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 <stdio.h>
#include <math.h>
typedef long long ll;

ll func(ll a, ll b, ll x)
{
    return floor(a * x / b) - a * floor(x / b);
}

int main()
{
    ll a, b, n, ans;
    scanf("%lld %lld %lld", &a, &b, &n);

    ans = func(a, b, 0);
    for (int i = 1; i <= n; i++)
    {
        if (ans < func(a, b, i))
            ans = func(a, b, i);
    }

    printf("%lld", ans);
    return 0;
}