+ 2

# What is the best(fast) way to get all divisors of a number?

3 Answers

+ 1

from functools import reduce
def factors(n):
# get factors and their counts
factors = {}
nn = n
i = 2
while i*i <= nn:
while nn % i == 0:
if not i in factors:
factors[i] = 0
factors[i] += 1
nn //= i
i += 1
if nn > 1:
factors[nn] = 1
return factors
def divisorGen(n):
factors = list(factorsGen(n))
nfactors = len(factors)
f = [0] * nfactors
while True:
yield reduce(lambda x, y: x*y, [factors[x][0]**f[x] for x in range(nfactors)], 1)
i = 0
while True:
f[i] += 1
if f[i] <= factors[i][1]:
break
f[i] = 0
i += 1
if i >= nfactors:
return
def factorsGen(n):
for k,v in factors(n).items():
yield k,v
s = int(input())
print('Divisors of number {}, is {}'.format(s, list(divisorGen(s))))

0

#include <vector>
#include <iostream>
using namespace std;
int main()
{
vector<int> tab;
int num;
cin>>num;
for(int i = 0; i<=num; i++)
{
if (num%i == 0)
{
tab.push_back(i);
}
}
system("pause>nul");
return 0;
}
You can check if it's working, because I wrote this code on train, so I couldn't check myself ^^
tab holds divisors

0

A simple answer in python would be looping all numbers,
def divisor(num):
return [i for i in range(1,num+1) if not num%i]
print(divisor(60))
for faster algorithm, you should update the maximum number of your loop.
def divisor(num_max, counter=1):
if counter < num_max:
if num%counter == 0:
return [counter, int(num/counter)] + divisor(int(num/counter), counter+1)
else:
return divisor(num_max, counter+1)
else:
return []
print(divisor(60))