A******o
发帖数: 231 | 1 1. programming
for an integer, find out all the prime factors whose product is the integer
itself, eg:
12 = 2*2*3
print out those factors in a list.
2. database
describe how you would design an inventory system for a library, what tables
would you create, what attributes would there be, etc.
答得一般，用wireless通话中间还断线了一下，ft |
p*****2
发帖数: 21240 | |
Q*******e
发帖数: 939 | 3 def prime_factor(num):
factor = 2
st = "%d 's factor:" % num
while factor <= num:
if num % factor == 0 :
st = st + ' ' + str(factor)
num = num / factor
else:
factor = factor + 1
print st
prime_factor(17)
prime_factor(20)
prime_factor(200)
prime_factor(217)
~ |
P**********m
发帖数: 213 | |
e******x
发帖数: 184 | |
i*********7
发帖数: 348 | 6 咋一眼看的确有不对劲，但仔细推敲后实际上是对的。
【在 P**********m 的大作中提到】
: 楼上的不太对吧...
|
l******n
发帖数: 9344 | 7 loop明显到sqrt(num)就可以了
这算常识吧
【在 Q*******e 的大作中提到】
: def prime_factor(num):
: factor = 2
: st = "%d 's factor:" % num
: while factor <= num:
: if num % factor == 0 :
: st = st + ' ' + str(factor)
: num = num / factor
: else:
: factor = factor + 1
: print st
|
l******n
发帖数: 1250 | |
w****f
发帖数: 684 | 9 Should be stop at num/2, not sqrt(num). right?
eg. num=26 =2*13 while sqrt(num) =5.
【在 l******n 的大作中提到】
: loop明显到sqrt(num)就可以了
: 这算常识吧
|
