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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

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

int main(void){
    long double a,b,n,tmp,max=floorl(a*n/b)-a*floorl(n/b);
    scanf("%Lf %Lf %Lf",&a,&b,&n);

    int x;
    for(x=n;1;x--){
        tmp=floorl(a*x/b)-a*floorl(x/b);
        if(tmp<max){
            break;
        }
        if(tmp>=max){
            max=tmp;
        }
    }
    printf("%ld\n",(long)max);

    return 0;
}