Offshores

Fix some bank $T$ and pretend this is the bank we want to maximize.

The obvious plan is: every other bank sends as many transfers as it can directly into $T$. Bank $i \ne T$ can send

$$\left\lfloor \frac{a_i}{x} \right\rfloor$$

transfers, and each one adds $y$ rubles to $T$. This gives

$$a_T + y\sum_{i\ne T}\left\lfloor \frac{a_i}{x} \right\rfloor.$$

The only real question is whether some clever chain of transfers can beat that by combining leftovers. That sounds plausible, but it is the trap.

For this fixed target $T$, define

$$P = a_T + y\sum_{i\ne T}\left\lfloor \frac{a_i}{x} \right\rfloor.$$

Think of $P$ as the current money in the target plus the value of all full transferable chunks outside it.

Now check what any operation can do to $P$:

• If some non-target bank sends to $T$, then $a_T$ increases by $y$, while that source bank loses exactly one full $x$-chunk. So $P$ changes by $+y-y=0$.
• If a non-target sends to another non-target, the source loses one full chunk. The destination receives only $y\le x$, so its number of full chunks can increase by at most $1$. Therefore the chunk sum outside $T$ never increases, so $P$ never increases.
• If $T$ sends money out, $a_T$ drops by $x$, and the recipient can gain at most one chunk, worth $y$. Since $y\le x$, this also cannot increase $P$.

So $P$ is an upper bound on the final amount in bank $T$. The direct strategy reaches that bound: make every bank $i\ne T$ send exactly $\left\lfloor a_i/x\right\rfloor$ transfers into $T$.

Therefore, if bank $T$ is chosen as the answer bank, the optimum is exactly

$$a_T + y\sum_{i\ne T}\left\lfloor \frac{a_i}{x} \right\rfloor.$$

Let

$$c_i=\left\lfloor \frac{a_i}{x}\right\rfloor, \qquad C=\sum_{i=1}^n c_i.$$

Then for each possible target $i$, the value is

$$a_i + y(C-c_i).$$

Take the maximum over all banks.

Edge cases are handled naturally:

• If $x=y$, transfers lose no money, and the formula can gather all transferable chunks exactly.
• If a bank has less than $x$, then $c_i=0$ and it simply cannot send anything.
• The target’s own chunks do not matter, because sending out of the target never helps.

The total sum of $n$ over all test cases is at most $2\cdot 10^5$, so one linear pass per test case is plenty.

Complexity:

$$O(n)$$

per test case, with $O(n)$ storage, or $O(1)$ extra storage besides the array.