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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System.Collections.Generic;
using static System.Console;

class Program {
	static void Main() {
		var X = NextInt();
		var Y = NextInt();
		var Z = NextInt();

		WriteLine((X - Z) / (Y + Z));
	}

	static int NextInt() {
		return int.Parse(NextString());
	}

	static string NextString() {
		var result = new List<char>();
		while (true) {
			int next = Read();
			if (next < 0)
				break;
			var nextChar = (char)next;
			if (!char.IsWhiteSpace(nextChar))
				result.Add(nextChar);
			else if (nextChar != '\r')
				break;
		}
		return string.Join("", result);
	}
}