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: Time Limit Exceeded)

#include<bits/stdc++.h>
#define _Max(a,b,c) Max(a,Max(b,c))
#define lop(i,a,b) for(int i=a;i<b; i++)
#define pb push_back
using namespace std;
/*
    *
    * Saroar Zahan Sojib
    * Green University - Bangladesh (GUB)
    *
*/
int main()
{
    set<long long>vec;
   long long a,b,n,i;
   cin>>a>>b>>n;
   for(i=1;i<=n;i++)
   {
       long long e,f,g;
       e=floor((a*i)/b);
       f=floor(i/b);
       g=e-(a*f);
       vec.insert(g);
   }
  long long max=0;
  for(auto x:vec)
  {
      if(max<x)
      {
          max=x;
      }
  }
  cout<<max<<endl;
    return 0;
}