Question Name: Chocolates-By-Numbers or ChocolatesByNumbers
When we met with an empty wrapper, we must have been this position for twice. We use i for the first time and j for the second time. Due to the modulo feature, there must be nature number, to say k, so that: i * M + k * N = j * M. Then we could easily prove that the smallest (earliest) i must be zero (for all i != 0, then (i-i) * M + k * N = (j-i) * M ). So the first eaten position would be first position that you meet again. Finally, the j would be the number of chocolates that you will eat.
Solution to Chocolates-By-Numbers by codility
def gcd(a, b):
# Get the greatest common divisor
if (a % b == 0):
return gcd(b, a % b)
def solution(N, M):
lcm = N * M / gcd(N, M) # Least common multiple
return lcm / M
You can just remove M from last two lines, so it becomes:
N / gcd(N, M) # this is the correct result
You are right! Thanks for your reminder!
First of all, congratulations for the blog, it is an incredible source of knowledge!
I came up with an alternative solution without using the gcd() function. It looks good but the performance for extremely large numbers not satisfied, I just can’t figure out why https://app.codility.com/demo/results/trainingXCB2M3-MYJ/
I leave it here in case someone finds it useful, its only 87%. Thanks again!
A java solution with 100/100
How did you come up with the following?
round <= (N/oneRoundMaxChocolate) +1
The above implies that the maximum value of "round" is "(N/oneRoundMaxChocolate)+1"