### Editorial for Equation, Modulo Equality, Long Beautiful Integer

Problem Discussed Today:
Editorial

This is a simple question, from the Codeforces editorial answer is 9n & 8n. Initially, I was not aware of this so I have tried the brute force as the value on n<=10^7 which will pass all the test cases in a given time limit.

I have taken b=1 and a=b+n and increment both a and b until both a and b are composite.
The complexity of this solution is O(n).

This is also a simple question as n<=2000 so we can go for a solution running in quadratic time complexity. So for this, I have sorted array b.

Now how should we choose x? Anyway, the x can take only the value from the (b[i]-a[i])%m as (a[i]+x)%m = b[i].
so for this, I have found all the difference (b[i]-a)%m as anyone value which we will get here must satisfy all other. I.e. for anyone x, (a[i]+x)%m will be equal to b[i]. For Comparision purpose I am using sort as this will make our comparison simple.

Sorting can be done in O(nlogn) time complexity. This should be done for all x, which at max can be n distinct number. So the total running time complexity of this solution is O(n^2logn). Which will pass in given time limit?

This is a little bit tricky problem, for which I have struggled a lot.
The problem simply says that find the number greater or equal to the given number X, such that a[i]=a[i+k].

So for this, the pattern should repeat, like if we that this example,
8 2
12345678

13131313 ( is the solution in which pattern is repeating )

So for this what we do, we take first k string and repeat it to get a string of size n, like here we will get 12121212 now check whether this is greater than or equal to given X if it is then done. Else get the last non-nine element in a string of size k from the beginning and increase it by 1 and all the 9 coming after that to be 0.
And using this string generate a pattern string of size n which we will get 13131313.

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