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 <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <bitset>    // bitset
#include <deque>     // deque
#include <iomanip>   //setprecsion
#include <iostream>  // cout, endl, cin
#include <map>       // map
#include <queue>     // queue, priority_queue
#include <set>       // set
#include <stack>     // stack
#include <string>    // string, to_string, stoi
#include <tuple>     // tuple, make_tuple
#include <utility>   // pair, make_pair
#include <vector>    // vector

#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <cmath>  //sin, cos, tan

using namespace std;

#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define rrep(i, n) for (ll i = (n - 1); i >= 0; --i)
#define blank(ans) cout << (ans) << " ";
#define mp(p, q) make_pair(p, q)
#define mt(p, q, r) make_tuple(p, q, r)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef string str;

typedef vector<ll> vll;
typedef vector<ld> vd;
typedef vector<bool> vb;
typedef vector<char> vc;
typedef vector<str> vs;
typedef vector<vector<ll>> vvll;
typedef vector<vector<ld>> vvd;
typedef vector<vector<bool>> vvb;
typedef vector<vector<char>> vvc;
typedef vector<vector<str>> vvs;
typedef vector<pair<ll, ll>> vpll;
typedef vector<tuple<ll, ll, ll>> vtlll;

const ld PI = acos(-1.0);
const ll MAX = 9000000000000000000;
const ll MIN = -9000000000000000000;
const ld DMAX = 4500;
const ld DMIN = -4500;
const ll MOD = 1000000007;

template <typename T>
void fin(T a)
{
  cout << a << endl;
  exit(0);
}

void Main()
{
  ll a, b, n; cin >> a >> b >> n;
  ll flag = 0;
  ll x = 0;
  while(1)
  {
    x += 25000;
    if(x > n||((a * x) / b) <= ((x / b) * a))
    {
      ll ans = 0;
      for(ll i = x - 25000 - 10;i <= n;i++)
      {
        if(i <= 0) continue;
        flag = max((a * i) / b - a * (i / b), flag);
      }
      fin(flag);
    }
  }
}

//前処理
int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20); //高精度少数表示
  //cout << setfill('0') << internal << setw(4); 0埋め４桁
  Main();
  return 0;
}