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>
using namespace std;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

int main() {
    ll a,b,n;
    cin>>a>>b>>n;
    ll ans = 0;
    ll tmp = 10000000;
    ll cnt = 0;
    if(n<=tmp){
        for(ll i=1;i<=n;i++){
            ll axb = (a*i)/b;
            ll xb = i/b;
            xb = xb*a;
            /*if(ans<axb-xb){
                cout<<i<<" "<<axb<<" "<<xb<<endl;
            }*/
            ans = max(ans, axb - xb);
        }
    }else{
        for(ll i=n-tmp;i<=n;i++){
            ll axb = (a*i)/b;
            ll xb = i/b;
            xb = xb*a;
            /*if(ans<axb-xb){
                cout<<i<<" "<<axb<<" "<<xb<<endl;
            }*/
            ans = max(ans, axb - xb);
        }
        for(ll i=1;i<=tmp;i++){
            ll axb = (a*i)/b;
            ll xb = i/b;
            xb = xb*a;
            /*if(ans<axb-xb){
                cout<<i<<" "<<axb<<" "<<xb<<endl;
            }*/
            ans = max(ans, axb - xb);
        }
        for(ll i=10000001;i<=n-10000000;i=i+100000){
            //cnt++;
            ll axb = (a*i)/b;
            ll xb = i/b;
            xb = xb*a;
            /*if(ans<axb-xb){
                cout<<i<<" "<<axb<<" "<<xb<<endl;
            }*/
            ans = max(ans, axb - xb);
            /*if(cnt>=10000000){
                i+=10000000000;
                cnt = 0;
            }*/
        }
    }
    cout<<ans<<endl;
}