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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	static Scanner scanner;

	public static void main(String[] args) {
	    scanner = new Scanner(System.in);
	    int x=gi();
	    int y=gi();
	    int z=gi();
	     
	    int n=Math.floorDiv(x, y+z);
	    int m=x%(y+z);
	    if (m<z)n--;

	    System.out.print(n);
//	    if (s.contains("AC")) {
//	    	System.out.print("Yes");
//	    }else {
//	    	System.out.print("No");
//	    }


	}

	// 文字列として入力を取得
	public static String gs() {
		return scanner.next();
	}

	// intとして入力を取得
	public static int gi() {
		return Integer.parseInt(scanner.next());
	}

	// longとして入力を取得
	public static long gl() {
		return Long.parseLong(scanner.next());
	}

	// doubleとして入力を取得
	public static double gd() {
		return Double.parseDouble(scanner.next());
	}
}