 = 1
print(2, )
num = num / 2
for i in range(3, int(math.sqrt(num)) + 1, 2):
# while i divides n , print i ad divide n
while num % i == 0:
added2 = 1
print(i, )
added2 += 1
num = num / i
if num > 2:
print(num)
added = added1 + added2
print(added)
return added
# calling function
input_numbers = []
for i in range(0,10):
input_numbers.append(int(input()))
contest = 1
for item in input_numbers:
if prime_factors(item) > contest:
contest += 1
contest2 = item
print(contest2)
Using an algorithm from this answer I wrote one to solve your problem.
def prime_factors(n):
i = 2
factors = []
while i * i <= n:
if n % i:
i += 1
else:
n //= i
factors.append(i)
if n > 1:
factors.append(n)
return factors
input_numbers = []
for i in range(0,10):
input_numbers.append(int(input()))
highest_num = 0
highest_amount = 1
for num in input_numbers:
factors = list(dict.fromkeys(prime_factors(num)))
factors_amount = len(factors)
if factors_amount > highest_amount:
highest_num = num
highest_amount = factors_amount
elif factors_amount == highest_amount:
if num > highest_num:
highest_num = num
print(highest_num,highest_amount)