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Solutions Codeforces Round #830 (Div. 2) Editorial

  Codeforces Round #830 (Div. 2) Editorial Thanks for the participation! 1732A - Bestie Let's make an important observation:  gcd ( n − 1 , n ) = 1 gcd ( n − 1 , n ) = 1  for any value of  n n . Moreover, choosing  i = n − 1 i = n − 1  and  i = n i = n  are the cheapest operations. From this we can conclude that the answer is  ≤ 3 ≤ 3 . Let  g g  be the  gcd gcd  of all numbers in the array. Then we have the following cases: If  g = 1 g = 1 , then the operation can be omitted and the answer is  0 0 , Otherwise, let's try the cheapest operation  i = n i = n . If  gcd ( g , n ) = 1 gcd ( g , n ) = 1 , then the answer is  1 1 . Otherwise, let's try the next cheapest operation, ie  i = n − 1 i = n − 1 . If  gcd ( g , n − 1 ) = 1 gcd ( g , n − 1 ) = 1 , then the answer is  2 2 . Otherwise, the answer is  3 3 , since  gcd ( g , n − 1 , n ) = 1 gcd ( g , n − 1 , n ) = 1 . 1732B - Ugu Let's mentally imagine the following array of length  n − 1 n − 1 :  a i = 0 a i = 0  if  s i

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