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 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		long b = sc.nextLong();
		long n = sc.nextLong();
		double bb = b;
		sc.close();
		double ans = 0;
		double inv = 1 / bb;
		int count = 1000;
		int j = 0;
		for (long i=n; i>=1; i--) {
			double inv2 = inv*i;
			double kai = Math.floor(a*inv2)-a*Math.floor(inv2);
			if (ans < kai) 	{
				ans = kai;
			} else {
				j++;
				if (j == count) {
					i = 10000;
				}
			}
		}
		/*
		j = 0;
		for (long i=0; i<=n; i++) {
			double inv2 = inv*i;
			double kai = Math.floor(a*inv2)-a*Math.floor(inv2);
			if (ans < kai) 	{
				ans = kai;
			} else {
				j++;
				if (j == count) {
					break;
				}
			}
		}
		*/
		System.out.println((int)ans);
	}
}