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: Wrong Answer)

import java.util.Scanner;
//import java.util.Arrays;
public class Main{
	public static void main(String[] args){
		Scanner scan = new Scanner(System.in);
		int x = scan.nextInt();
		int y = scan.nextInt();
		int z = scan.nextInt();
		int n = 0;
		for(int i=x/y; 0<=i; i++){
			if(x > y * i + z * (i+1)){
				n = i;
				break;
			}
		}
		System.out.println(n);
	}
}