Factor for Python
Factor some numbers…
"""
Factor a number
Wj.
"""
# Simple factoring
def factor(n, noduplicates = False):
intn = int(n)
factors = {}
lastfactor = n
i = 0
# 1 is a special case
if n == 1:
return {1: 1}
while 1:
i += 1
# avoid duplicates like {1: 3, 3: 1}
if noduplicates and lastfactor <= i:
break
# stop when i is bigger than n
if i > n:
break
if n % i == 0:
factors[i] = n / i
lastfactor = n / i
return factors
if __name__ == "__main__":
import sys
print "Enter an integer:"
number = sys.stdin.readline()
print "Factors: " + str(factor(int(number), True))
I think that you can make it a bit faster by stopping when i is greater than n**0.5. You could also then eliminate the check for duplicates.
Variable “intn” created at line 9 is never used.
Perhaps the correct line is:
“n = int(n)”
?
A bit slow, and such a long code, here is one i just wrote:
im not sure if the indentation will be right after i comment it, but if you know python you can figure it out:
#011011000110111101101100
x= int(raw_input (“Enter a number to get its factors: “))
factorlist=[]
for a in range (1,x):
for b in range (1,x):
if a*b==x:
if [b,a] not in factorlist:
factorlist.append([a,b])
if factorlist==[]:
print “This Number has no factors ”
print factorlist
raw_input()
Sorry but none of your code do what they should.
lets say factoring number 8. The right answer is 2x2x2 = 2^3 = 8
so the right answer always has to contain prime numbres.
Take a look a Euclides algoritm !!!!
Sorry for this news from Barcelona.