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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;

int main()

{
    ll a, b, n;
    scanf("%lld%lld%lld", &a, &b, &n);
    if(b == 1)
    {
        printf("0\n");
        return 0;
    }
    ll ans = -1e18;



    // if(n < b)
    // {
    //     cout << ans << endl;
    //     return 0;
    // }
    for(ll i = b; i <= n; i += b)
    {
        //if(i % b && (i * a ) % b == 0)
        ans = max(ans, (i * a) / b - i / b);
    }
    for(ll i = 1; i <= n; i *= 5)
    {
        ans = max(ans, (i * a) / b - i / b);
    }
    for(ll i = b - 1; i <= n; i += b)
    {
        ans = max(ans, (i * a) / b - i / b);
    }

    for(ll i = 0; i * a <= n; i++)
    {
        ans = max(ans, (i * a) / b - i / b);
    }
    ans = max(ans, a * n / b - a * (n / b));
    cout << ans << endl;
}