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 rep(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define xrep(i, a, b) for (int i = int(a); i < int(b); ++i)
int main() {
    int A, B, N;
    cin >> A >> B >> N;
    vector<int> Nums;
    int num;
    rep(x, N) {
        Nums.push_back((A*x)/B-(A*(x/B)));
    }

    int max = 0;
    rep(i, Nums.size()) {
        if(max<Nums[i]) {
            max = Nums[i];
        }
    }

    cout << max << endl;

    return 0;
}