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>
using namespace std;

#define each(i, c) for (auto& i : c)
#define mkp(a, b) make_pair(a, b)

typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> Pll;
const ll MOD = 1e9+7;

template<typename P, typename Q> ostream& operator << (ostream& os, pair<P, Q> p) { os << "(" << p.first << ": " << p.second << ")"; return os; }
template<typename T> ostream& operator << (ostream& os, vector<T> v) { os << "("; each (i, v) os << i << ", "; os << ")"; return os; }
template<typename K, typename V> ostream& operator << (ostream& os, map<K, V> m) { os << "{"; each (i, m) os << i << ", "; os << "}"; return os; }

int main() {
  ll a, b, n;
  cin >> a >> b >> n;

  ll lb = 0;
  ll ub = n+1;
  while (ub-lb > 1) {
    ll mid = (lb+ub)/2;
    ll x = (a*mid)/b - (a * (mid/b));

    if (x > n) {
      ub = mid;
    } else {
      lb = mid;
    }
  }

  ll ans = (a*lb)/b - (a * (lb/b));
  cout << ans << endl;

  return 0;
}