Getting timelimit errors

#1

i think my code is performing too many iterations
quest link-https://online.codingblocks.com/player/16974/content/5167?s=3847
code link-https://ide.codingblocks.com/s/69547

one more thing some number have more than one triplets set like(39,52,65),(39,80,89),(39,252,255)…these are all pythagors triplets of 39 ,so do i have to print all these triplets or only one triplet will fetch the answer?

#2

@shubhamshagy hey shubham you don’t need to use loops I am going to discuss a naive approach for this pythagorus triplet problem
take a input n
if n>0
if input n is even use this formula
x=((n x n)/4)-1;
y=((n x n)/4)+1;
if n is odd
x=(n x n-1)/2;
y=(1+n x n)/2;
else
cout<<-1;

#3

yeah i agree that this step is fast and simple but u cannot know the formula of all problem ,so i would like to go with the logical approach,so can you plz suggest any change in my code to remove that timelimit error yet giving correct solution.

#4

now ,talking about your solution any number we input >0,will be definitely even or odd so it will surely give output according to the formula,so in what case it will return -1?

#5

Hey, in a lot of ad-hoc problems, You need to use the problem specific approach (like formulas and identifying patterns etc.) to come up with a fast and effective solution.

Now in this particular question, every number greater than 2 would be a part of at least one triplet…
Check this link for the proof:

Also, showing any one triplet of that particular number would do, if it is a part of more than 1 triplet.

#6

whats wrong in this code-https://ide.codingblocks.com/s/69628

#7

@shubhamshagy hey shubham I have corrected your code you can refer this.

#8

Hey Shubham,
As you are not responding to this thread, I am marking your doubt as Resolved for now. Re-open it if required.

Please mark your doubts as resolved in your course’s “ Ask Doubt ” section, when your doubt is resolved.