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 Rust (1.42.0) (judge verdict: Accepted)

extern crate core;

use std::io::{self, Read};

#[macro_export]
macro_rules! input {
  ($s:expr=>$($t:tt)*) => {
    let mut lines=$s.split("\n");
    $(
        line_parse!(lines,$t);
    )*
  };
}

macro_rules! line_parse {
  ($lines:expr,($($name:ident:$t:tt)*)) => {
    let mut line=$lines.next().unwrap().split_whitespace();
    $(value_def!(line,$name,$t);)*
  };

  //複数行
  ($lines:expr,{$n:expr;$name:ident:$t:tt}) => {
    values_def!($lines,$n,$name,$t);
  };
}

macro_rules! value_def {
    ($line:expr, $name:ident, $t:tt) => {
        let $name = value!($line, $t);
    };
}

macro_rules! values_def {
    ($lines:expr, $n:expr, $name:ident, $t:tt) => {
        let $name = {
            let mut vec = Vec::new();
            for i in 0..$n {
                let mut next = $lines.next().unwrap().split_whitespace();
                vec.push(value!(next, $t));
            }
            vec
        };
    };
}

macro_rules! value {
  //配列
  ($line:expr,[$t:tt]) => {
    $line.map(|x|{
      let mut iter=::std::iter::once(x);
      value!(iter,$t)
    }).collect::<Vec<_>>()
  };
  //タプル
  ($line:expr,($($t:tt),*)) => {
    ($(value!($line,$t),)*)
  };
  //文字列
  ($line:expr,#) => {
    $line.next().unwrap()
  };
  //インデックス(-1)
  ($line:expr,@) => {
    $line.next().unwrap().parse::<usize>().unwrap()-1
  };
  //単一値
  ($line:expr,$t:ty) => {
    $line.next().unwrap().parse::<$t>().unwrap()
  };
}

fn main() {
    let mut input = String::new();
    io::stdin().read_to_string(&mut input).unwrap();
    let output = solve(input.trim().to_string());
    println!("{}", output);
}

fn solve(input: String) -> String {
    input!(input=>(a:i64 b:i64 n:i64));
    // floor(a*x/b)-a*floor(x/b)
    /*
    a*x/bの切り捨てが小さい, x/bの切り捨てが多い
    ex1:5 7 4. x=3
    5*3/7=2.14285714286, 3/7=0.42857142857

    x/bはp/aより少し小さい値？

    x<pb/a
    */
    let x = std::cmp::min(n, b - 1);
    (((a * x) / b) - a * (x / b)).to_string()
}

macro_rules! tests {
    ($($name:ident: $input:expr=>$output:expr,)*) => {
        mod tests {
            $(
                #[test]
                fn $name() {
                    assert_eq!($output.trim().to_string(),super::solve($input.trim().to_string()));
                }
            )*
        }
    }
}
tests! {
    test1: "5 7 4\n" => "2\n",
    test2: "11 10 9\n" => "9\n",
}