text: # SUMAGCD easiest Editorial
  
  Editorial:
  
  1. Use set to store the unique elements from all the given elements.
  2. Find the GCD of all the unique elements(say X) except the last two elements i.e. maximum and second maximum elements.
  3. now find max from GCD (X, second maximum)+ maximum and GCD (X, maximum)+ second maximum.
  4. That’s it, you have got the maximum element as the ans.
  8 Likes
  
  And that’s not an easy editorial at all cuz you never explained why are you doing those steps
  
  27 Likes
  
  How yoy reached to the above conclusion
  
  Even i followed the same approach, but i understood the logic after trying to find the max GCD of various different problems, but i still was in doubt as to why that approach worked in the first place. If anyone here could provide a concrete proof as to why the following approach works, it would be great!
  
  2 Likes
  
  This proof is not very rigorous but I hope it gives some intuition. So the first thing is to realize that the highest (let’s call its value max) and the second highest unique elements cannot be in the same subsequence. Let’s say they were in the same subsequence, then the GCD of the group containing both of them can be at most the max / 2. This is because the prime factorization of the highest element might look p1 * p2 * … pk, and the GCD of the subsequence would be maximized if the second highest contained all these primes p1…pk except the smallest one (they have to be unique). In the best case that prime is 2.
  
  Now, in the best case the other subsequence just has one element, the third highest, but since that is less than the second highest it can also only be max / 2. So overall the total can only be max, but in that case we are just better off putting the max alone in one subsequence, and the other subsequence will have gcd at least 1, which will give us max + 1.
  
  By the same logic, we can also see that both subsequences cannot have more than 1 unique element, because if they did the highest gcd for both would be max / 2, which only yields a total of max. So, you see which one is better, putting the highest separately or the second highest and then output your solution.
  
  9 Likes
  
  first of all GCD of two no.s is always a no. less than or equal to the smaller no.
  After doing some observations i got to know the exact steps.
  And it is an easy editorial because complexity of a soln doesn’t depend upon its explanation.
  The steps are easy to understand and might not be difficult for a top 100 coder of INDIA.
  
  2 Likes
  
  Absolutely correct explained.
  
  1 Like
  
  Bro but your editorial is not required for top coders…they solve by their own…only new coders need it so you need to explain well …
  I got it and hv no issues but saying all this just to realize you what you said…
  
  7 Likes
  
  Could you please help me to find out a test case where my code for SUMAGCD fails?
  
  code
  
  What observation? Hit and trial is not observation. You gotta prove it bro.
  
  3 Likes
  
  please give some input to fail my code
  https://www.codechef.com/viewsolution/24828838
  
  In one subsequence,we have put either the max element or the second max element…Why are we not checking all the elements in that subsequence?
  
  Yeah that easy but not an editorial
  
  I have a solution which take one element in one set and rest of them in another and it is O(n) also but still what is the proof that one set will only have one value elements ?
  
  what’s wrong here
  give me runtime error
  
  import math
  t = int(input())
  while t>0 :
  n = int(input())
  l = set(map(int,input().split()))
  l = list(l)
  l.sort()
  #print(l)
  if(len(l)==1):
  print(2*l[0])
  elif(len(l)==2):
  print(l[0]+l[1])
  elif(len(l)>2):
  fmax = l[-1]
  smax = l[-2]
  gcd3 = l[0]
  for i in range(0,n-2):
  gcd3 = math.gcd(gcd3,l[i])
  gcd1 = math.gcd(gcd3,smax) + fmax
  gcd2 = math.gcd(gcd3,fmax) + smax
  if(gcd1>gcd2):
  print(gcd1)
  else:
  print(gcd2)
  t-=1
  
  I used the same approach . But yours seems cleaner
  
  Read the official editorial . The reason to put a single element in one set is explained clearly.
  
  I know but it’s not in this editorial !!
  It’s incomplete !!
  
  1 Like
  
  Wish !!! had a haha react option…
  
  Solution Approach :
  
  ==> Reason for two maximum distinct number not being in same set :
  
  The easiest maximum you will get is (maximum number + 1).
  Now the GCD of two distinct numbers is always <= (maximum of two numbers/ 2)
  
  if i did so, the maximum value i can now reach is (Third maximum + 1 :: In case max two number’s gcd goes to zero) or (maximum of the given values :: How -> see this example 70,35,35,35)
  
  But i have (maximum + 1) in my pocket already with me.
  
  ==> Reason for taking GCD of rest of the numbers and putting it in only one set :
  GCD of two numbers is always <= (maximum of two numbers/ 2)
  
  Now instead of making both the maximum number be divided by 2 halves we should make only one which suits the best for the maximum answer because again i have (maximum + 1) in my pocket already with me.
  
