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 C (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <math.h>

int main(int argc, const char **argv)
{
    double a = 0, b = 0, n = 0, loop = 1, max = 0, calc = 0;

    fscanf(stdin, "%lf %lf %lf", &a, &b, &n);

    for ( loop = 1; loop <= n; loop++ )
    {
        calc = floor( a * loop / b ) - a * floor( loop / b );
        if ( calc > max ) max = calc;
    }

    fprintf(stdout, "%ld\n", (long)max);
    return 0;
}