I'm sorry, but we can't move the contest time. Maybe you can virtual-participate in the Opencup.

Clashes with innopolis-open finals as well

Edit:doesn't clash(I thought innopolis is 5 hours instead of 4)

Yes.

C was easy than B but solution of B was on internet so lot of people solved it very quickly .

I found A the hardest in A,B,C

Who solved F are in almost 2000-st place because F costs 110min...

Killing a mosquito with a bazooka.

Why edge cases though? Couldn't you hardcode the periods of powers of 0-9, and use that?

You can also notice that all the numbers 0-9 will repeat the same ones digit every 4 powers.

For B, here's a simpler solution.

We know that $$$A, B, C > 1$$$ and the modulus is 10. The period is always a divisor of 4, so it suffices to find $$$B^C$$$ modulo $$$4$$$, say $$$x$$$, and find $$$a^{x + 4}$$$ modulo $$$10$$$.

Let $$$n>2,m>2$$$. Then observe that the only constraints on $$$A$$$ and $$$B$$$ are that $$$\forall i, B_i > max(A)$$$. So, we can find for a given maximum $$$x \in [1,k]$$$ the number of different sequences $$$A$$$ such that $$$max(A) = x$$$, which is $$$x^{n} - (x-1)^{n}$$$. Then, the number of sequences $$$B$$$ such that $$$B_i \geq x$$$ will be $$$(k-x+1)^m$$$. Multiply these values and add them for all $$$x$$$, and you have your answer.

Handle $$$n=1$$$ and $$$m=1$$$ separately.

Complexity is $$$O(k(\log m + \log n))$$$.

The answer turns out to be the number of monotone $$$k-$$$weightings of a complete bipartite digraph with $$$n$$$ vertices on the left and $$$m$$$ on the right for $$$n, m > 1$$$. For $$$\min(n, m) = 1$$$ the answer is $$$k^{\max(n, m)}$$$.

Toooooooooooooo complicated!

Edit:Brute force is actually $$$O(n \ log \ n)$$$ (?) but i'm not sure... It is $$$\sum_{i=1}^n \frac{n}{i}$$$

A can also be done with 3 forloops from 1 to n if you just break the loop when i*j>n or i*j*k>n.

    ll ans=0;
    for(ll i=1;i<=n;i++){
        for(ll j=1;j<=n;j++){
            if(i*j>n) break; 
            for(ll k=1;k<=n;k++){
                if(i*j*k<=n) ans++;
                else break;
            }
        }
    }
    cout<<ans;

why are u choosing <= i as maximum for m but exact i for choosing minimum in D??

O(n) solution

not so difficult to understand （ actually too lazy to explain ）

There are so many situations to consider: note that we can almost keep all b's except some special cases, and cancel all the a's to 0 or 1 if we want, the problem becomes "how to arrange the rest of a's" and "how many a's we can keep at the end of the final string".

1. All the a's are at the beginning: we cannot do anything to make b at the beginning, so we should keep 0 or 1 a at the beginning(depending on the parity of number of a's) and keep all the b's.

2. There are some a's at the end: we should move the continuous a's to the end by an operation, and by one operation with $$$k$$$ continuous a's we can increase the length of suffix with all a's by $$$k - 2$$$, so we only move those larger than 2 to the end.

3. No a's at the end: if the parity of total number of a's is even, we just cancel all the a's, otherwise, we must keep some a's between b's, or we need to remove (at most 2) b's to make a's at the end. When the suffix of b's is no longer than 2, we can keep the last a and remove any other a's, otherwise, try to keep the longest suffix of a's by removing 2 b's.

I haven't found a "clean" solution yet...

what if power(b,c,4) = 0

check for TC 2,2,2