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>
using namespace std;

#define mp make_pair
#define pb push_back

#define int int64_t
#define ld long double

const int MOD = 1e9+7;
const int N = 1e6+5;

int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int a,b,n;
    cin>>a>>b>>n;
    int t = a*n;
    int lol = t/b;
    int sed = lol*b;
    int gg = sed/a;
    int f = gg*a;
    f+=a;
    if(f>t)
        f-=a;
    f/=a;
    cout<<(f*a/b - a*(int(f/b)));

    return 0;
}