How it works in python? ruby code

Question

I don't understand what it code doing, please help.

How it will be working in python or in another simple language?

a = []
a << [1]

for i in 2..10001
    f = 0
    a.each{ |group|
        m = 1
        group.each { |c|
            m *= i % c
        }
        f += m
        if m > 0
            group << i
            break
        end
    }
    a << [i] if f == 0
end

p a
p a.size

Can you be more specific about what question you would like answered? — Linuxios
– Linuxios, Commented Oct 28, 2016 at 18:52
@Linuxios would be nice if someone show same program on another language, more simply — Dima Kabaruhin
– Dima Kabaruhin, Commented Oct 28, 2016 at 18:54
@DanielRoseman sorry if I broke the rules of stackoverflow can please you mention a few services that may help me — Dima Kabaruhin
– Dima Kabaruhin, Commented Oct 28, 2016 at 18:59

Adam Van Prooyen · Accepted Answer · 2016-10-28 18:57:04Z

2

Literally translated to python this is:

a = []
a.append([1])

for i in range(2,10001 + 1):
    f = 0
    for group in a:
        m = 1
        for c in group:
            m *= i % c
        f += m
        if m > 0:
            group.append(i)
            break
    if f == 0:
        a.append([i])

print a
print len(a)

answered Oct 28, 2016 at 18:57

Adam Van Prooyen

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Comments

Patrick Haugh · Accepted Answer · 2016-10-28 18:57:14Z

0

In python:

a = []
a.append(1)

for i in range(2, 10000):
    f = 0
    for group in a:
        m = 1
        for c in group:
            m *= i%c
        f += m
        if m>0:
            group.append(i)
            break
    if not f:
        a.append(i)

print(a)
print(len(a))

answered Oct 28, 2016 at 18:57

Patrick Haugh

61.3k13 gold badges94 silver badges101 bronze badges

4 Comments

Jared Goguen Over a year ago

Presumably the down-vote was for the wrong upper limit of range?

Patrick Haugh Over a year ago

@JaredGoguen dunno. It's been a while since i did any ruby, so i'm not surprised i misremembered something, but i am surprised that someone would choose a loop bound of 10001

Jared Goguen Over a year ago

Even so, the upper bound on range would then be 10001 as it isn't inclusive.

Adam Van Prooyen Over a year ago

In ruby a..b is a to b inclusive and a...b is a to b exclusive.

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