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 <vector>
#include <map>
#include <set>
#include <cmath>

typedef long long int ll;

using namespace std;
long double a,b,c,n,m,q,aa[100000],bb[100000];
bool ra,rb,rc;
vector<ll> va(20),vb;

#define CE cout<<endl
#define C cout<<
#define E <<endl

int main(){
    cin>>a>>b>>n;

    
/*
    for(double i=1;i<100;i++){
        c = i/b;
        C i<<":"<<c<<"::"<<(floor(a*c)-a*floor(c)) E;
    }
*/
    if( (b-1)<=n) c = b-1/b;
    else c = n/b;
    C c E;

    C (floor(a*c)-a*floor(c)) E;

}