Vfrdtyky

The task is to write function that foldes number into prime factors. By given number 'n' this function should return a list of tuples p_i,c^i, for example if input is 100, the output is (2,2),(5,2).
So, here is how i try to write it:

def factor(n):

c = 1

pre_ans = list()

temp_n=n

for i in range(2,temp_n+1):

    if (is_prime(i) == True) and (temp_n % i == 0):

        for j in range (2,temp_n+1):

            if (temp_n % (i ** j) == 0):

                pre_ans.append((i,j))

                temp_n /= (i **j)

        pre_ans.append((i,c))

        temp_n /= i

print(pre_ans)

It works wrong but I can't find a mistake :(

Your general idea is ok. However there are some minor issues with the following part of your code:

for j in range (2,temp_n+1):

    if (temp_n % (i ** j) == 0):

        pre_ans.append((i,j))

        temp_n /= (i **j)

pre_ans.append((i,c))

temp_n /= i

In fact the main problem is that you need to iterate in the other direction in this statement for j in range (2,temp_n+1). If you rewrite it like

def factor(n):

    c = 1

    pre_ans = list()

    temp_n=n

    for i in range(2,temp_n+1):

        if (is_prime(i) == True) and (temp_n % i == 0):

            for j in range (temp_n+1, 0,-1):

                if (temp_n % (i ** j) == 0):

                    pre_ans.append((i,j))

                    temp_n /= (i **j)

    print(pre_ans)

it will work. The whole code can also get written a bit shorter:

from collections import Counter



def factor(n):

    lst = 

    for i in range(2, n+1):

        while n % i == 0:

            lst.append(i)

            n = n / i

    return Counter(lst).items()



print(factor(100))

Fixed the code. This is working version

def factor(n):

c = 1

pre_ans = list()

temp_n=n

for i in range(2, n // 2 + 1):

    if (is_prime(i) == True):

        k = 1

        while temp_n % (i ** k) == 0:

            if temp_n % (i ** (k + 1)) == 0:

                k += 1

            else:

                k += 1

                break

        if k > 1:

            pre_ans.append((i, k - 1))

return pre_ans