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>
#define ll long long
using namespace std;

// GCD, LCM
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}

ll f(ll x,ll A,ll B){
    return floor(A*x/B) - A*floor(x/B);
}

int main()
{
    ll A,B,N,M=0,temp;
    cin>>A>>B>>N;

    
    M=max(f(N,A,B),(ll)0);

    M=max(f(N-B,A,B),M);
    if((N-ceil(B/A))>=0){
        M=max(f(N-(ceil(B/A)),A,B),M);
    }
    if(N-(ceil(A/B))>=0){
        M=max(f(N-(ceil(A/B)),A,B),M);
    }
    if((N-A)>=0){
        M=max(f(N-A,A,B),M);
    }
    
    cout<<M<<endl;

    return 0;
}