To solve this problem, we need to determine the maximum number of squares of side length s that can fit inside a rectangle of dimensions L (length) and W (width). The key insight here is to calculate how many squares can fit along each dimension of the rectangle and then multiply these values to get the total number of squares.

Approach

1. Calculate Squares Along Length: Determine how many squares of side s can fit along the length L using integer division (floor value of L/s). This gives the count of squares along the length.
2. Calculate Squares Along Width: Similarly, find the number of squares that can fit along the width W using integer division (W/s).
3. Total Squares: Multiply the counts from the two dimensions to get the total number of squares that can fit inside the rectangle.

Solution Code

L, W, s = map(int, input().split())
count_length = L // s
count_width = W // s
print(count_length * count_width)

Explanation

• Integer Division: Using integer division (//) ensures we only consider complete squares (since partial squares cannot be counted). For example, if L=5 and s=2, 5//2=2 (two full squares fit along the length).
• Product of Counts: The total number of squares is the product of the counts along the length and width because each row of squares along the length can be repeated along the width. For example, if L=6 (3 squares of s=2), W=4 (2 squares of s=2), the total is 3*2=6 squares.

This approach efficiently computes the result with a time complexity of O(1) since it involves basic arithmetic operations. It is optimal and works for all valid input values.

Answer: The code will output the maximum number of squares as required. For example, if input is 5 3 2, the output is 2. If input is 6 4 2, output is 6. If input is 10 10 3, output is 9. And so on.

The final answer is the product of the integer divisions of L and W by s, which is computed and printed by the code. So the answer for any input is given directly by the code's output. For example, if the input is L W s, the answer is (L//s)*(W//s).

Thus, the solution is as provided in the code above.

作者声明:本文包含人工智能生成内容。