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;

int main(){
  uint64_t A,B,N;
  uint64_t ans=0;
  uint64_t num;
  cin >> A >> B >> N;
  if(N<=1000000){
    for(uint64_t i=1; i<=N; i++){
      num = (A*i)/B - A*(i/B);
      ans = max(ans,num);
    }
  } else {
    if(B<=1000000){
      for(uint64_t i=1; i<=B; i++){
      num = (A*i)/B - A*(i/B);
      ans = max(ans,num);
    }
    } else {
      if(B>N){
        ans = (A*N)/B - A*(N/B);
      } else {
        // ここが違う
        ans = (A*(B-1))/B - A*((B-1)/B);
      }
    }
  }
  cout << ans << endl;
  return 0;
}