@sagijagadish when you write cin.tie(NULL) at the beginning of the code, compiler doesn’t clear the output buffer every time it prints something. Whatever you try to print is stored in the output buffer and printed only once, when the execution of code is finished.
Bonus : Try using endl instead of ‘\n’ in your code. ‘endl’ clears the buffer and add new line at the end.
for the CHEFHPAL problem I tried writing the possible palindrome of length 16 and found a pattern
A short Pythonic code
My code
Nice way. I had said there are many solution to this problem.
Sorry, but I’ll see solutions later mate. Writing second part (even longer than the first. Though i hope anyone else on forum might be able to help you even before me.
It Took me 2 days to solve CHEFHPAL…and find the pattern for A = 2 and N > 8.
Here’s my soln - Soln
And Nice Editorial Bro!!!
Thanks Mate. Glad u liked.
Bonus: endl slows down execution a bit, try to use \n if possible xD
Hahaha I thought the same thing too ⌈Log2N⌉… I got irritated and randomly kept pressing ‘a’ and ‘b’ and I got a huge string with maximum value four …XD
You were lucky. I decided spending half an hour messing with random string till i accepted defeat and made tester program 
I couldn’t solve the CHEFHPAL problem, I read your editorial and now I understand it!!
And @taran_1407 thank you so much for spending time to write these helpful editorials!!
For CHEFHPAL, the proof of correctness is by exhaustion. For any value of a but 2, the answer is trivially "aaa..." or "abcabc...".
For a = 2, let’s look at the possibilities. The length-two palindromes are aa and bb. All eight three-letter strings contain at least one two-letter palindrome, or are length-three palindromes themselves, so two is minimal. We can therefore return the strings aab and aabb.
The three-letter palindromes are aaa, bbb, aba and bab. All five-letter strings contain at least one of these, as we can see by noticing that any three-letter string that contains no three-letter palindrome is aab, either with as and bs switched, backwards, both or neither. So we can just look at one and our impossibility proof will apply to all four. (aab is to bba as aabb is to bbaa, to baa as aabb is to bbaa, etc.) We have to start from a string symmetric to that and try to add two more letters to it. If we add a, we get aaba, if we add ba, we get aabba. If we add bb, we get aabbb. Therefore, three is minimal, and we can return aaabb, aaabab, aaababb, and aaababbb.
For n > 8, there is no way to fit three more as or bs before, after, or between aaa and bbb without getting a palindrome at least four letters long. So, four is minimal. The way to verify that "abaabbabaabb..." is optimal is to list all the substrings of length five and length six, and verify that they are not palindromes. Any odd palindrome of length greater than five would need to have a palindrome of length five in the center, wrapped by a pair of symmetric segments like a...a or ab...ba. Any even palindrome of length greater than six would need to have a palindrome of length six in the center. Since no such segment of length five or six exists in the string, it cannot contain any palindromes longer than four letters. So it is optimal.
Here is my simple approach to PERPALIN.
if(P>N || P<3){
“impossible”.
}
else{
1. generate a string of length P with all characters as same(let ‘b’). // string s(P,‘b’)
2. replace first and last characters with another character (let ‘a’). // s[0]=s[P-1]=‘a’
//Now we have a palindrome with length P;
3. print string s , N/P times. /* int r=N/P; for(int i=0;i<r;i++) cout<<s;*/
4. done // we generated string with period P… 
}For PERPALIN if(N % P != 0) impossible…
Glad u found them helpful!!.
I did it in complete lazy way just find out the pattern for the length of string 11,12,13,14,15 and 16 then i continued it then i noticed there was repetition of some pattern i.e., for 11 to 15,17 to 21,23 to 27 and so on add “bbaa” next 12 to 16,18 to 22 and so on add “baab” next 13 to 17,19 to 23 and so on add “aaba” next 14 to 18 ,20 to 24 and so on add “abab” similarly for 15 to 19 add “babb” and last 16 to 20 and so on add “abba”.you will get always ans 4 for string having only a & b for length having greater than 8.
Here is my solution link text
if you like please upvote me.
Nice approach mate. I said there are many approaches, but saw first submission, that didn’t use a string precomputed.
It is given in problem, that N%P = 0