Cyclists

Key Observation

The deck operation looks like it wants simulation pain. It doesn’t. The only important thing is the position of the win-condition.

Suppose the win-condition is currently at position $x>k$. Since only the first $k$ cards are playable, every playable card is before the win-condition. Playing one of them moves it to the back, so the win-condition shifts forward by exactly $1$.

That means we need exactly

$$x-k$$

ordinary card plays before the win-condition becomes playable.

Also, before the win-condition is played, a card before it cannot be played twice: once played, it goes behind the win-condition. So those $x-k$ cards are distinct cards chosen from the $x-1$ cards before it.

Therefore, the cost before the next win-condition play is at least the sum of the $x-k$ cheapest costs among those $x-1$ cards.

Why Greedy Works

Mark the $x-k$ cheapest cards before the win-condition. There are exactly

$$(x-1)-(x-k)=k-1$$

unmarked cards before it.

While the win-condition is still not playable, the first $k$ positions are all before it. Since there are only $k-1$ unmarked cards total, at least one marked card must appear in the first $k$ positions. So if we always play the cheapest playable non-win-condition card, we can always play one of the marked cards.

After $x-k$ such moves, we have played exactly those cheapest marked cards. Greedy hits the lower bound, which means it is optimal. Nice. The queue tried to be scary and failed.

First Play vs Later Plays

For the first win-condition play, it starts at position $p$.

Let $A$ be the cost needed before the first play:

$$A = \text{sum of the } \max(0,p-k) \text{ cheapest costs among } a_1,\dots,a_{p-1}.$$

Then the first win-condition play itself costs $a_p$, so the first successful play costs

$$A+a_p.$$

After the win-condition is played, it is moved to the bottom, position $n$. Now all other $n-1$ cards are before it. To make it playable again, we need $n-k$ ordinary plays.

Let $B$ be

$$B = \text{sum of the } n-k \text{ cheapest costs among all cards except } p.$$

So every additional win-condition play costs

$$B+a_p.$$

Answer Formula

If

$$A+a_p>m,$$

we cannot even play the win-condition once, so the answer is $0$.

Otherwise, after the first play we have $m-A-a_p$ energy left, and each extra play costs $B+a_p$. Thus the answer is

$$1 + \left\lfloor \frac{m-A-a_p}{B+a_p} \right\rfloor.$$

Complexity

We just sort two small lists of costs:

$$O(n\log n)$$

per test case, with $O(n)$ memory. Since $\sum n\le 5000$, this is hilariously safe.

Source note: I cross-checked the greedy proof against the official Codeforces Round 1086 editorial and rewrote the explanation here.