Problem:
Score : 400 points

Problem Statement

Given are integers A, B, and N.

Find the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N.

Here floor(t) denotes the greatest integer not greater than the real number t.

Constraints

1 ≤ A ≤ 10^{6}

1 ≤ B ≤ 10^{12}

1 ≤ N ≤ 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B N

Output

Print the maximum possible value of floor(Ax/B) - A × floor(x/B) for a non-negative integer x not greater than N, as an integer.

Sample Input 1

5 7 4

Sample Output 1

2

When x=3, floor(Ax/B)-A×floor(x/B) = floor(15/7) - 5×floor(3/7) = 2. This is the maximum value possible.

Sample Input 2

11 10 9

Sample Output 2

9

Solution in Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)

import java.util.HashSet;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		long A = sc.nextLong();
		long B = sc.nextLong();
		long N = sc.nextLong();

		long max = 0;

		HashSet<Long> nSet = new HashSet<Long>();
		for (long i = 1; i <= N; i++) {
			nSet.add(i);
		}

		for (long i : nSet) {
			long tmp = floor(A * i, B) - A * floor(i, B);
			max = Math.max(max, tmp);
		}

		System.out.println(max);
	}

	private static long floor(long a, long b) {
		return a / b;

	}
}

class Const {
	final static char[] alp = "abcdefghijklmnopqrstuvwxyz".toCharArray();

}