You guys getting lives?

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So you're going to say the opposite thing when there is no contest in 3 consecutive days?

Actually, while this is generous of CodeForces, I think it would be ideal if contests were spread out a little more evenly. I had prior commitments for the weekend and Monday and will miss 3 contests (that would typically be spread over 2-3 weeks instead of 3 days).

same here

You are giving the contest that's the best thing.

Yes, it was a typo in the notes, but it was fixed just about 5 minutes into the contest. We are sorry for that issue, but sometimes typos happen, and we almost always fix them quickly.

Eulerean tour of the letters

a + ab + ac + ad .. + az + b + bc + bd + be + ...

if k = 7, we can concan strings below aabacadaeafag bbcbdbebfbg ccdcecfcg ddedfdg eefeg ffg g and then append strings repeat.

you want to minimize occurrence of each distinct pair of letters (adjacent letters) that appears in the string , add an edge between each pair of usable letters .Then our answer is minimized if we take euler cycle of this graph ,euler cycle always exists since its a directed complete graph, each letter has equal indegree and outdegree.

I did brute force over all pairs ,$$$"aa","ab","ac"\cdots\ "zz"$$$ and greedily selected the best pair with minimum occurence.

Compleixty; $$$O(n*26)$$$

https://mirror.codeforces.com/contest/1511/submission/112864604

Problem D: There can be $$$k*k$$$ possible pairs of characters. Now consider each of these pairs as a vertex of a graph. Then add a directed edge from $$$c_1c_2$$$ to $$$c_2c_3$$$ for all possible $$$c_1$$$, $$$c_2$$$ and $$$c_3$$$. Now do topological sorting to find a sequence of pairs. This sequence has each of the pairs exactly once, so the cost is $$$0$$$. So, you get the maximum lengthed string which has $$$0$$$ cost ($$$c_1c_2 \rightarrow c_2c_3$$$ will become $$$c_1c_2c_3$$$). Now just keep concatanating this string until its size becomes n.

Pay attention to the Example 1's output.

I have not given the contest. But just saw B. I think you can have many possible answers . One possible answer can be like this :

X = 100...000(a-1 zeroes) Y = 77...7000...00( (b-c+1) 7s & c-1 zeroes).

This will always give you gcd as 10..00(c-1 zeroes)

For B, I decided to assign a unique prime factor to each of a (2), b (3) and c (7).
As all the numbers (2, 3, 4) are lesser than 10, there will always be a power of {2, 3 and 7} whose decimal representation is of any given number of decimal digits.
First of all, I find c by multiplying it with 7 till I get the required number of digits.
Then, I multiply both a and b by c so that their GCD becomes c.
Finally, I multiply a and b with their respective prime factors till their representations reach the required number of digits.
Eg: 2, 2, 1
Output: 14 (2 * 7), 21 (3 * 7), 7
My Submission

OEIS

I just used dynamic programming (didn't even know there is such thing as OEIS) to find the sequence of the 1D o's and the equation I found was f(k) = f(k-1) + 2f(k-2)+ 2^(k-2) where f(k) denotes the total sum of the maximum tiling for a length k vertical or horizontal row of o's, and then multiplied that by 2^(nbtotal_white-k)

another way to solve for some n is to iterate on len of the consecutive blocks which have same color this solves for some n in O(N). my submission

This submission should also work for bigger C than 50.

D could be solved by the pattern like a+ab+ac+b+bc+c for k=3 , and then just repeat it until you reach n . Can you tell how to solve E

first of all given condition is nothing but checking substrings of length two which are repeated how many times.If you are able to observe it then your task is simply to minimize the Cost which can be minimized if two length substrings are repeated minimum times hence try to make all distinct possible two length substrings which wont be repeated if you add it to end your resulting string continue this and make your resulting string

This Video might help you. Video Explanation

Second solution is not straightforward $$$O(nq)$$$. It is auto-vectorized with GCC and has same logic as your. See assembly here. Click on cycle for and click on "reveal linked code".

I tried several variations of your code and only a few of them passed, you can check my submission history. I guess it's easy to miss that gcc might optimize this particular problem.

How can we do C problem if color of cards is of order 10^5(instead of 50)?

Fairly SImple way to do C.

cin>>n>>k;
    vector<ll> mp(51,-1),m1(51);
    for(ll i=0;i<n;i++) {
        cin>> a[i];
        if(mp[a[i]]==-1)
            mp[a[i]]=i;
    }
    ll tot=0;
    for(ll i=0;i<k;i++) {
        ll j;
        cin>>j;
        cout<<mp[j]+1<<" ";
        for(ll i=1;i<=50;i++) {
            if(mp[i]!=-1 && mp[i]<mp[j]) {
                mp[i]++;
            }
        }
        mp[j]=0;
    }
    cout<<endl;

The second one

Linked lists :)...Lol, these people are dumb!

Yes my submission is also similar to you https://mirror.codeforces.com/contest/1511/submission/112834762

yeah , i got the idea and implemented it. It shows wrong answer i spend approx 10 min to find that order matters.

Just put all the 1st and 3rd type no. in one server and 2nd type no.s in another server