finally ,, thank you :) ; i think this is my chance to become a specialist after 4 years in codeforces
really i solve 2 problem in Dev2 .. so i will try my best here

is it rated for unrated accounts?

All I want is a smooth and queue free contest today

How to become tester for CF round? Are testers author's friends or something else?

For testers this problems is easy. Maybe before the round gives this problems to grey people?

Hello!

If i have 1400+ and only 1 rated round and i'll write this div.4 will the rating be recalculated?

Thanks for the new contest :)

GOOD LUCK TO EVERYBODY!

Affect of Div.4 contests on us:

I am looking forward to the problem A. How much difficult it is!

Will it be unrated? Like last round?

Hope Not to See Atcoder's ABC's A and B problem :)

Today ,we are going to witness a historical moment and be part of it .

Upcoming generation will witness this post and will be enlightened by our high intellectual comments and will praise our intelligence. (First div 4 round on CF)

What is the meaning of D3A-D3D problems?

By seeing the upvotes of this blog we can imagine how codeforces comunity got disappoint after div4 got introduced :(

There seems to be an unwritten rule to not post an editorial before end of hacking phase. I do not see the point. People discuss problems in the comments anyway after the coding phase.

Is there an official opinion?

Thank you MikeMirzayanov for running an awesome platform not only for advanced users! Maybe it will be helpful for newbies like us.. We will try to give our best shot :)

Is it rated?

this(division 4) will be helpfull. great idea..... :)

All time high.. 26K+ participants ..wow

Still long queue?

is this a really Div.4 are you sure???

Well, usually I hate problems with test cases, but not in this case.
Samples are so strong so I can just submit the solution without carefully checking the answer, since WA 1 doesn't count.
I really appreciate it. Thanks, Mike.

Nice contest!! (Although there can be a slight increase in difficulty of F or G as also in Div3 end questions are around 1700-1800 rating.)

The queue time almost does not exist! Thanks for the great round!

What div.1 and div.4 have in common for me: having nothing to do for about half an hour before the end.

Very good work of the codeforces administrators, the queue time was significantly reduced. The queue time for round 639 was around 40 minutes while in this round the cap was 2 minutes. I think codeforces is ready for div 1 and div 2. Thank you for your wonderful work.

I solved all problems in this first contest of Div 4!!!

I'm so happy about it!!!

Officially, I have the 12 place in official list of participants!

Predictor says that I will have +330 rate!!!

Amazingg~!!!!!

Love you, Mike Mirzayanov!!!!!

P.S. Mike made my dreams come true!!! I really want T-shirt with your autograph, cause you are my superstar!!!

I think the order was a little unbalanced. It should be

A C B D E G F

OR

A C B D G E F

Problem of long queue has improved a lot wrt last div1/2 contest!

When you solve 4 problems in div3, but here you solve one......

I like problem A. It is easy but interesting and has an educational value.

Unofficial editorial in Python 3. For more editorials see my discord http://interview.solutions

Problem A

for testcase in range(int(input())):
    s = input()  # input as string
    ans = []
    for i, c in enumerate(s):
        if c != '0':
            ans.append(int(c) * (10 ** (len(s) - 1 - i)))

    print(len(ans))
    print(*ans)

Problem B

for testcase in range(int(input())):
    n, k = map(int, input().split())

    # 1, 1, ..., 1, last
    last = n - (k - 1)
    ans = []
    if last > 0 and last & 1 == 1:
        ans = [1] * (k - 1) + [last]

    # 2, 2, ..., 2, last
    last = n - 2 * (k - 1)
    if last > 0 and last & 1 == 0:
        ans = [2] * (k - 1) + [last]

    if ans:
        print("YES")
        print(*ans)
    else:
        print("NO")

Problem C

for testcase in range(int(input())):
    n, k = map(int, input().split())

    k -= 1
    ans = 0
    for bit in range(30, -1, -1):
        take = (n - 1) * (1 << bit)
        # Let's look at the first n * (1 << bit) numbers.
        # (n - 1) * (1 << bit) of them are not divisible
        # by n.  If we could take them, then we do so.
        if k >= take:
            k -= take
            ans += n * (1 << bit)
    print(ans + k + 1)

