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: Accepted)




#include <bits/stdc++.h>
#define ll long long
#define PB push_back
#define MP make_pair

#define REP(i, n) for(int i=0;i<(int)(n);++i)
#define FOR(i,l,h) for(int i=(int)(l);i<=(int)(h);++i)
#define FORD(i,h,l) for(int i=(int)(h);i>=(int)(l);--i)

#define max(a,b) a>b?a:b
#define min(a,b) a>b?b:a

ll gcd(ll a, ll b) { return !b ? a : gcd(b, a % b); }
ll lcm(ll a, ll b) { return (a / gcd(a, b)) * b; }

#define dump(x)  cerr << #x << " = " << (x) << endl
#define ALL(t) (t).begin(),(t).end()

using namespace std;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll int a,b,n;
    cin>>a>>b>>n;
    ll int x;
    if(n < b){
        x = n;
    }
    else if(n >= b){
        x = b-1;
    }
    cout<<floor(((ll)x*a)/(double)b);
    return 0;
}