Problems from ACM ICPC 2013

I would like to ask you if you will post the problem from ACM ICPC 2013 on codechef? (post = put them on practice session).


I doubt it, because adding problems to practice will require test files. And if test files are not available then recreating them would be too much work. But wait for the official word.

Anyhow problems can always be discussed on Discuss!

You could practice the problems at this link :slight_smile:


thanks @deepankarak
what kind of problem is to think about it?

just print “Hello World!” -_-

are u sure???

If we get the test data, we will be happy to put it in our practice section.

views on this question???

were u able to do hello???

Yes I have, though I thought the answer might be really a secret it was not.

what was it???

@ka4tik has already hinted, its just a cakewalk I think they might have put it for imposing some penalty or testing purpose, rest I don’t know

Official discussion of all problems on YouTube :slight_smile:


seems the link isnt working? here it is

@ka4tik no this is not correct!!

Hello @devanshug,

About that question you asked about, which is very interesting, I guess it’s just a maths question right? Although I don’t know how to solve it yet, I guess all we need to do is to find out the number of distinct factors of a given number and add up all the permutations of the remaining factors… So for 20 = 5*2^2, we have two distinct factors, 2 and 5, so we can start arrangements with two different numbers 2 and 5. So we can have:

2 _ _

5 _ _

So we get:

2 2 5

2 5 2

5 2 2

This is because having the 5 and 2 switched on the two free places counts as a different arrangement…

So my guess is to count all distinct arrangements that begin with a new factor and sum everything up… Of course, the possible number of distinct arrangements will be conditioned by the number which we are factoring, but, finding out how to do the “reverse work” and actually implementing this or prove that it is correct, is, for me, a totally impossible task so far, but this is the line of thought I have developed so far for this problem… It might not me correct and for sure it is not complete… But I’m clueless… :frowning:

Best regards,


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You can see the official discussion on channel link posted by sunny_patel. Here is the link for this particular problem

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COunting the number of ways I think is “Permutations with Repetition of Indistinguishable Objects” = n!/(f1!*f2!..)

Thanks for this link :smiley: