Problem:
Score : 200 points

Problem Statement

We have a long seat of width X centimeters.
There are many people who wants to sit here. A person sitting on the seat will always occupy an interval of length Y centimeters.

We would like to seat as many people as possible, but they are all very shy, and there must be a gap of length at least Z centimeters between two people, and between the end of the seat and a person.

At most how many people can sit on the seat?

Constraints

All input values are integers.

1 \leq X, Y, Z \leq 10^5

Y+2Z \leq X

Input

Input is given from Standard Input in the following format:

X Y Z

Output

Print the answer.

Sample Input 1

13 3 1

Sample Output 1

3

There is just enough room for three, as shown below:

Figure

Sample Input 2

12 3 1

Sample Output 2

2

Sample Input 3

100000 1 1

Sample Output 3

49999

Sample Input 4

64146 123 456

Sample Output 4

110

Sample Input 5

64145 123 456

Sample Output 5

109

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>(
    stdin_lock: &mut std::io::StdinLock,
    buf: &mut Vec<u8>,
    delimiter: u8,
) -> T {
    buf.clear();
    let l = std::io::BufRead::read_until(stdin_lock, delimiter, buf).unwrap();
    buf.truncate(l - 1); // remove delimiter
    let s = unsafe { std::str::from_utf8_unchecked(&buf) };
    s.parse().unwrap_or_else(|_| panic!("read"))
}

fn main() {
    let stdin = std::io::stdin();
    let mut stdin_lock = stdin.lock();
    let mut buf: Vec<u8> = Vec::new();
    let x: i32 = read(&mut stdin_lock, &mut buf, b' ');
    let y: i32 = read(&mut stdin_lock, &mut buf, b' ');
    let z: i32 = read(&mut stdin_lock, &mut buf, b'\n');

    let ans = (x - z) / (y + z);
    println!("{}", ans);
}