Yes.
All beginner, regular and grand contests are rated

Yes, they are usually rated (AtCoder Regular Contests are rated for ratings up to 2799 in their platform).

Around the end of last year, 2021, AtCoder has started providing options to register for a Rated contest either as Rated Participant or Unrated Participant. If you do a Rated Registration against a competition, be sure to participate in it because it will affect your rating whether or not you participate in that contest, or unregister at least 5 minutes before the contest begins.

More information regarding it here, in their post on it: https://atcoder.jp/posts/745

What is 'is begun'?

Once I thought that there's only SpeedForces.

And now I learn there's also SpeedCoder.

wtf dude

imagine crying over difficulty distribution on a contest full of amazing problems.

non-negligible number of rated people solved E today

The main observation is that the second XOR operation kind of negates the first. So actually we do only one XOR operation, with one of the a[i].

Knowing this we only want to find which one. That can be done by counting the number of set bits at each bit-position, and then try each a[i] as the candidate for the XOR operation.

Just do recursion with memoization. Submission

I solved this task with a memorized dfs.We can see that if there is an integer x(x<=4),we can't get a better result by breaking it down.

That also means if there is an integer x(x>4) ,we can get a greater product by change it to x/2 and (x+1)/2.So I used kind of memorized searching.

Here is my code

Hope it will help you.

For n=1 , The answer would be Directly s[0] , 0 , 0 .

s[i] = a[i] + a[i+1] + a[i+2]
s[i+1] = a[i+1] + a[i+2] + a[i+3]
Thus a[i+3] - a[i] = s[i+1] - s[i] 

Thus once we fix the values of a[0] , a[1] and a[2] the other elements of a get fixed. Try to find minimum a[0],a[1],a[2]>=0 such that for all remaining i a[i]>=0 if s[0]>= sum of those minimum values of a[0] a[1] and a[2] then The solution is possible else not

I have solved it as follows:

If we have some $$$i$$$ where $$$S[i]=0$$$, this uniquely determines the answer, where $$$A[i]=A[i+1]=A[i+2]=0$$$, and we propagate other values accordingly in $$$A$$$ to satisfy those in $$$S$$$. Suppose the minimum value in $$$S$$$ is $$$min$$$, we can subtract $$$min$$$ from all values in $$$S$$$ and calculate an answer $$$AnsTmp$$$.

Now we want to change $$$AnsTmp$$$ in some way to account for the $$$min$$$ we subtracted earlier. This is like calculating answer $$$AnsTmp2$$$ for an array $$$S$$$ whose all values are all $$$min$$$. In this case, if the first $$$2$$$ elements in $$$AnsTmp2$$$ are $$$a$$$ and $$$b$$$, the third value will be $$$min-a-b$$$, the fourth will be $$$min-(min-a-b)-b=a$$$, the fifth will be $$$min-a-(min-a-b)=b$$$, and so on, this means $$$AnsTmp2$$$ will be similar for all $$$i$$$ which are equal $$$mod$$$ $$$3$$$.

Let's see the minimum non-negative integer for each $$$i$$$ from $$$0$$$ to $$$2$$$ which if we added to $$$AnsTmp[j]$$$ ($$$j$$$ $$$mod$$$ $$$3=i$$$) will make $$$AnsTmp[j]$$$ non-negative for all $$$j$$$ and add them to $$$AnsTmp2$$$. If $$$AnsTmp2[0]+AnsTmp2[1]+AnsTmp2[2]>min$$$, then there is no answer. Else we can just add $$$min-AnsTmp2[0]-AnsTmp2[1]-AnsTmp2[2]$$$ for $$$AnsTmp2[i]$$$ ($$$i$$$ $$$mod$$$ $$$3=0$$$) and each $$$ans[i]$$$ will be just $$$AnsTmp[i]+AnsTmp2_[i]$$$.

Note that in max(val,ch). U hv taken the mod of ch in step ll ch = (recurse(op1)%mod * recurse(op2)%mod)%mod;. So even if ch would hv been actually big than val, it would hv become less than val after mod. The observation is that, u only hv to avoid splitting the numbers if they are less than or equal to 3. Handle that seperatly, and keep dp[val]=ch;

The idea with mod is to mod the result of computations, but your code is MODing the input numbers, which will of course not yield the correct answer. Consider n = MOD * 2. Your code obtains p1 = p2 = MOD, which immediately just becomes 0.

Maybe we can notice the increasing rate of A[], which is of $$$O(n^2)$$$, so we have got an upper boundary of A[]. Then we divide it by $$$i$$$ (according to B[]'s definition) to see what happens on B[].

