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: Accepted)

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long int
#define endl "\n"
#define f(i, a, b) for(ll i=a; i<b; i++)
#define fb(i, a, b) for(ll i=a; i>=b; i--)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector <int>
#define vl vector <long long>
#define pi pair <int, int>
#define sz(x) (int) x.size()
#define M 100005
#define mod 1000000007
using namespace std;

bool comp(tuple <int, int, int, int>& a, tuple <int, int, int, int>& b)
{
	return get<3>(a) > get<3>(b);
}

void solve()
{
	double a, b, n, x;
	cin>>a>>b>>n;
	if(n >= b-1) x = b-1;
	else x = n;
	cout<<floor((a*x)/b) - a*floor(x/b);
}

int main()
{
    IOS
    int t = 1;
    //cin>>t;
    f(tno, 1, t+1)
    {
        solve();
    }
    return 0;
}