PROBLEM CODE:
ZIO05001
PROBLEM LINK:
Editorialist:
Darshan Bari (https://www.codechef.com/users/buzz_95)
DIFFICULTY:
SIMPLE
PREREQUISITES:
Math
PROBLEM:
Given N, break it into some m integers P_1,P_2,…,P_M such that the LCM of these m integers is maximized. Answer for three values of N: {9,12,15}.
QUICK EXPLANATION:
Since the values of N are small, one can brute with optimizations such as avoiding repeated values and keeping the sequence increasing. For more explanation, read below.
EXPLANATION:

Since there are N circles and N arrows are supposed to be drawn between them such that for each circle, exactly one arrow begins at a circle and exactly one arrow ends at that circle. So
the problem breaks down to breaking the given number N into m integers P_1, P_2,…, P_m such that
P_1 + P_2 + … + P_m equals N. This means we are creating m loops of sizes P1, P2,…,
Pm. 
Also, because of “No arrow can both begin and end at the same circle.”, any loop cannot be
of size 1.(P_i \neq 1). 
The dance ends when all the monkeys have simultaneously returned to the circles where
they initially started. Hence a monkey from loop 1 will return to its initial position after
every P_1 steps, a monkey from loop 2 will return to its initial position after every P_2 steps,
and so on. Generalizing a monkey from loop i will return to its initial position after every P_i
steps. Hence the first instant when every monkey will return to its initial position will be the
LCM of P_1, P_2,…, P_m.
Hence we have to maximize lcm of P_1, P_2, P_3,…, P_m. 
One way to do this is to break the given N into m distinct integers considering that the lcm is
to be maximized. For this, it is greedy to consider m integers such that each pair is a
coprime integers that add up to N. Also, to minimize the number of duplicate sequences
keep the next term of the sequence greater than the current term. 
For example, consider N = 15. Start with 15 = 2 + 3 + 5 + 5, but since this yields the answer
as 30 and 5 is repeated, we try for another such sequence P (15 = 3 + 4 + 8), the answer is 
Next sequence can be
3 4 8
3 5 7
3 12
4 5 6
4 11
5 10
6 9
7 8
15 
For example: if N (Number of circles) = 5, it is profitable to make two loops of sizes 2 and

Because of this, after 6 steps all the monkeys return back to their initial positions
simultaneously.
Simulation:
Initial Positions: 1 2 , 3 4 5
Step 1: 2 1 , 5 3 4
Step 2: 1 2 , 4 5 3
Step 3: 2 1 , 3 4 5
Step 4: 1 2 , 5 3 4
Step 5: 2 1 , 4 5 3
Step 6: 1 2 , 3 4 5
For 9, breaking 9 into 4 and 5 gives 20 as the maximum number of steps.
For 12, breaking 12 into 3, 4 and 5 gives 60 as the maximum number of steps.
For 15, breaking 15 into 3, 5 and 7 gives 105 as the maximum number of steps.