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: Wrong Answer)

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

int main(void){
    long long int A,B,N,max=0,i,pre,c,d;
    
    scanf("%lld",&A);
    scanf("%lld",&B);
    scanf("%lld",&N);
    
    if(A>=B){
        i=N;
        c=floor(A*i/B);
        d=floor(i/B);
        max=c-A*d;
    }else if(A<B && B<=N){
        i=B-1;
        c=floor(A*i/B);
        d=floor(i/B);
        max=c-A*d;
    }else if(A<B){
        i=N;
        c=floor(A*i/B);
        d=floor(i/B);
        max=c-A*d;
    }else{
        for(i=0;i<=N;i++){
            c=floor(A*i/B);
            d=floor(i/B);
            
            pre=c-A*d;
            
            if(pre>max){
                max=pre;
            }
        }
    }
    
    printf("%lld\n",max);
    
    return 0;
}