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 <iostream>
#include <queue>
#include <map>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstdarg>
#include <set>
#include <limits>
#include <iomanip>

using namespace std;
#define INF 2e9 + 9
#define MOD 1000000007
#define ALL(v) v.begin() , v.end()
#define Int int64_t
#define pll pair<Int,Int>
template<class T> using pql = priority_queue<T, vector<T>>;
template<class T> using pq = priority_queue<T, vector<T>, greater<T>>;
#define ll(n) Int n; cin >> n;
#define LL(...) Int __VA_ARGS___; cin >> __VA_ARGS___;
#define OUT(...) cout << __VA_ARGS___;
#define RP(n, i, s) for(Int i = s ;i < n ;i++)
#define REP(n, i) for (Int i = 0; i < n; i++)
#define REPe(n, i) for (Int i = 1; i <= n; i++)
#define RPe(n, i, s) for (Int i = s; i <= n; i++)
#define upd_max(maxer, updater) if (maxer < updater) maxer = updater;
#define upd_min(miner, updater) if (miner > updater) miner = updater;

void ansYes(bool flag){
    if (flag) cout << "Yes" << endl; else  cout << "No" << endl;
}

void ansYES(bool flag){
    if (flag) cout << "YES" << endl; else  cout << "NO" << endl;
}

struct node {
    Int to;
    Int length;
};


vector<Int> divisor(Int n) {
    vector<Int> ret;
    for (Int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
            ret.push_back(i);
            if (i * i != n) ret.push_back(n / i);
        }
    }
    return (ret);
}

/*
vector<vector<int>> to;
vector<int> flag;
vector<int> c;


Int dfs(int from, int mae) {
    flag[from] = 1;
    Int mini = 0;
    for (auto next : to[from]) {
        if (next == mae) continue;
        if (flag[next] == 1) {
            return -1;
        } else {
            mini = min(mini, dfs(next, from));
        }
    }
    return mini;
}*/

int dx8[8] = {0, -1, -1, -1, 0, 1, 1, 1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};

int nanko(string s,Int n,Int i){
    Int length = min(n - i - 1,i);
    Int ans = 0;
    for (int k = i-length;k < i;k++){
        if (s.at(k) == 'R' && s.at(i +  (i - k) == 'B')) ans++;
    }
    return ans;
}

struct datas{
    Int x;

};

int main(){
    double a, b , n;
    cin >> a >> b >> n;
    Int max = 0;
    for (double x = 1; x <= min(a,n);x++){
        upd_max(max,floor(a*x/b) - a * floor(x /b));
    }
    cout << max << endl;
}