After trying so hard, I could not solve problem A or problem B today, they were both so hard, help me how could i have solved it.
I thought there would be some logical solution for A, i thought of using STack, binary search in problem A.
But the solution is loops for 100 times? I dont understand what is purpose of such problems.
I want to quit cp after today.
ANy tips on how to solve problems will be appreciated.
Thanks
udgm should help you
Solving problems like A and B can actually be disheartening to face at times. Note that problem difficulties are not precisely consistent across contests (i.e. div2 A and B in contest 1 may be much easier than A and B in div2 contest 2). If you look at the standings, people generally struggled with A and B for a little while -- more than usual. I also did. I recommend 2 things then:
Systematic process of thinking. This refers to solving problems by repeating a 2 step process again and again: (1) make observations, (2) what is the next intuitive step from here (e.g. different cases, something needs to be optimized). This applies to B in particular from today's contest, but all problems roughly require this pattern of thought.
Contest strategy. This refers to jumping between problems depending on your strengths and weaknesses in problem topics or perhaps depending on the particular problem. I spent 30 mins on A and B without solving them, then jumped to C and solved that in a few minutes since I know binary search very well. Then I gained some confidence in the time I had left to take the time to solve A and B.
In all cases, keep practicing -- I know losses can be disheartening.
In Problem B, observe that the array a would be a arithmatic progression so you may notice one thing that all the unwanted numbers are to be converted to wanted numbers for example=> The n=7,b=2,c=1 then array would be a=[1,3,5,7,9,11,13] you can see that the unwanted numbers are [7,9,11,13] so you will have to eliminate them which would take 4 operations. So the answer would be the number of unwanted elements which could be easily calculated by using the formula n-floor((n-1-c)/b)+1 (Derived from the ap formula for nth term a+(n-1)*d=nth term). Then there were a few edge cases if all numbers are same that is b=0 then if c=n-2&&c<n then you would need n-1 operations else in other cases you will take n operations (you may take examples and see this).
And don't quit brother, sometimes it happens keep trying
pretty much a womp-womp situation, seeing you submitting problems once in different languages with AI-lookalike variable names and all other stuff
Good for u