  Hope this would help
explanation: The provided text is a discussion thread about a competitive programming problem, specifically the "SUMAGCD" problem. It includes an editorial explaining the solution approach and a series of comments from users discussing the solution, asking for clarifications, and sharing their thoughts. The text contains multiple questions and answers related to the problem and its solution.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # SUMAGCD easiest Editorial
  
  Editorial:
  
  1. Use set to store the unique elements from all the given elements.
  2. Find the GCD of all the unique elements(say X) except the last two elements i.e. maximum and second maximum elements.
  3. now find max from GCD (X, second maximum)+ maximum and GCD (X, maximum)+ second maximum.
  4. That’s it, you have got the maximum element as the ans.
  8 Likes
  
  And that’s not an easy editorial at all cuz you never explained why are you doing those steps
  
  27 Likes
  
  How yoy reached to the above conclusion
  
  Even i followed the same approach, but i understood the logic after trying to find the max GCD of various different problems, but i still was in doubt as to why that approach worked in the first place. If anyone here could provide a concrete proof as to why the following approach works, it would be great!
  
  2 Likes
  
  This proof is not very rigorous but I hope it gives some intuition. So the first thing is to realize that the highest (let’s call its value max) and the second highest unique elements cannot be in the same subsequence. Let’s say they were in the same subsequence, then the GCD of the group containing both of them can be at most the max / 2. This is because the prime factorization of the highest element might look p1 * p2 * … pk, and the GCD of the subsequence would be maximized if the second highest contained all these primes p1…pk except the smallest one (they have to be unique). In the best case that prime is 2.
  
  Now, in the best case the other subsequence just has one element, the third highest, but since that is less than the second highest it can also only be max / 2. So overall the total can only be max, but in that case we are just better off putting the max alone in one subsequence, and the other subsequence will have gcd at least 1, which will give us max + 1.
  
  By the same logic, we can also see that both subsequences cannot have more than 1 unique element, because if they did the highest gcd for both would be max / 2, which only yields a total of max. So, you see which one is better, putting the highest separately or the second highest and then output your solution.
  
  9 Likes
  
  first of all GCD of two no.s is always a no. less than or equal to the smaller no.
  After doing some observations i got to know the exact steps.
  And it is an easy editorial because complexity of a soln doesn’t depend upon its explanation.
  The steps are easy to understand and might not be difficult for a top 100 coder of INDIA.
  
  2 Likes
  
  Absolutely correct explained.
  
  1 Like
  
  Bro but your editorial is not required for top coders…they solve by their own…only new coders need it so you need to explain well …
  I got it and hv no issues but saying all this just to realize you what you said…
  
  7 Likes
  
  Could you please help me to find out a test case where my code for SUMAGCD fails?
  
  code
  
  What observation? Hit and trial is not observation. You gotta prove it bro.
  
  3 Likes
  
  please give some input to fail my code
  https://www.codechef.com/viewsolution/24828838
  
  In one subsequence,we have put either the max element or the second max element…Why are we not checking all the elements in that subsequence?
  
  Yeah that easy but not an editorial
  
  I have a solution which take one element in one set and rest of them in another and it is O(n) also but still what is the proof that one set will only have one value elements ?
  
  what’s wrong here
  give me runtime error
  
  import math
  t = int(input())
  while t>0 :
  n = int(input())
  l = set(map(int,input().split()))
  l = list(l)
  l.sort()
  #print(l)
  if(len(l)==1):
  print(2*l[0])
  elif(len(l)==2):
  print(l[0]+l[1])
  elif(len(l)>2):
  fmax = l[-1]
  smax = l[-2]
  gcd3 = l[0]
  for i in range(0,n-2):
  gcd3 = math.gcd(gcd3,l[i])
  gcd1 = math.gcd(gcd3,smax) + fmax
  gcd2 = math.gcd(gcd3,fmax) + smax
  if(gcd1>gcd2):
  print(gcd1)
  else:
  print(gcd2)
  t-=1
  
  I used the same approach . But yours seems cleaner
  
  Read the official editorial . The reason to put a single element in one set is explained clearly.
  
  I know but it’s not in this editorial !!
  It’s incomplete !!
  
  1 Like
  
  Wish !!! had a haha react option…
  
  Solution Approach :
  
  ==> Reason for two maximum distinct number not being in same set :
  
  The easiest maximum you will get is (maximum number + 1).
  Now the GCD of two distinct numbers is always <= (maximum of two numbers/ 2)
  
  if i did so, the maximum value i can now reach is (Third maximum + 1 :: In case max two number’s gcd goes to zero) or (maximum of the given values :: How -> see this example 70,35,35,35)
  
  But i have (maximum + 1) in my pocket already with me.
  
  ==> Reason for taking GCD of rest of the numbers and putting it in only one set :
  GCD of two numbers is always <= (maximum of two numbers/ 2)
  
  Now instead of making both the maximum number be divided by 2 halves we should make only one which suits the best for the maximum answer because again i have (maximum + 1) in my pocket already with me.
  
  Hope this would help
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The provided text is a discussion thread about a competitive programming problem, specifically the "SUMAGCD" problem. It includes an editorial explaining the solution approach and a series of comments from users discussing the solution, asking for clarifications, and sharing their thoughts. The text contains multiple questions and answers related to the problem and its solution.
  
  Label: multiple-question-and-answer