# Alternate Solution
for testcase in range(int(input())):
    n, k = map(int, input().split())

    def f(x):
        # Among {1, 2, ..., x}, how many are
        # not divisible by n?
        return x - x // n

    # f is a monotone increasing function, and
    # we want to know the first x so that f(x) = k.
    lo, hi = 0, 2 * k + 1
    while lo < hi:
        mi = lo + hi >> 1
        if f(x) < k:
            lo = mi + 1
        else:
            hi = mi
    print(lo)

# Alternate Solution
for testcase in range(int(input())):
    n, k = map(int, input().split())
    q, r = divmod(k, n - 1)
    print(n * q + r - (r == 0))

Problem D

for testcase in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))

    i, j = 0, n - 1
    turns = alice = bob = last = 0
    while i <= j:
        turns += 1
        eat = 0
        if turns & 1:  # Alice's turn
            while i <= j:
                eat += arr[i]; i += 1
                if eat > last: break
            alice += eat
        else:  # Bob's turn
            while i <= j:
                eat += arr[j]; j -= 1
                if eat > last: break
            bob += eat

        last = eat

    print(turns, alice, bob)

Problem E

for testcase in range(int(input())):
    n = int(input())
    arr = list(map(int, input().split()))

    special = [False] * (n + 1)
    for i in range(n):
        s = arr[i]
        for j in range(i + 1, n):
            s += arr[j]
            if s > n:
                break
            special[s] = True

    print(sum(1 for x in arr if special[x]))

Problem F

for testcase in range(int(input())):
    n0, n1, n2 = map(int, input().split())
    if n1 == 0:
        if n0:
            ans = "0" * (n0 + 1)
        else:
            ans = "1" * (n2 + 1)
    else:
        ans = ["0" * (n0 + 1), "1" * (n2 + 1)]
        for i in range(n1 - 1):
            ans.append(str(i & 1))
        ans = "".join(ans)

    print(ans)

Problem G

for testcase in range(int(input())):
    n = int(input())

    if n <= 3:
        ans = [1 if n == 1 else -1]
    else:
        ans = list(range(2, n + 1, 2))
        if n & 1:  # Eg. n = 7, ans = [2, 4, 6]
            ans.append(n - 4)
            ans.append(n)
            ans.append(n - 2)
            ans.extend(range(n - 6, 0, -2))
        else:  # Eg. n = 6, ans = [2, 4, 6]
            ans.append(n - 3)
            ans.append(n - 1)
            ans.extend(range(n - 5, 0, -2))

    print(*ans)


# Alternate Solution

def solve(n):
    answers = {
        1: [1],
        2: [-1],
        3: [-1],
        4: [2, 4, 1, 3],
        5: [2, 4, 1, 3, 5],
        6: [2, 4, 6, 3, 5, 1],
        7: [2, 4, 6, 3, 7, 5, 1],
        8: [2, 4, 6, 8, 5, 7, 3, 1],
    }

    if n <= 8:
        return answers[n]

    ans = []
    floor = 0
    while floor + 8 < n:
        ans.extend(floor + x for x in (1, 4, 2, 5, 3))
        floor += 5
    ans.extend(floor + x for x in solve(n - floor))
    return ans

for testcase in range(int(input())):
    print(*solve(int(input())))

I loved Problem G! It reminded me of one of my favorite problems on Codeforces: 24 Game

Hi everyone! I uploaded screencast of this contest by this link (Solved all problems in 48 minutes!).

Hey which approach in problem E would cause MLE?

problem set was great. and I think your experiment has become successful. Today There is no Inqueue problem.

Since I was tester in this round I was not able to participate. But I used the time to write an unofficial editorial

The problems were easy but very interesting!

I think there are some problems with that. According to the prediction results of CF preidcitor, the div4 rank1 will increase 341 score after this round. So, his new rank will be 1733 (1392+341). I think it unbelievable. Maybe his new rank is too high.

It seems that we can't view other's source after the contest?

UPD: it's ok now

I'm not sure if it's only me, but while submitting a solution I kept receiving a message saying "Unexpected error". This happened for about 5 minutes and then everything was normal again. What could be the reason?

I think it could've been div3 if one more difficult problem was added, and 1-2 current problems removed?

how is it possible to have runtime error on c++ 11, but not on c++ 17, when someone uses only vector?