The $$$O(n^2)$$$ thing: notice that A[i+1]-A[i] <= i+1, so A[i] — X = A[i] — A[0] <= (i+2)(i-1)/2.

Sadly, I didn't solve the problem during the contest, but I upsolved it afterwards (with editorial).

I'll try to explain. If I can do it, that means I understand how to solve this problem. :-)

First, you forget the condition $$$0 \le A_i$$$ for a moment. Just choose some random $$$A_1 = a$$$ and $$$A_2 = b$$$.

Doing some math, we have $$$S_i = A_i + A_{i+1} + A_{i+2}$$$, and $$$A_{i+2} = S_i - A_{i+1} - A_i$$$. So, if we know $$$a$$$ and $$$b$$$, we can determine all other terms of sequence.

Now we continue our math (with the formula for $$$A_{i+2}$$$ above):

$$$A_1 = (0) + a$$$ (by definition)

$$$A_2 = (0) + b$$$ (by definition)

$$$A_3 = (S_1) - a - b$$$

$$$A_4 = (S_2 - S_1) + a$$$

$$$A_5 = (S_3 - S_2) + b$$$

$$$A_6 = (S_4 - S_3 + S_1) - a - b$$$

$$$\dots$$$

We can now see the pattern:

$$$A_i = (x_i) + a$$$, for $$$i = 1, 4, 7, \dots$$$

$$$A_i = (x_i) + b$$$, for $$$i = 2, 5, 8, \dots$$$

$$$A_i = (x_i) - a - b$$$, for $$$i = 3, 6, 9, \dots$$$

for some mystery constants $$$x_i$$$.

But how do we know these constants $$$x_i$$$? Well, we can choose $$$a$$$ and $$$b$$$ to be equal to $$$0$$$ and compute them directly.

We now recall the condition $$$0 \le A_i$$$. To fulfill this condition, we need the following satisfied:

$$$A_i = (x_i) + a \ge 0$$$, for $$$i = 1, 4, 7, \dots$$$

$$$A_i = (x_i) + b \ge 0$$$, for $$$i = 2, 5, 8, \dots$$$

$$$A_i = (x_i) - a - b \ge 0$$$, for $$$i = 3, 6, 9, \dots$$$

Rearranging, we get:

$$$-x_i \le a$$$, for $$$i = 1, 4, 7, \dots$$$

$$$-x_i \le b$$$, for $$$i = 2, 5, 8, \dots$$$

$$$a + b \le x_i$$$, for $$$i = 3, 6, 9, \dots$$$

Let's take a closer look. The first two inequalities tells us that we want to maximize $$$a$$$ and $$$b$$$. The third inequality tells us that we want to minimize $$$a + b$$$. So, we can try to take the minimum possible $$$a$$$ and $$$b$$$.

The minimum possible $$$a$$$ is $$$m_a = max(-x_1, -x_4, -x_7, \dots)$$$.

The minimum possible $$$b$$$ is $$$m_b = max(-x_2, -x_5, -x_8, \dots)$$$.

The maximum possible $$$a + b$$$ we allowed to have is $$$m_s = min(x_3, x_6, x_9, \dots)$$$, otherwise we can't satisfy the condition for at least one of $$$x_3, x_6, x_9, \dots$$$

So if $$$m_a + m_b \gt m_s$$$, that means we can't satisfy all three conditions above for at least one of $$$x_i$$$ and the answer is NO. Otherwise the answer is YES and we can take $$$a = m_a$$$ and $$$b = m_b$$$.

Now we can easily restore the whole answer.

Think of 2 operations, selecting number $$$B_0$$$ and $$$B_1$$$. $$$B_0$$$ is selected first, it's in the original array. Then consider $$$B_1$$$, there must be some $$$A_0$$$ in original array such that $$$B_1=B_0\oplus A_0$$$, because you must choose it in the new array you got after selecting $$$B_0$$$. Then we discover that the operation sequence $$${B_0,B_1}$$$ is totally equivalent to one single operation of choosing $$$A_0$$$.

If there's more than 2 operations, we compress the first 2 into one single operation $$$O$$$, then do so to $$$O$$$ and the 3rd operation, until there's only one operation in the queue.

you can think like this,

v = A1 A2 A3 A4

after performing operation with x = v[1] = A1

0 A1^A2 A1^A3 A1^A4

after performing operation with x = v[4] = A1^A4

A4^A1 A4^A2 A4^A3 0

which is same as if you do operation on A1 A2 A3 A4 and x=v[4]=A4

Ah, it's actually easy to solve, me dumb.