Though I have solved 1st problem in 22 min in 1st attempt yet why I'm not ratted?

Though, not too long ago.

Just solve more and it wont affect you.

Sorry to say but your rank in last round look suspisious .

UPD :- There is nothing wrong in your rank and performance , I misunderstood your rank which you acheived in Technocup round 3 as your rank in CF round 759 . I recently came across blog stating 22-40 rank in previous contest are all cheaters and lot of cheating happend in last contest and then I saw you rank and it look suspisious and I didn't saw the contest name . I apologize for that .

My rating was 1199. I remember. Techno cube gave me 1199 r

maybe you could turn green today itself after cheaters are removed.

Upd: sadly you lost 1 point

Wow, I am very impressed by your persistence on cp(by looking at your graph). Keep it up!

BledDest and awoo have written about a hundred educational rounds and they are pretty good at their work. So I think (and hope) we can expect a great round :))

You can copy 'p' from here and paste it.

Also me when it's accepted

Increase by how much ?

Depends on rank not no. Of problems

I think three or four will be enough

solve just like my alt hi123bye he is again pupil now

We decided that the contest was difficult enough with 7 problems, so one problem was excluded.

Why?

just read the sample explanation. I didn't understand the statement either.

Your solution was difficult to understand. But basically, you want to match the last $$$2k$$$ elements, $$$a_n$$$ with $$$a_{n-k}\ $$$, $$$a_{n-1}$$$ with $$$a_{n-k-1}\ $$$ and so on..

sum(a) = sum(b) / (n * (n — 1) / 2) b[i] — b[i — 1] = sum(a) — n * a[i]

I love it, the we have two main observation, first is:

Let S(b) sum of all bi and S(a) sum of all ai

the total time of every singer is ai + 2*ai + ... + n*ai = ai*n*(n+1)/2

S(b) = sum of total time of every singer -> S(b) = a1*n*(n+1)/2 + ... an*n*(n+1)/2 -> S(b) = S(a)*n*(n+1)/2

Using 0 indexed, the second observation is:

• bi = 1*ai + 2*a(i-1)%n + ... n*a(i+1)%n

• b(i+1) = 2*ai + 3*a(i-1)%n + ... 1*a(i+1)%n

• b(i+1) — bi = ai + a(i-1)%n + ... — (n-1)*a(i+1)%n

• b(i+1) — bi = ai + a(i-1)%n + ... + a(i+1)%n — n*a(i+1)%n

• b(i+1) — bi = S(a) — n*a(i+1)%n

• a(i+1)%n = (b(i+1) — bi — S(a))/n

To check if the answer is YES or NO, you have to be sure that all divisions leave 0 remainder and all ai > 0.

My code: https://pastebin.com/iaxwvTSq

yes. sorting in decrease order then greedy use a[i + k] / a[i] for each i from 1 to k

We must/want remove the k*2 biggest numbers.

So sort descending, and pair a[i],a[i+k] for cost of 0 if these both numbers differ, or cost=1 if same. Then add the other (n-k*2) smallest numbers.

Take the GCDs of all the elements at odd and even indices, let's say G1 and G2. If any element at an odd index is divisible by G2 and any element at even index by G1, the answer does not exist. Else answer is either G1 or G2.

get gcd of all odd-index elements, then check if any even-index element is divisble by it. Then do reversely.

It's O(nlogn). I am also confused.

I generated all possible binary numbers as strings up to length 62. Thats fairly quick since we basically can only append a '1' at begin and reverse.

at first operation, if you add 0, then all the 0s at the end of the original number is removed. Then you can only add 1 to both sides of the original number to get the target number. Add 0 is no-op.

for F it's kind of understandable if you can break it down. how i did it was:

If you add a zero, essentially you are saying, I can flip the string and trim all zeroes at the front and back. So essentially I can always reverse it whenever I want.

If you add a one, you can think of it as I can add a one on the left or right of the string without much thought because I can flip whenever I want

So the only caveat left is, did I have to trim the zeroes to do this reversing operation in the first place?

So there are 2 cases:

A) String is [1?????0].

B) String is [1?????1].

For case A: I can add a 1 at the back, and only then can I add the 1s however I want. Or, I can trim the zeroes, and then add the 1s however I want.

For case B: I can add 1s at the front or back, however I want.

So your final answer will look like this:

Case A:

Possible if string itself is [1?????0]
Possible if string is of the form ...1[1?????01]1...
Possible if string is of the form ...1[10?????1]1...
Possible if string is of the form ...1[1????1(trimmed)]1...
Possible if string is of the form ...1[1????1(trimmed)(flipped)]1...

Case B:

Possible if string is of the form ...1[1?????1]1...
Possible if string is of the form ...1[1?????1(flipped)]1... 

and thats the final answer, maybe a little complicated now that i type it out

edit: removed redundant statement for clarity

very educational channel

I think it is because you are taking maximum of the two gcds. Instead you should try both.

sorry for my bad case

First you have a string s equal to the first bigram. for B, if one bigram's second character and the next bigram's first character is the same, you can always treat them as adjacent, add the next bigram's last character to s. Otherwise, add the next bigram to s. After you have iterated all the bigrams, if length of s is smaller than N, add one 'a'.

After you read a bigram into a string, you can delete all spaces with a consecutive letter if that space has the same letters on both sides. After that you delete the last space left or add any letter otherwise. Regex code: 139328475

no. only one for each operation.

DO practice my friends !!!!!

not min(a), it's gcd(a).

35 6 14 9. No element divides any other, but $$$d=3,7..$$$ are answers.

1 2 4 4, I think.

Think about this case:

1 4 3 6 5 The answer is 2 but your code says that there is no answer. Divisor can be not presented in array, you should check gcd of all odd and then for even positions as a d.

Check ur solution for value of a=[ 1 2 4 ]