I was about 30 seconds short or else I would have officially done that last problem...
Anyways I enjoyed this round and it really for first time in life taught me that i should have good typing speed
I encourage to have more div.4 rounds
It motivated me a lot

Not satisfied with the order of problems!!!

The first time I solved all of the problems. What an amazing first experence to me. I'm so happy to attend this round even though I'm out of competiton.

Thanks MikeMirzayanov for providing such contests. Finally I have solved a problem in an online contest. Thank you for providing such contest . Also, introducing Div4. contest was an awesome initiative. Keep continuing such kind of contests.

Why am I in practice mode, and how do I hack?

Good time , why my code for problem A with g++11 compiler got wrong but with g++17 got Accepted? from the begining I try to change my code and fix it , after the contest I got problem is the compiler I choose for code :(((

Simple solution for problem G: if n<4 clearly no answer exists else first keep 3,1,4,2 in a deque and alternate appending elements between left and right. My submission link

Мне кажется, меня обмануло "С целью отсечения эффективных решений от неэффективных в этой задаче ограничение времени довольно строгое. Постарайтесь написать в самом деле эффективное решение." в Е — особые элементы.

Я полчаса думал и пытался придумать что-то эффективное. Не придумал. Написал самое простое решение на C++, работающее за O(n^2) и использующее O(1) дополнительной памяти — просто для каждого из чисел в массиве проверяю, может оно быть составлено или нет. И получил полное решение.

Simple solution for G again!

Put odd elements first in decreasing order then from even elements first put 4 then 2 then same order from 6 till last even element in increasing order.

Although it's div4, I still think it's very difficult. Maybe I'm too bad.

Did anyone else feel that the level of Div 4 was somewhat similar to Div 3? I mean, there is a noticeable difference between Div 2 and Div 3 but I don't think i can say the same about Div 3 and Div 4. (In the previous three Div 3 contests, I could comfortably solve 2 questions well within 30 mins but could only solve A and C today). Anyways, the problems were good, hope to see more such contests.

Why does not O(n^2) solution for E TLE? I thought they reduced constraints to keep those away, and tried to do that in LogN time. Still, 8000^2/2 is a lot of time for one second, and solution passes in 70 ms. What about cases with all 1's?

I solved just one problem. It was still fun because I could finally read the problem statements and know what was expected of me. I hope to get better in the upcoming rounds

A very cool round. Keep the good work going, Codeforces administration team.

Nice problem set

Поделюсь здесь решением задачи G.

Заметим, что решений для n <= 3 не существует, потому в таком случае выводим "-1". Иначе давайте ответ будем хранить в двухсторонней очереди (в с++ для этого можно использовать deque<>). В самом начале давайте будем сзади добавлять все нечетные числа в промежутке [1..n]. Затем давайте добавим в начало "4" (поскольку мы рассматриваем решения для n > 3, всё законно) и "2". Затем будем в начало добавлять все четные числа в промежутке [6..n].

Конечным ответом является получившаяся очередь.

В справедливости решения убедиться довольно легко — разница между любыми соседними чётными или соседними нечётными числами в точности равна двум, разница между "1" и "4" равна 3, между "2" и "4" равна 2 и, наконец, между "2" и "6" равна 4.

#include <bits/stdc++.h>

using namespace std;

void Solve(){
    int n;
    cin >> n;

    if(n <= 3){
        cout << -1 << endl;
        return;
    }

    deque<int> deq;
    for(int i = 1; i <= n; i += 2) deq.push_back(i);
    deq.push_front(4);
    deq.push_front(2);
    for(int i = 6; i <= n; i += 2) deq.push_front(i);
    for(int i : deq) cout << i << ' ';
    cout << endl;
}
 
signed main(){
    int test = 1;
    cin >> test;
    for(int i = 1; i <= test; i++)
        Solve();
}

Screencast of this round. Click here , I was able to solve upto F.

I doubt,that one test included n=0

Nice problems! My screencast of the round in Chinese bilibili BV1tT4y137wS, don't forget to 一键三连!

Interestingly, the Div4 seems to be harder than the first few Div3 rounds (although Div3s are relatively harder now). Do you have the same feeling?

Could someone try and hack my D? PS : I think the worst case is with 200 test cases each having 1000 elements. Also I didn't use 2 pointers, and it is most possibly O(n^2) per test case.