The problem actually asks how many different sums can be got from all subsets.

Let $$$Y = X - D$$$. Then $$$A_i=Y+iD$$$. We can divide $$$gcd(Y, D)$$$ from $$$Y$$$ and $$$D$$$, now let's assume $$$gcd(Y,D)=1$$$.

We can maintain n maps. Enumerate how many elements we choose from the sequence, now assume it's k.

So the sum of elements we picked from sequence should be $$$kY+bD$$$, you can store all the sums possible if we pick k elements into map $$$(k\mod D)$$$.

If there are different ways to choose elements with the same sum, then they will be stored into same map.

Now let's consider $$$k$$$-th maps, it stores sum like $$$(k+tD)Y+bD$$$, if we subtract $$$kY$$$ from them, they will be multiple of $$$D$$$.

Consider all sums picked $$$k$$$ elements, actually the sum should be

You can subtract $$$(k\mod D)Y$$$ from them, then divide $$$D$$$, you'll find all sums contribute from picked $$$k$$$ elements will be compressed as a continuous interval $$$[l,r]$$$, you can maintain them with map.

At last, count the total length of all intervals stored in maps. The total time complexity is $$$O(n\log_2n)$$$.

F:

Let $$$AP = A + (A + 1\cdot D) + (A + 2\cdot D) + \dots + (A + (N - 1)\cdot D$$$)
The problem essentially asks to count unique sums of every subset of $$$AP$$$.

Let's iterate over $$$l$$$, in which we consider subsets of size $$$l$$$.
For each $$$l$$$, we can find a range $$$[L, R]$$$, such that each sum of subsets of size $$$l$$$, lies in $$$[L, R]$$$.
It is easy to see that $$$L = A\cdot l + \sum\limits_{i=0}^{l-1}{i\cdot D}$$$
Similarly, $$$R = A\cdot l + \sum\limits_{i=n-l}^{n-1}{i\cdot D}$$$
By taking $$$l$$$ leftmost, and $$$l$$$ righmost numbers respectively.
It can be proven that for each such range, all sums with intervals of $$$D$$$ are possible to obtain, i.e. $$$\left(L, L + D, \dots, R\right)$$$
Since, there can overlapping of sums in ranges for those $$$l$$$'s for which $$$L \pmod D$$$ are same, we merge such ranges before adding there contribution.
Also, handle the case for $$$D = 0$$$ separately.

Link

The editorial is out! Link

You can use Google translate for the pdf file

C. If you decide whether each person is honest or unkind, you can easily check in O (N2) whether the condition is consistent with the testimony. Since there are only 2N ways to determine whether or not each person is honest, it is sufficient to investigate whether or not there is a contradiction in all these cases, and to answer the largest number of honesty before the contradiction occurs. The above shows that an integer j between 0 and 2N is for an integer k between 1 and N, person k is honest and that the k-1 digit is 1 when j is represented in binary. By defining it as equivalent state, you can easily try all cases, this method is called bit traversal.

Since N is at most 15, the main idea is to generate all 2^15 = 32768 possible combinations of whether one is honest, and test whether there are any contradictions. I did this using bit operations, but of course there are other ways of doing that.

My submission: https://atcoder.jp/contests/abc147/submissions/8866706

Isn't it that we need to divide by 2 somewhere?

The above calculates sum of all pairs, not only the ones where j>i. Or am I wrong?

When you paste the whole explanation from geeksforgeeks !!! the whole solution was available on gfg They should not add such question the contest when the solution is already available on such educational websites

here for D solid explanation link

I was also waiting for his editorial. His explanation of the problems are best

I had the same problem! Where I was making a mistake was calculating 2^60, I wasn't taking the modulo 1e9+7 during its computation assuming it to fit in a long long.

editorial is already published, here. First 6 pages are in japanese, but last few pages are in english!