We want the minimum cost to buy exactly $$$n$$$ yogurts when:

• A single yogurt costs $$$a$$$ burles.
• A promotion lets you buy two yogurts for $$$b$$$ burles.

You may mix buying single yogurts and buying pairs. Let's reason through the options, compare costs, and derive a compact formula.

Key Observations

1. Always-available baseline:
   • You can always buy all yogurts one-by-one. That gives the cost.
     • Cost = n.a

2. When is the promotion useful?

   • Buying two yogurts with the promotion costs $$$b$$$. Compare this to buying those two separately:$$$2a$$$.
   • If $$$b \geq 2a$$$, the promotion is not beneficial (or is equal). Then buying every yogurt individually is at least as cheap:
     • $$$b \geq 2a \quad \Rightarrow \quad \text{min cost} = n \cdot a$$$
   • If $$$b \lt 2a$$$, the promotion saves money for pairs, so we should use as many promotions as possible.

Case Analysis

Case A: Promotion is not better ($$$b \geq 2a$$$)

• Optimal strategy is to buy everything one by one:
  • $$$\text{min cost} = n \cdot a$$$

Case B: Promotion is better ($$$b \lt 2a$$$)

• We should buy as many pairs as possible:
  • If $$$n$$$ is even ($$$n = 2k$$$), buy $$$k$$$ pairs: -$$$\text{cost} = k \cdot b = \frac{n}{2} \cdot b$$$
  • If $$$n$$$ is odd ($$$n = 2k+1$$$), buy $$$k$$$ pairs for $$$2k$$$ yogurts and one extra single:
    • $$$\text{cost} = k \cdot b + a = \left\lfloor \frac{n}{2} \right\rfloor \cdot b + a$$$
• We can combine both cases into a single formula:
  • $$$\text{Cost}_{\text{promo}} = \left\lfloor \frac{n}{2} \right\rfloor \cdot b + (n \bmod 2) \cdot a$$$

Final Formula

Now, the answer is the minimum between the baseline and the promotion strategy:

• $$$\text{min cost} = \min\Big( n \cdot a,\ \left\lfloor \frac{n}{2} \right\rfloor \cdot b + (n \bmod 2) \cdot a \Big)$$$

This handles both cases because:

• If $$$b \geq 2a$$$, then $$$n \cdot a$$$ will be cheaper.

• If $$$b \lt 2a$$$, then the promotion plan will be cheaper.

Example Walkthrough

Let $$$n=5,\ a=4,\ b=7$$$.

• All singles:

• Promotion strategy:

Minimum cost = 18

Complexity

• Each test case is solved in $$$O(1)$$$ time (just arithmetic).
• For $$$t$$$ test cases, total complexity is $$$O(t)$$$.

Key Takeaways

• Always consider the baseline: $$$n \cdot a$$$.
• Promotion is only useful if $$$b \lt 2a$$$.

• Compact implementation:

  • $$$ \text{pairs} = \left\lfloor \frac{n}{2} \right\rfloor,\quad \text{remainder} = n \bmod 2$$$
  • $$$\text{Answer} = \min(n\cdot a,\ \text{pairs}\cdot b + \text{remainder}\cdot a) $$$