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CodeChef October Lunchtime Problem Solution | codechef Lunctime Problem solution

 CodeChef October Lunchtime Problem Solution | codechef Lunctime Problem solution - Codechef is one of the best coding website for competitive programming. Every month CodeChef conduct many contest so CodeChef lunchtime is one of a popular contest of CodeChef.  Codechef October Lunchtime 2020 is on 31st October. Lunchtime is a rated contest so you should must part on this contest. Contest Link- https://www.codechef.com/LTIME89/?itm_medium=hpbanner&itm_campaign=LTIME89 Problem 1- AND plus OR Sulution  till then you can  visit-  Codechef october cook off Contest Problem solution  and-  CodeChef Long challenge editorial | codechef long challenge solutions Join Telegram for Code Discussion and more editorial of upcoming coding contest - https://t.me/competitiveProgrammingDiscussion 

Decreasing Srrnmieeda October Cook off challenge Solution | Easy Explain Decreasing Srrmieeda

 Decreasing Srrnmieeda October Cook off challenge Solution | Easy Explain Decreasing Srrmieeda- Easy Explanation of Decreasing Srrnmieeda October Cook-off challenge Solution  With Code.Here Is Problem Statement. Problem Statement- You are given two integers  L L  and  R R . Find the smallest non-negative integer  N N  such that N % L > N % ( L + 1 ) > … > N % ( R − 1 ) > N % R . N % L > N % ( L + 1 ) > … > N % ( R − 1 ) > N % R . Here,  % %  is the modulo operator, so  A % B A % B  is the remainder of  A A  after division by  B B . For example,  11 % 3 = 2 11 % 3 = 2 . Input The first line of the input contains a single integer  T T  denoting the number of test cases. The description of  T T  test cases follows. The first and only line of each test case contains two space-separated integers  L L  and  R R . Output For each test case, print a single line containing one integer ― the smallest possible  N N , or  − 1 − 1  if no solution exists. It is guaranteed