Subtask 1

We can solve Subtask 1 if we search all possible ways of remittance. In fact, if ~A_i \le 5~, it can be proved that the amount of money in every house does not exceed ~9~ yen if we use the remittance service in any way. Since ~N \le 7~, there are at most ~10^N~ possible states of the amount of money in each house. We can search all of them in time.

Subtasks 2 and 3

In order to solve Subtask 2, we shall solve equations. Let us consider the process where the amount of money in a house becomes ~B~ yen. Let ~X_i~ be the amount of money sent by House ~i~. House ~i~ pays ~2X_i~ yen, and receives ~X_{i-1}~ yen. Here we put ~X_0 = X_N~. Then we have the equation

$$\displaystyle B_i-A_i = X_{i-1}-2X_i.$$

Let us solve it.

Taking the sum of the equation times ~2^{i-1}~ for every House ~i~, we get

$$\displaystyle \sum_{i=1}^N 2^{i-1}(B_i-A_i) = (1-2^N)X_N.$$

From this, we get ~X_N = X_0~. Then, solving the equation for each House ~i~ in order, we get ~X_i~.

Each ~X_i~ must be a non-negative integer. If this condition is not satisfied, the answer is No. In fact, if every ~X_i~ is a non-negative integer and at least one of ~B_i~ is non-zero, the answer is Yes. We shall prove this.

We consider the operations in the opposite direction. We shall transform ~B~ into ~A~.

If some entry of the sequence ~B~ is non-zero and the sequence ~A~ is different from ~B~, we have ~B_{i+1} > 0~ and ~X_i > 0~ for some ~i~. In fact, if this condition is not satisfied, we have ~B_{i+1} = 0~ for some ~i~ with ~X_i > 0~. Since ~A_{i+1} \ge 0~, we have ~X_{i+1} > 0~ by the above equation. Hence we have ~X_i > 0~ for every ~i~, and ~B_i = 0~, which is absurd.

For such ~i~, we apply the operation which is opposite to sending ~1~ yen from House ~i~ to House ~i+1~ to the sequence ~B~. Then ~B_{i+1}~ decreases by ~1~, ~B_i~ increases by ~2~, and ~X_i~ decreases by ~1~. Repeating this operation, we can transform ~B~ into ~A~.

Therefore, if all entries of ~B~ are ~0~, the answer is Yes if all entries of ~A~ are ~0~. Otherwise, the answer is No.

In Subtask 2, the value of

$$\displaystyle \sum_{i=1}^N 2^{i-1}(B_i-A_i)$$

can be calculated as a 64 bit integer. In Subtask 3, this value can be very large. Even in such cases, we can solve the task in ~\mathcal{O}(N)~ time using multiple precision integers.

Another solution

In fact, we can solve this task by the following simulations.

For each House ~i~, if it has ~a~ yen, send ~\left\lceil\frac{a-B_i}{2}\right\rceil~ yen if ~a > B_i~. Repeat this for each ~i~. We increase ~i~ by ~1~ in each step. When ~i~ becomes ~N~, we return to ~i = 1~. Note that the amount of money in the House can be ~-1~ in this process.

If the answer is Yes, after repeating this process in ~N + \mathcal{O}\left(\log \sum_{i=1}^N A_i\right)~ steps, the amount of money in every House becomes ~B~. Let us prove this fact. During this simulation, the amount of money sent from House ~i~ does not exceed ~X_i~ yen. Thus, from the intermediate status, there is a way to make the amount of money in every House be ~B~. After one cycle of ~i~, the amount of money in every House does not exceed ~B~ except possibly for one House. Moreover, the excess for that House becomes half after each step. Therefore, after ~N + \mathcal{O}\left(\log \sum_{i=1}^N A_i\right)~ steps, the amount of money in every House becomes ~B~.

By the above argument, we can also show that the answer is Yes if we can make the amount of money in every House be ~B~ and the sequence ~B~ contains a non-zero value.