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: Wrong Answer)

#include <bits/stdc++.h>
//#include "stdc++.h"
using namespace std;

int lfloor(int t){
    
    return t ;
}

int main() {
    int a ,b ,n ;
    cin >> a ;
    cin >> b ;
    cin >> n ;
    int max , maxIndex ;
    for (int i = 0 ; i <= n ;i++){
        int value = lfloor(a*i/b) - a*lfloor(i/b);
        if (i == 0 ){
            max = value ;
            maxIndex = 0;
        }else{
            if ( value > max){
                max = value ;
                maxIndex = i ;
            }
        }
    }
    cout << max ;
}