# Find the maximum possible value for the given periodic function

Given three numbers A, B, and N, the task is to find the maximum possible value of floor(A * x / B) – A * floor(x / b) where x is a non-negative integer less than or equal to N. Here floor(T) = denotes the greatest integer not greater than the real number T (G.I.F function).

Constraints: 1 ≤ A ≤ 106, 1 ≤ B ≤ 1012, 1 ≤ N ≤ 1012. All values in input are integers.

Input: A = 5, B = 7, N = 4
Output: 2
Explanation:
The maximum value is obtained for the value x = 3. On substituting this value in the equation:
floor((5 * 3)/7) – (5 * floor(3 / 7)) = floor(2.1) – 0 = 2.

Input: A = 11, B = 10, N = 9
Output: 9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: The naive approach for this problem is to consider all the possible numbers from 1 to N and compute the maximum possible value.

Time Complexity: O(N).

Efficient Approach: The idea is to make an observation on the function f(x) = floor(A * x / B) – A * floor(x / B).

• We can observe that the given function is a periodic function. This can be proved by:

f(x + B) = floor(A * (x + B)/B) – A * floor((x + B)/B)
=> f(x + B) = floor((A * x / B) + A) – A * floor((x /B) + 1)

By floor-function property, floor(x + Integer) = Integer + floor(x).
=> f(x + B) = floor(A * x / B) – A * floor(x / B) = f(x)

• Hence, we can conclude that 0 ≤ x ≤ B. However, if x = B, f(x) = 0. So, we exclude it and get 0 ≤ x ≤ B-1.
• However, we must also consider the condition x ≤ N. Since floor(x) is a monotonically non-decreasing function, we must incorporate the best of both the ranges.
• Hence, the maximum value of f(x) is obtained when x = min(B – 1, N).

Below is the implementation of the above approach:

## C++

 `// C++ Program to find the maximum  ` `// possible value for the given function ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the maximum  ` `// value of f(x) ` `int` `floorMax(``int` `A, ``int` `B, ``int` `N) ` `{ ` `    ``int` `x = min(B - 1, N); ` ` `  `    ``return` `(A * x) / B; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A = 11, B = 10, N = 9; ` ` `  `    ``cout << floorMax(A, B, N); ` `    ``return` `0; ` `} `

Output:

```9
```

Time Complexity: O(1)

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Improved By : abhasbind, Akanksha_Rai